Class 12 Commerce Maths - Probability

Given that $E$ and $F$ are events such that $P\left(E\right) = 0.6 , P\left(F\right) = 0.3$ and $P\left(E \cap F\right) = 0.2$ , find $6P\left(F | E\right)$ .

It is given that

$P\left(E\right) = 0.6 , P\left(F\right) = 0.3$ and

$P\left(E \cap F\right) = 0.2$

$\Rightarrow P\left(E | F\right) = \displaystyle\frac{P\left(E \cap F\right)}{P\left(F\right)} = \displaystyle\frac{0.2}{0.3} = \displaystyle\frac{2}{3}$

$\Rightarrow P\left(F | E\right) = \displaystyle\frac{P\left(E \cap F\right)}{P\left(E\right)} = \displaystyle\frac{0.2}{0.6} = \displaystyle\frac{1}{3}=0.33$

Compute $100\times P\left(A | B\right)$ , if $P\left(B\right) = 0.5$ and $P\left(A \cap B\right) = 0.32$ .

It is given that

$P\left(B\right) = 0.5$ and

$P\left(A \cap B\right) = 0.32$

$\Rightarrow P\left(\displaystyle\frac{A}{B}\right) = \displaystyle\frac{P\left(A \cap B\right)}{P\left(B\right)} = \displaystyle\frac{0.32}{0.5} = \displaystyle\frac{16}{25}=0.64$ .

If $P\left( A \right) = \displaystyle\frac { 3 }{ 5 }$ and $P\left( B \right) = \displaystyle\frac { 1 }{ 5 }$ , find $100P\left( A\cap B \right)$ if $A$ and $B$ are independent events.

It is given that

$P\left( A \right) = \displaystyle\frac { 3 }{ 5 }$ and

$P\left( B \right) = \displaystyle\frac { 1 }{ 5 }$

$A$ and

$B$ are independent events. Therefore,

$P\left( A \cap B \right) = P\left( A \right) \cdot P\left( B \right) = \displaystyle\frac { 3 }{ 5 } \cdot \displaystyle\frac { 1 }{ 5 } = \displaystyle\frac { 3 }{ 25 }=0.12$

$100P\left( A \cap B \right) =12$

A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. If the probability of the lost card is a diamond is $p$ enter $100p$

Let

${ E }_{ 1 }$ and

${ E }_{ 2 }$ be the respective events of choosing a diamond card and a card which is not diamond.
Let

$A$ denote the lost card.
Out of

$52$ cards,

$13$ cards are diamond and

$39$ cards are not diamond.

$\therefore P\left( { E }_{ 1 } \right) =\displaystyle\frac { 13 }{ 52 } =\displaystyle\frac { 1 }{ 4 }$

$P\left( { E }_{ 2 } \right) =\displaystyle\frac { 39 }{ 52 } =\displaystyle\frac { 3 }{ 4 }$
When one diamond card is lost, there are

$12$ diamond cards out of

$51$ cards.
Two cards can be drawn out of

$12$ diamond cards in

$_{ }^{ 12 }{ { C }_{ 2 } }$ ways.
Similarly,

$2$ diamond cards can be drawn out of

$51$ cards in

$_{ }^{ 51 }{ { C }_{ 2 } }$ ways. The probability of getting two cards, when one diamond card is lost, is given by

$P\left( A|{ E }_{ 1 } \right)$ .

$P\left( A|{ E }_{ 1 } \right) =\displaystyle\frac { _{ }^{ 12 }{ { C }_{ 2 } } }{ _{ }^{ 51 }{ { C }_{ 2 } } } =\displaystyle\frac { 12! }{ 2!\times 10! } \times \displaystyle\frac { 2!\times 49! }{ 51! } =\displaystyle\frac { 11\times 12 }{ 50\times 51 } =\displaystyle\frac { 22 }{ 425 }$
When the lost card is not a diamond, there are

$13$ diamond cards out of

$51$ cards.
Two cards can be drawn out of

$13$ diamond cards in

$_{ }^{ 13 }{ { C }_{ 2 } }$ ways whereas

$2$ cards can be drawn out of

$51$ cards in

$_{ }^{ 51 }{ { C }_{ 2 } }$ ways.
The probability of getting two cards, when one card is lost which is not diamond, is given by

$P\left( A|{ E }_{ 2 } \right)$ .

$P\left( A|{ E }_{ 2 } \right) =\displaystyle\frac { _{ }^{ 13 }{ { C }_{ 2 } } }{ _{ }^{ 51 }{ { C }_{ 2 } } } =\displaystyle\frac { 13! }{ 2!\times 11! } \times \displaystyle\frac { 2!\times 49! }{ 51! } =\displaystyle\frac { 12\times 13 }{ 50\times 51 } =\displaystyle\frac { 26 }{ 425 }$
The probability that the lost card is diamond is given by

$P\left( { E }_{ 1 }|A \right)$ .
By using Baye's theorem, we obtain

$P\left( { E }_{ 1 }|A \right) =\displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( A|{ E }_{ 2 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 4 } \cdot \displaystyle\frac { 22 }{ 425 } }{ \displaystyle\frac { 1 }{ 4 } \cdot \displaystyle\frac { 22 }{ 425 } +\displaystyle\frac { 3 }{ 4 } \cdot \displaystyle\frac { 26 }{ 425 } }$

$=\displaystyle\frac { \displaystyle\frac { 1}{425} \left( \displaystyle\frac { 22 }{ 4 } \right) }{ \displaystyle\frac { 1}{425} \left( \displaystyle\frac { 22 }{ 4 } +\displaystyle\frac { 26\times 3 }{ 4 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 11 }{ 2 } }{ 25 }$

$=\displaystyle\frac { 11 }{ 50 }$

A fair coin is tossed $8$ times. Find the probability that it shows heads at least once.

$p=$ probability of a head on a fair coin

$p=\cfrac { 1 }{ 2 } ,q=1-\cfrac { 1 }{ 2 } =\cfrac { 1 }{ 2 }$
If head don't come any time
Probability

$={ _{ }^{ 8 }{ C } }_{ 0 }{ p }^{ 0 }{ q }^{ 8 }={ _{ }^{ 8 }{ C } }_{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 }$

$=1\times \cfrac { 1 }{ { 2 }^{ 8 } } =\cfrac { 1 }{ 256 }$

$\therefore$ shows heads atleast once

$=1-\cfrac { 1 }{ 256 } =\cfrac { 255 }{ 256 }$

From a bag containing 2 one rupee and 3 two rupee coins a person is allowed to draw 2 coins indiscriminately; find the value of his expectation.

1455916_778971_ans_9ec3551ef3a24fa187d68a3acab8571e.jpeg

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die

When a die is tossed two times, we obtain

$\left( 6\times 6 \right) =36$ number of observations.
Let

$X$ be the random variable, which represents the number of successes.
(i) Here, success refers to the number greater than 4.

$P\left( X=0 \right) =P\left(\text{number less than or equal to 4 on both the tosses}\right) =\dfrac { 4 }{ 6 } \times \displaystyle\frac { 4 }{ 6 } =\displaystyle\frac { 4 }{ 9 }$

$P\left( X=1 \right) =P\left(\text{number less than or equal to 4 on first toss and greater than 4 on second toss} \right) +$

$P\left(\text{number greater than 4 on first toss and less than or equal to 4 on second toss}\right)$

$=\displaystyle\frac { 4 }{ 6 } \times \displaystyle\frac { 2 }{ 6 } + \displaystyle\frac { 4 }{ 6 } \times \displaystyle\frac { 2 }{ 6 } =\displaystyle\frac { 4 }{ 9 }$

$P\left( X=2 \right) =P\left(\text{number greater than 4 on both the tosses} \right) =\displaystyle\frac { 2 }{ 6 } \times \displaystyle\frac { 2 }{ 6 } =\displaystyle\frac { 1 }{ 9 }$
Thus, the probability distribution is as follows.
(ii) Here, success means six appears on at least one die.

$P\left( Y=0 \right) =P\left(\text{six does not appear on any of the dice} \right) =\displaystyle\frac { 5 }{ 6 } \times \displaystyle\frac { 5 }{ 6 } =\displaystyle\frac { 25 }{ 36 }$

$P\left( Y=1 \right) =P\left(\text{six appears on at least one of the dice} \right) =\displaystyle\frac { 11 }{ 36 }$
Thus, the required probability distribution is as follows.

$8p(\bar {A})$

Given:

$p(A)=\dfrac 38, p(B)=\dfrac 12, p(A\cap B)=\dfrac 14$

To find:

$8p(\overline A)$

We know,

$p(\overline A)=1-p(A)=1-\dfrac 38=\dfrac 58$

Hence,

$8p(\overline A) = 8\times \dfrac 58=5$

Find $P(B/A)$ when $P(A)=\dfrac {5}{11},P(B)=\dfrac {6}{11}$ and $P(A \cup B)=\dfrac {7}{11}$ .

Given:-

$P(A) = \cfrac{5}{11}$

$P(B) = \cfrac{6}{11}$

$P(A \cup B) = \cfrac{7}{11}$

To find:-

$P\left( {B} / {A} \right) = ?$

As we know that,

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$\therefore \; \cfrac{7}{11} = \cfrac{5}{11} + \cfrac{6}{11} - P(A \cap B)$

$P(A \cap B) = \cfrac{5}{11} + \cfrac{6}{11} - \cfrac{7}{11} = \cfrac{5 + 6 - 7}{11} = \cfrac{4}{11}$

$\therefore \; P\left({B} / {A} \right) = \cfrac{P(A \cap B)}{P(A)} = \cfrac{\left( \cfrac{4}{11} \right)}{\left( \cfrac{5}{11} \right)}$

$\Rightarrow \; P \left( {B} / {A} \right) = \cfrac{4}{5}$

The probability that $A$ hit a target is $1/4$ and the probability that $B$ hits the target is $1/3$ . If each of them fired once, what is the probability that the target will be hit atleast once?

$P$ =

$A$ hits

$B$ missed+

$B$ hits

$A$ missed+Both hits

$\begin{array}{l} = \cfrac{1}{4} \times \cfrac{2}{3} + \cfrac{1}{3} \times \cfrac{3}{4} + \cfrac{1}{3} \times \cfrac{1}{4}\\ = \cfrac{2}{{12}} \times \cfrac{3}{{12}} \times \cfrac{1}{{12}} = \cfrac{1}{2}\end{array}$

Find the probability of successes in toss of a die, where a success is defined as
(i) Number greater than $4$
(ii) Six appears on die

$i)$ The favourable condition is

$5,6$

The possible no.of outcomes is

$2$

Probability is

$\dfrac{2}{6}=\dfrac{1}{3}$

$ii)$ The favourable condition is

$6$

The possible no.of outcomes is

$1$

Probability is

$\dfrac{1}{6}$

Given $3$ identical boxes $1,2,$ & $3$ . Each containing $2$ coins. In box $1$ , both are gold coins & in box $2$ both are silver coins and in box 3 there is $1$ silver and $1$ gold coin. A person chooses a box at random and takes out the coin. If the coin is gold then what is the probability that it is coming from the first box?

Let

$E_{1},E_{2},E_{3}$ be the event that boxes 1, 2 and 3 are chosen respectively.

Then,

$P(E_{1})=P(E_{2})=P(E_{3})=\dfrac{1}{3}$

Let A be the event that the coin drawn is of gold.

Then,

$P(A|E_{1}) = \dfrac{2}{2}=1$

$P(A|E_{2})=0$

$P(A|E_{3})=\dfrac{1}{2}$

Probability that other gold coin is drawn from Bag 1 =

$P(E_{1}|A)$

By Baye's theorem,

$P(E_{1}|A)=\dfrac{P(E_{1})P(A|E_{1})}{P(E_{1})P(A|E_{1})+P(E_{2})P(A|E_{2})}$

$=\dfrac{\dfrac{1}{3}\times 1}{\dfrac{1}{3}\times 1 + \dfrac{1}{3}\times 0+\dfrac{1}{3}\times \dfrac{1}{2}}=\dfrac{2}{3}$

Hence, required probability is

$\dfrac{2}{3}$ .

A black and a red die are rolled together. Find the conditional probability of obtaining the sum $8$ , given that the red die resulted in a number less than $4$ .

Let the outcomes be represented as ordered pairs of

$(B,R)$ where

$B$ is the value obtained by rolling the black die and

$R$ is the value obtained by rolling the red die in the same roll. Since the dies are fair, each outcome has an equal probability of occurring. So the probability of any outcome

$(B,R)$ is

$\dfrac{1}{36}$ as there are

$6\times 6 = 36$ possible outcomes.

The event of obtaining a sum of

$8$ is represented as

$E_1 := B+R = 8$

The event of obtaining a number less than

$4$ in the red die is represented as

$E_2:= R<4$ .

We have to find

$P(E_1|E_2)$

By the formula for conditional probability, we have:

$P(E_1|E_2) = \dfrac{P(E_1\cap E_2)}{P(E_2)}$

The outcomes defining the set

$E_1$ are

$(2,6),(3,5),(4,4),(5,3),(6,2)$

The outcomes defining the set

$E_2$ are

$(1,1),(2,1),...,(6,1),(1,2),(2,2),...,(6,2),(1,3),(2,3),...,(6,3)$ So clearly, we have

$P(E_2) = \dfrac{18}{36} = \dfrac{1}{2}$ . So clearly,

$P(E_1\cap E_2) = \dfrac{2}{36}$

The outcomes defining the set

$E_1\cap E_2$ are

$(5,3),(6,2)$

So, the required probability is

$P(E_1|E_2) = \dfrac{2/36}{1/2} = \dfrac{1}{9}$

Of the students in a school in a school , it is known that 30% have 100% attendance and 70% students are irregular . Previous year results report that 70% of all students who have 100% attendance attain A grade and 10% irregular students attain A grade in their annual examination . At the end of the year,one student is chosen at random from the school and he was found to have an A grade ,what is the probability that the student has 100% attendance?

Let ${E_1}:$ Students having 100% attendance,

${E_2}:$ Students not having 100% attendance

And so $p\left( {{E_1}} \right) = {{30} \over {100}}$

$p\left( {A/{E_1}} \right){{70} \over {100}}$

$P\left( {A/{E_1}} \right) = \frac{{10}}{{100}}$

By boy’s theoxm,

$p\left( {{E_1}/A} \right) = {{p\left( {{E_1}} \right)p\left( {A/{E_1}} \right)} \over {p\left( {{E_1}} \right)p\left( {A/{E_1}} \right) + p\left( {{E_2}} \right)p\left( {A/{E_2}} \right)}}$

$= {{{{30} \over {100}} \times {{70} \over {100}}} \over {{{30} \over {100}} \times {{70} \over {100}} + {{70} \over {100}} \times {{10} \over {100}}}}$

$= \dfrac{21}{28}$

$= \dfrac{3} {4}$

For two events $A$ and $B$ with $P(B)\ne 1$ prove that $P(A|B')=\dfrac{P(A)-P(A\cap B)}{1-P(B)}$ .

Now,

$P(A|B')=\dfrac{P(A\cap B')}{P(B')}$ [ According to the definition of conditional probability]

or,

$P(A|B')=\dfrac{P(A)-P(A\cap B)}{1-P(B)}$ [ Since

$P(B')=1-P(B)$ and

$P(A\cap B')=P(A)-P(A\cap B)$ ] [Given

$P(B)\ne 1$ ]

A n insurance company insured 2000 scoo terdrivers. 4000 car drivers and 6000 truck drivers , 7 he probeblitey of an accident involving a scooter. a and truck is $\frac{1}{100},\frac{3}{100}and\frac{3}{20}$ respectively. one of the insured persons meets with an accident. what is the probability that he is a scooter driver?

According to the problem,

Let

${E_1}, {E_2}\, \& \, {E_3}$ be the insured person of scooter, car and truck

Therefore,

$\begin{array}{l} P\left( { { E_{ 1 } } } \right) =\dfrac { { 2000 } }{ { 2000+4000+6000 } } =\dfrac { { 2000 } }{ { 12000 } } \\ =\dfrac { 1 }{ 6 } \\ P\left( { { E_{ 2 } } } \right) =\dfrac { { 4000 } }{ { 2000+4000+6000 } } =\dfrac { { 4000 } }{ { 12000 } } \\ =\dfrac { 1 }{ 3 } \\ P\left( { { E_{ 3 } } } \right) =\dfrac { { 6000 } }{ { 2000+4000+6000 } } =\dfrac { { 6000 } }{ { 12000 } } \\ =\dfrac { 1 }{ 2 } \end{array}$

Given

$E_1$ be the person who insured with an accident.

Hence,

$\begin{array}{l} P\left( { E/{ E_{ 1 } } } \right) =\dfrac { 1 }{ { 100 } } =0.01,\, \, \, P\left( { E/{ E_{ 2 } } } \right) =\dfrac { 3 }{ { 100 } } 0.03,\, \, \, P\left( { E/{ E_{ 3 } } } \right) =\dfrac { 3 }{ { 20 } } =0.15 \\ P\left( { { E_{ 1 } }/E } \right) =\dfrac { { P\left( { { E_{ 1 } } } \right) P\left( { E/{ E_{ 1 } } } \right) } }{ { P\left( { { E_{ 1 } } } \right) P\left( { E/{ E_{ 1 } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { E/{ E_{ 2 } } } \right) +P\left( { { E_{ 3 } } } \right) P\left( { E/{ E_{ 3 } } } \right) } } \\ =\dfrac { { \left( { \dfrac { 1 }{ 6 } } \right) \left( { 0.01 } \right) } }{ { \left( { \dfrac { 1 }{ 6 } } \right) \left( { 0.01 } \right) +\dfrac { 1 }{ 3 } \left( { 0.03 } \right) +\dfrac { 1 }{ 2 } \left( { 0.15 } \right) } } \\ =\dfrac { { \dfrac { 1 }{ 6 } } }{ { \dfrac { 1 }{ 6 } +\dfrac { 3 }{ 3 } +\dfrac { { 15 } }{ 2 } } } =\dfrac { 1 }{ { 52 } } \end{array}$

A card from pack of 52 cards is lost. From the remaining card of the pack, one card is drawn and is found to be heart, find the probability of missing card to be (I) heart (II) club.

Let ${E_1}$ be the event that the lost card is a diamond. Given that there are $13$ diamonds in the deck, $P({E_2}) = \dfrac{{13}}{{52}} = \dfrac{1}{4}$

Let ${E_2}$ be the event that the lost card is not a diamond. $P({E_2}) = 1 - \dfrac{{13}}{{52}} = \dfrac{{39}}{{52}} = \dfrac{3}{4}$

Let A be the event that the two cards drawn are found to be both diamonds.

We need to calculate P ( $2$ cards are both diamond w the lost card being a diamond) next.

Given $13$ diamond cards, if one is lost, then we have $12$ remaining diamonds out of $51$ total cards now.

The two cards that are both diamond can be drawn in $^{12}{C_2} = \dfrac{{12 \times 11}}{{1 \times 2}} = \dfrac{{132}}{2}$ = $66$ ways.

Likewise, the two diamonds can be drawn from the pack of $51$ remaining cards in $^{51}{C_2} = \dfrac{{51 \times 50}}{{1 \times 2}} = \dfrac{{2550}}{2}$ = 1275 ways.

Therefore, the P ( $2$ cards drawn are diamond given one is lost) = $P(A|{E_1})\dfrac{{66}}{{1275}}$

Now, consider the event where the two cards drawn are both diamonds but the lost card is not a diamond.

The two cards that are both diamond can be drawn in $^{13}{C_2} = \dfrac{{13 \times 12}}{{1 \times 2}} = \dfrac{{156}}{2}$ = 78ways.

Likewise, the two diamonds can be drawn from the pack of 51 remaining cards in $^{51}{C_2} = \dfrac{{51 \times 50}}{{1 \times 2}} = \dfrac{{2550}}{2}$ = 1275 ways.

Therefore, the P (2 cards drwan are diamond given one card which is not a diamond is lost) = $P(A|{E_1})\dfrac{{78}}{{1275}}$

We need to calcuate the probability that the lost card is a diamond. $P({E_1}|A)$ .

We can use Baye's theorem, according to which $P({E_1}|A) = \dfrac{{\dfrac{{66}}{{1275}}.\dfrac{1}{4}}}{{\dfrac{{66}}{{1275}}.\dfrac{1}{4} + \dfrac{{78}}{{1275}}.\dfrac{3}{4}}} = \dfrac{{66}}{{66 + 78 \times 3}} = \dfrac{{66}}{{300}} = \dfrac{{11}}{{50}}$

$P({E_1}|A) = \dfrac{{\dfrac{{66}}{{1275}}.\dfrac{1}{4}}}{{\dfrac{{66}}{{1275}}.\dfrac{1}{4} + \dfrac{{78}}{{1275}}.\dfrac{3}{4}}} = \dfrac{{66}}{{66 + 78 \times 3}} = \dfrac{{66}}{{300}} = \dfrac{{11}}{{50}}$

On a Saturday night $20\%$ of all divers is U.S.A. are under the influence of alcohol. The probability that a driver under the influence of alcohol will have an accident is $0.001$ . The probability that a sober driver will have an accident is $0.0001$ . If a car on a Saturday night smashed into a tree, the probability that the driver was under the influence of alcohol, is?

Hence, Probability that the diver was under the influence of alcohol when the accident happened-

$0.001\times 0.20 = 0.0002$
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A man is know to speak truth $4$ out of $5$ times. He tosses a coin and reports that it is head. Find the probability that it is actually head.

probability of man speaking truth=

$\dfrac{4}{5}$

probability of actually its head is

$\dfrac{4}{5}.\dfrac{1}{2}=\dfrac{2}{5}$

An urn contains 5 red and 2 black balls . Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ? Is X a random variable ?

These two balls may be selected as

$RR, RB, BR, BB$ where

$R$ represents red and

$B$ represents black ball, variable

$X$ has the value

$0, 1, 2,$ i.e., there may be no black balls, may be one black ball, or both the black balls.

Yes, X is a random variable.

A bag contains $10$ white and $15$ black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

${\textbf{Step - 1: Consider the first draw and second draw}}{\textbf{.}}$

${\text{This bag have 10 white and 15 black balls}}{\text{. So total balls}} = 10 + 15 = 25{\text{ balls}}{\text{.}}$

${\text{Suppose the event of getting white balls in first draw is A}}{\text{.}}$

${\text{So P(A)}} = \dfrac{{10}}{{25}}$

${\text{Without replacement of first draw , there will be 25}} - 1 = 24{\text{ balls}}{\text{.}}$

${\text{Suppose the event of getting black ball in second draw is B}}{\text{.}}$

${\text{so , probability will be , P(B)}} = \dfrac{{15}}{{24}}.$

${\textbf{Step - 2: Calculate the total probability}}{\textbf{.}}$

${\text{we know as there is no replacement ; so , trial will be independent of each other}}{\text{.}}$

${\text{so probability of being first ball white and second one black is}} =$ ${\text{P(A)}} \times {\text{P(B)}}$

$= \dfrac{{10}}{{35}} \times \dfrac{{15}}{{24}}$

$= \dfrac{1}{4}$

${\textbf{ So , the required probability is }}\mathbf{\dfrac{1}{4}.}$

An urn contains $5$ red and $2$ black balls. Two balls are randomly selected. If $X$ represents the number of black balls, what are the possible values of $X$ ?

$X$ can take the values

$0,1,2$ It is a Random variable because the experiment is a random experiment.

$X=x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X=x)$ $0.1$ $K$ $0.2$ $2K$ $0.3$ $K$
is the probability distribution of random variable $X$ . Find the value of $K$ and variance of $X$ .

$\begin{array}{l}\sum {P\left( {X = x} \right) = 1} \\ \Rightarrow 0.1 + k + 0.2 + 2k + 0.3 + k = 1\\ \Rightarrow k = 0.1\end{array}$
Therefore,

$X=x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.1$	$0.1$	$0.2$	$0.2$	$0.3$	$0.1$

Variance (

$X$ )

$= {\left( {x - \mu } \right)^2}P\left( {X = x} \right)$

$\begin{array}{l}\mu = - 2 \times 0.1 + \left( { - 1} \right)\left( {0.1} \right) + 0\left( {0.2} \right) + 1\left( {0.2} \right) + 2\left( {0.3} \right) + 3\left( {0.1} \right)\\\mu = 0.8\end{array}$
Therefore,
Var(

$X$ )

$= {\left( { - 2 - 0.8} \right)^2}\left( {0.1} \right) + {\left( { - 1 - 0.8} \right)^2}\left( {0.1} \right) + {\left( {0 - 0.8} \right)^2}\left( {0.2} \right) + {\left( {1 - 0.8} \right)^2}\left( {0.2} \right) + {\left( {2 - 0.8} \right)^2}\left( {0.3} \right) + {\left( {3 - 0.8} \right)^2}\left( {0.1} \right)=2.05$

A box contain 10 red balls , 20 yellow balls and 50 blue balls. If a ball is drwan at random from the box , find the probability that it will be
(i) a blue ball (ii) neither yellow nor blue

Total no of balls

$=10+20+50=80$

$i.e,n(S)=80$

$(i)$

$P(E)=$ getting a blue ball

$=$

$\dfrac{n(E)}{n(S)}$

$=$

$\dfrac{50}{80}=\dfrac{5}{8}$

$(ii) P(F)=$ getting neither yellow nor blue

$=$ getting red

$=$

$\dfrac{n(F)}{n(S)}=\dfrac{10}{80}$

$=$

$\dfrac{1}{8}$

If two different numbers are taken from the set $\{0,1,2,3............,10\}$ ; then the probability that their sum as well as absolute different are both multiple of $4$ , is :

The sample space here is equal to $55$ .

Now the possible combinations are $\{0,4\},\{0,8\},\{2,6\},\{2,10\},\{4,8\},\{6,10\}$

so the answer is $6/55$ .

Compute $P(A|B)$ if $P(B)=0.5$ and $P(A \cap B)=0.32$ .

$\textbf{Hint: Use the formula of conditional probability}$

Solution:

$\textbf{Use the formula of conditional probability}$

we have given,

$P(B)=0.5$ and

$P(A \cap B)=0.32$

$\because P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$\therefore P(A|B)=\dfrac{0.32}{0.5}$

$\Rightarrow P(A|B)=0.64$

$\textbf{Hence,}$

$\boldsymbol{ P(A|B)}$

$\textbf{ is 0.64.}$

50 plants were planted each school out of 10 schools. After a month, the number of planets that survived are given below:Find Mean and Variance
School 1 2 3 4 5 6 7 8 9 10
Number of planet survived 40 35 32 45 35 30 45 40 35 25

Solution:-

IF $P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{12}$ and $P(A \cap B)=\dfrac{1}{36}$ , are the events $A$ and $B$ independent ?

Given,

$\begin{array}{l} P\left( A \right) =\dfrac { 1 }{ 3 } \\ P\left( B \right) =\dfrac { 1 }{ { 12 } } \\ P\left( { A\cap B } \right) =\dfrac { 1 }{ { 36 } } \end{array}$

Now,

$P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)$

Hence,

Events

$A$ and

$B$ are independents.

If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is :

Total outcomes =

$\{ (12,0,0), (11,1,0), (10,1,1), (10,2,0)$

$(9,3,0), (9,2,1), (8,4,0), (8,3,1)$

$(8,2,2),(7,5,0), (7,4,1)$

$(7,3,2),(6,6,0),(6,5,1),$

$(6,4,2)(6,3,3)(5,5,2)(5,4,3)$

$(4,4,4)\} = 19$

Favorable cost = 4

$\{ (3,9,0), (3,8,1), (3,7,2), (3,5,4)\}$

Probability =

$4/19$

1116912_1201728_ans_d838a0834ad445549bd2133f35318f06.jpg

A number is chosen at random from numbers -3,-2, -1,0,1,2,What is the probability that |x| < 2.

Total no.

$=7$

$|x| < 2=-1, 0, 1=3$

$\therefore p=\dfrac{3}{7}$ .

Probability of getting 2 when we roll a dice.

If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is

$\dfrac{1}{6}$ .

So,

The probability of rolling a

$2$ is

$\dfrac{1}{6}$ .

Hence, this is the answer.

If $p \in = 0.6$ and $p ( E \cap F ) = 0.2$ then find $p ( F / E )$

Given

$P(E)=0.6,P(E\cap F)=0.2$

$P(F/E)=\dfrac{P(E\cap F)}{P(E)}=\dfrac{0.2}{0.6}=\dfrac{1}{3}=0.33$

Bag $B_{1}$ contains $4$ white and $2$ black balls. Bag $B_{2}$ cantains $3$ white and $4$ black balls. A bag is drawn at random and a ball is chosen at random from it. What is the probability that the ball drawn is white?

Let E denote the event of drawing while ball

$\Rightarrow P(E)=P(B_1)P(E/B_1)+P(B_2)P(E/B_2)$

$=\dfrac{1}{2}\times \dfrac{4}{6}+\dfrac{1}{2}\times \dfrac{3}{7}=\dfrac{1}{2}\left(\dfrac{2}{3}+\dfrac{3}{7}\right)=\dfrac{1}{2}\left(\dfrac{14+9}{21}\right)=\dfrac{23}{42}$ .

1214862_1396584_ans_633f43e6dce54d488ee1c76a2753b2ed.jpg

For the following probability density function (p. d. f.) of $X$ , find: (i) $P(X < 1)$ , (ii) $P (|X| < 1)$ ,
if $\left \{\begin{matrix}f(x) = \dfrac{x^2}{18}, & -3 < x < 3 \\= 0, & \text{ otherwise} \end {matrix} \right.$

We have,

$f(x)=\dfrac{x^2}{18}, \,\,\,\,\,\, -3<x<3$

$=0$ , otherwise

$Part\ (i)$

$P(X<1)=\displaystyle \int_{-3}^1 f(x) dx$

$P(X<1)=\displaystyle \int_{-3}^1 \dfrac{x^2}{18}dx$

$P(X<1)=\dfrac{1}{18}\left[\dfrac{x^3}{3}\right]_{-3}^1$

$P(X<1)=\dfrac{1}{54}\left[{1^3-(-3)^3}\right]$

$P(X<1)=\dfrac{1}{54}\left[{1+27}\right]$

$P(X<1)=\dfrac{28}{54}$

$P(X<1)=\dfrac{14}{27}$

$Part\ (ii)$

$P(|X|<1)=P(-1<x<1)$

$P(|X|<1)=\displaystyle \int_{-1}^1 f(x) dx$

$P(|X|<1)=\displaystyle \int_{-1}^1 \dfrac{x^2}{18}dx$

$P(|X|<1)=\dfrac{1}{18}\left[\dfrac{x^3}{3}\right]_{-1}^1$

$P(|X|<1)=\dfrac{1}{54}\left[{1^3-(-1)^3}\right]$

$P(|X|<1)=\dfrac{1}{54}\left[{1+1}\right]$

$P(|X|<1)=\dfrac{2}{54}$

$P(|X|<1)=\dfrac{1}{27}$

Hence, this is the answer.

A bag contains $10$ red balls and $8$ green balls. $2$ balls are drawn at random one by one with replacement. Find the probability that both the balls are green that a second year student is chosen.

In a conditional probability, if

$A$ and

$B$ are independent events, then

$P\left(A\cap B\right)=P\left(A\right)P\left(B\right)$

Given

$:$ Total number of red balls

$=10$ and that of green balls

$=8$

Probability of drawing a red ball

$=\dfrac{10}{18}=\dfrac{5}{9}$

Probability of drawing a green ball

$=\dfrac{8}{18}=\dfrac{8}{9}$

Probability of both the balls are green

$=P$ (a green is drawn first and green is drawn second time also )

$=P$ (drawing a green)

$\times P$ (drawing a green)

$=\dfrac{4}{9}\times\dfrac{4}{9}=\dfrac{16}{81}$

A bag contains 6 red balls and 8 green balls. A ball is drawn at random . What is the probability that the ball is green

No .of green balls

$8$

No.of balls

$6+8=14$

Probability is

$\dfrac {8}{14} =\dfrac 47$

Chance that Sheela tells truth is 35% and for Ramesh is 75%. In what percent they likely to contradict each other in the same question?

Probablity (

$Sheela\rightarrow Truth$ ) =

$\dfrac{7}{20}$

Probablity (

$Sheela\rightarrow Lie$ ) =

$\dfrac{13}{20}$

Probablity (

$Ramesh\rightarrow Truth$ ) =

$\dfrac{3}{4}$

Probablity (

$Ramesh\rightarrow Lie$ ) =

$\dfrac{1}{4}$

Probablity that they contradict each other =

$\dfrac{7}{20}$

$\times$

$\dfrac{1}{4}$ +

$\dfrac{13}{20}$

$\times$

$\dfrac{3}{4}$ =

$\dfrac{46}{80}$

Probablity that they contradict each other =

$57.5$

A die is thrown once. If A is the event "the number appearing is a multiple of $3"$ and B is the event "the number appearing is even" Are the event A and B independent ?

Two event A and B are independent if

$P(A\cap B)=P(A).P(B)$ A die is thrown.

$S=\left \{ 1,2,3,4,5,6 \right \}$

Let two events be

A: the number appear is a multiple of 3.

B: the number appearing is even.

A:[3,6]

$P(A)=\frac{2}{6}=\frac{1}{3}$

$B:[2,4,6]$

$P(B)=\frac{3}{6}=\frac{1}{2}$

$A\cap B$ = the number appearing is even multiple of 3 = [3]

So,

$P(A\cap B)=\frac{1}{6}$

Now,

$P(A).P(B)=\frac{1}{3}.\frac{1}{2}=\frac{1}{6}$

Since,

$P(A\cap B)=P(A).P(B)$

Therefore A and B are independent events.

1201468_1409731_ans_19af21daffa6460698b421bc4bee4492.jpg

A bag contains 4 balls.Two balls drawn at random without replacement and are found to be white. What is the probability that all balls are white?

Let A be the event of drawing 2 white balls.

from the bag containing 4 balls.

The remaining 2 balls of the bag has three options.

Let

$E_{1}$ be the event that the remaining 2 balls of the bag are not white

Let

$E_{2}$ be the event that are remaining 2 balls of the bag are one white and one not white.

Let

$E_{3}$ be the event that rae remaining 2 balls of the bag are white.

$\therefore P(E_{1})=P(E_{2})=P(E_{3})=\frac{1}{3}$

$P(A/E_{1})=P$ (drawing 2 white balls from the bag contains 2 white and 2 not white)

$=\frac{^{2}C_{2}}{^{4}C_{2}}=\frac{1}{6}$

$P(A/E_{2})=P$ (drawing 2 white balls from the bag containing 2 white and 1 non-white)

$=^{3}.C_{2}/^{4}.C_{2}=\frac{3}{6}=\frac{1}{2}$

$P(A/E_{3})=P$ (drawing 2 white ball from bag containing 4 whire balls)

= 1.

$P(E_{2}/A)=\frac{P(E_{2})(PA/E_{2})}{P(E_{1})P(A/E_{1})+P(E_{2})+P(E_{3})P(A/E_{3})}$

$=\frac{1}{3}\times \frac{1}{6}+\frac{1}{3}\times \frac{1}{2}+\frac{1}{3}.\frac{1}{1}$

$=\frac{1}{3}(\frac{1}{6}+\frac{1}{2}+1)$

$=\frac{1}{3}$

$P(E_{2}/A)=\frac{\frac{1}{3}.\frac{1}{2}}{\frac{1}{3}.\frac{5}{3}}=\frac{\frac{1}{6}}{\frac{5}{9}}$

$=\frac{1}{6}\times \frac{3}{5}=\frac{3}{10}$

1201389_1405757_ans_f9aaf32c154a45e685a7be40a60c5491.jpg

Compute $P ( A | B ) ,$ if $P ( B ) = 0.25$ and $P ( A \cap B ) = 0.18$ .

$P(B)=0.25$

$P(A\cap B)=0.18$

$\Rightarrow$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$=\dfrac{0.18}{0.25}$

$=\dfrac{18}{25}$

$\therefore$

$P(A|B)=\dfrac{18}{25}$

Evaluate $P(A \cup B)$ , if $2P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$ .

$2P(A)=P(B)=\dfrac{5}{13}$

$P(A)=\dfrac{5}{26}$ and

$P(B)=\dfrac{5}{13}$

Now,

$P(A/B)=\dfrac{P(A \cap B)}{P(B)}$

$\dfrac{2}{5}=\dfrac{P(A \cap B)}{\dfrac{5}{13}}$

$P(A \cap B)=\dfrac{2}{13}$

Therefore,

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$

$=\dfrac{5}{26}+\dfrac{5}{13}-\dfrac{2}{13}=\dfrac{11}{26}$

Find the expected number of boys in a family with 8 children,assuming the sex distribution to be equally probable.

Let n and p be the parameters of binomial distribution.

Let p = Probability of having a boy in the family

Given,

$p=q$

Since,

$p+q=1$

$p+p=1$

$p=\dfrac{1}{2}$

$q=\dfrac{1}{8}$

$n=8$

The expected number of boys =

$np=8 \times \dfrac{1}{2}=4$

Compute $P ( A | B ) ,$ if $P ( B ) = 0.5$ and $P ( A \cap B ) = 0.32$ .

$P(B)=0.5$

$P(A\cap B)=0.32$

$\Rightarrow$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$=\dfrac{0.32}{0.5}$

$=\dfrac{32}{50}$

$=\dfrac{16}{25}$

$\therefore$

$P(A|B)=\dfrac{16}{25}$

The probability distribution function of a random variable $X$ is given by
$x_{i}$ : 0 1 2
$p_{i}$ : $3c^{}$ $10c$ $5c-1$
Find $c$ .

In a probability Distribution The Sum of all Probabilities is equal to

$1$

$3c+10c+5c-1=1$

$\\18c=2$

$\\c=\dfrac 19$

The probability of student $A$ passing an examination is $\dfrac{2}{9}$ and of students, $B$ passing is $\dfrac{5}{9}$ . Assuming the two events: $A$ passes'. $B$ passes' as independent, find the probability of both passing the examination.

Given

Probability of

$A$ is

$P(A)=\dfrac 29$

Probability of

$B$ is

$P(B)=\dfrac 59$

Required probability

$P( A\cap B)=P( A)P( B)$

$=\dfrac 29\times \dfrac 59=\dfrac {10}{81}$

An urn contains $10$ black and $5$ white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

${\textbf{Step - 1: Finding probability of individual events}}$

$\text{Given an urn contains 10 black and 5 white balls.}$

$\text{So, total number of balls = 10+5 = 15}$

${\text{Let A be the event of first ball being black and B denote the event of second ball being black}}$

${\text{P(A) = }}\dfrac{{{\text{Number of black balls}}}}{{{\text{Total number of balls}}}}$

$\Rightarrow {\text{ P(A) = }}\dfrac{{{\text{10}}}}{{{\text{15}}}}{\text{ = }}\dfrac{{\text{2}}}{{\text{3}}}$

${\text{P(B) = }}\dfrac{{{\text{Number of remaining black balls}}}}{{{\text{Total number of remaining balls}}}}{\text{ (as ball is not replaced)}}$

$\Rightarrow {\text{ P(B) = }}\dfrac{{\text{9}}}{{{\text{14}}}}{\text{ }}$

${\textbf{Step - 2: Finding total probability}}$

${\text{Probability that both the balls drawn are black, P = P(A) }} \times {\text{ P(B)}}$

$\Rightarrow {\text{ P = }}\dfrac{{\text{2}}}{{\text{3}}}{\text{ }} \times {\text{ }}\dfrac{{\text{9}}}{{{\text{14}}}}$

$\Rightarrow {\text{ P = }}\dfrac{{\text{3}}}{{\text{7}}}$

$\mathbf{{\text{Hence, the probability that both the balls drawn are black is }}\dfrac{{\text{3}}}{{\text{7}}}}$

Two numbers are selected at random from integers $1$ through $9$ . If the sum is even, find the probability that both the numbers are odd.

Sum will be even if either both numbers are odd or both numbers are even .

So, Total no. of way to select numbers for making sum even =

$^5C_2+^4C_2=10+6=16$

And, Total numbers of ways to select two odd numbers such that sum is even =

$^5C_2=10$

Probability that both numbers are odd =

$\dfrac{10}{16}=\dfrac{5}{8}$

Kamal and Monika appeared for an interview for two vacancies. The probability of Kamal's selection is $1/3$ and that of Monika's selection is $1/5$ . Find the probability that at least one of them will be selected.

Consider the following events:

$K=$ Kamal is selected

$M=$ Monika is selected.

Then,

$P(K)=\dfrac{1}{3}, P(M)=\dfrac{1}{5}$

$P(\bar K) =1-\dfrac{1}{3}=\dfrac {2}{3}$ .... Probablity that Komal is not selected

$P(\bar M)=1- \dfrac {1}{5}=\dfrac {4}{5}$ .... Probablity that Monika is not selected

Required probability

$=1-P(\bar K \cap \bar M)=1-P(\bar K)\ P(\bar M)$ ....... As K and M are independent events

$=1-\dfrac {2}{3}\times \dfrac {4}{5}$

$=\dfrac {7}{15}$

Find the value of $k$ in the given probability distribution:
$x_{i}$ : 2 3 4
$p_{i}$ : 0.2 $k$ 0.3

In probability distributions , Sum of all Probabilities is given as

$1$

$\implies 0.2+k+0.3=0.5+k=1$

$\implies k=1-0.5=0.5$

The probability that it will rain tomorrow is $0.85$ . What is the probability that it will not rain tomorrow ?

$\textbf{Step 1: Declaring the events}$

$\text{Let P(A) be the event that it is going to rain tomorrow}$

$\text{Let P(B) be the event that it is not going to rain tomorrow}$

$\textbf{Step 2: Finding the complementary}$

$\text{P(A) is the event that it is going to rain tomorrow}$

$\Rightarrow\text{ P(A)= 0.85}$

$\text{P(B) is the event that it is not going to rain tomorrow}$

$\therefore\text{ P(B)= 1 - P(A)}$

$= 1-0.85$

$=0.15$

$\textbf{Thus, the probability that it will not rain tomorrow is 0.15.}$

To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:
Opinion: Like Dislike
Number of students: 135 65
Find the probability that a student chosen at random likes Mathematics

$Given$

$Total\quad number\quad of\quad students\quad =\quad 200$

$Number\quad of\quad students\quad who\quad likes\quad mathematics\quad =\quad 135$

$Probability\quad that\quad a\quad student\quad choosen\quad at\quad random\quad likes\quad mathematics$

$=\dfrac { Number\quad of\quad students\quad who\quad likes\quad mathematics }{ Total\quad number\quad of\quad students }$

$=\dfrac { 135 }{ 200 } \quad =\quad 0.675$

A school gives awards to the student each class $5$ for bravery, $3$ for punctuality, $3$ for full attendance, $4$ for social services $5$ self confidence. An awarded student is selected at random. What is the probability that he /she is being awarded for self confidence.

Total awards given to each class

$=5+3+3+4+5=20$

$P$ (Self-confident students)

$=\dfrac {5}{20}=\dfrac {1}{4}$

A school gives awards to the student each class $5$ for bravery, $3$ for punctuality, $3$ for full attendance, $4$ for social services $5$ self confidence. An awarded student is selected at random. What is the probability that he /she is being awarded for punctuality

Total awards given to each class

$=5+3+3+4+5=20$

$P$ (punctual students)

$=\dfrac {3}{20}$

Match column I with column II

${ E }_{ 1 }$ and

${ E }_{ 2 }$ are two mutually exclusive events, then

they won't have anything in common. So,

${ E }_{ 1 }\cap { E }_{ 2 }=\phi$

If

${ E }_{ 1 }$ and

${ E }_{ 2 }$ are the mutually exclusive and

exhaustive events, then

they won't have anything

in common and their total will be the universal set. So,

${ E }_{ 1 }\cap { E }_{ 2 }=\phi ; { E }_{ 1 }\cup { E }_{ 2 }=S$

If

${ E }_{ 1 }$ and

${ E }_{ 2 }$ have common outcomes, then

$({ E }_{ 1 }-{ E }_{ 2 })\cup ({ E }_{ 1 }\cap { E }_{ 2 })={ E }_{ 1 }$

If

${ E }_{ 1 }$ and

${ E }_{ 2 }$ are two events such that

${ E }_{ 1 }\subset { E }_{ 2 }$ then

${ E }_{ 1 }\cap { E }_{ 2 }= { E }_{ 1 }$

So,

$A - 4, B - 3, C - 2, D - 1$

In a class $30$ % students fail in English; $20$ % students fail in Hindi and $10$ % students fail in English and Hindi both. A student is chosen at random, then the probability that he will fail English if he has failed in Hindi is $\dfrac { k }{ 2 }$ .The value of $k$ is

Event of failing in English is denoted by 'E' and for Hindi it is denoted by 'H'.
Conditional probability states that

$P\left( \dfrac { E }{ H } \right) =\dfrac { P\left( E\cap H \right) }{ P(H) } =\dfrac { 0.1 }{ 0.2 } =\dfrac { 1 }{ 2 }$
Hence,

$k=1$

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is along with another ball of same colour. The process is repeated. the probability that the third ball drawn is black is a/b (no common divisor), then b-a =

Let

${ B }_{ i }=$ ith ball drawn is black

${ W }_{ i }=$ it ball drawn is white;

$i=12$
and

$A=$ third ball drawn is black.
We observe that the black ball can be drawn in the third draw in one of the following mutually exclusive ways.
1) Both first and second balls drawn are white and third ball drawn is black,
i.e

$\left( { W }_{ 1 }\cap { W }_{ 2 } \right) \cap A$
2) Both first and second balls are black and third ball drawn is black
i.e.

$\left( B_{ 1 }\cap { B }_{ 2 } \right) \cap A$
3) The first ball drawn is white, the second ball drawn is black and the third ball drawn is black
i.e

$\left( { W }_{ 1 }\cap B_{ 2 } \right) \cap A$
4) The first ball drawn is black, the second ball drawn is white and the third ball drawn is black
i.e

$v$

$\therefore P\left( A \right) =p\left[ \left\{ \left( { W }_{ 1 }\cap { W }_{ 2 } \right) \cap A \right\} \cup \left\{ \left( { B }_{ 2 }\cap { B }_{ 2 } \right) \cap A \right\} \cup \left\{ \left( { W }_{ 1 }\cap { B }_{ 2 } \right) \cap A \right\} \cup \left\{ \left( { B }_{ 1 }\cap { W }_{ 2 } \right) \cup A \right\} \right] \\ =P\left\{ \left( { W }_{ 1 }\cap { W }_{ 2 } \right) \cap A \right\} +P\left\{ \left( { B }_{ 1 }\cap { B }_{ 2 } \right) \cap A \right\} +P\left\{ \left( { W }_{ 1 }\cap B_{ 2 } \right) \cap A \right\} +P\left\{ \left( { B }_{ 1 }\cap { W }_{ 2 } \right) \cap A \right\}$

$\displaystyle =P\left( { W }_{ 1 }\cap W_{ 2 } \right) .P\left( \frac { A }{ \left( { W }_{ 1 }\cap { W }_{ 2 } \right) } \right) +P\left( { B }_{ 1 }\cap { B }_{ 2 } \right) .P\left( \frac { A }{ \left( { B }_{ 1 }\cap { B }_{ 2 } \right) } \right)$

$\displaystyle +P\left( { W }_{ 1 }\cap { B }_{ 2 } \right) .P\left( \frac { A }{ \left( { W }_{ 1 }\cap B_{ 2 } \right) } \right) +P\left( B_{ 1 }\cap { W }_{ 2 } \right) .P\left( \frac { A }{ \left( { B }_{ 1 }\cap { W }_{ 2 } \right) } \right)$

$\displaystyle =\left( \frac { 2 }{ 4 } \times \frac { 1 }{ 3 } \right) \times 1+\left( \frac { 2 }{ 4 } \times \frac { 3 }{ 5 } \right) \times \frac { 4 }{ 6 } +\left( \frac { 2 }{ 4 } \times \frac { 2 }{ 3 } \right) \times \frac { 3 }{ 4 } +\left( \frac { 2 }{ 4 } \times \frac { 2 }{ 5 } \right) \times \frac { 3 }{ 4 }$

$\displaystyle =\frac { 1 }{ 6 } +\frac { 1 }{ 5 } +\frac { 1 }{ 4 } +\frac { 3 }{ 20 } =\frac { 23 }{ 30 }$

Two sets of candidates are competing for the positions on the board of directors of a company. The probabilities that first and second sets will win are 0.6 and 0.4 respectively. If the first set wins the probability of introducing a new product is 0.8 and the corresonding probability if the second set wins is 0.The probability that the new product will be introduced is $0.1k$ . Find the value of $k$ .

Let

$\displaystyle P\left ( A_{1} \right )=$ Probability that the first set wins

$= 0.6$

$\displaystyle P\left ( A_{2} \right )=$ probability that the second set wins

$=0.4 \displaystyle P\left ( B\right )=$ probability that a new product is introduced

$\displaystyle \therefore P\left ( B \right )=P\left ( B\cap A_{1} \right )+P\left ( B\cap A_{2} \right )=P\left ( A_{1} \right )P\left ( B\mid A_{1} \right )+P\left ( A_{2} \right );P\left ( B\mid A_{2} \right ) ...\left ( 1 \right )$

Now as per hypothesis

$\displaystyle P\left ( B\mid A_{1} \right )=0.8,P\left ( B\mid A_{2} \right )=0.3$ Hence from

$\displaystyle \left ( 1 \right ),P\left ( B \right )=0\cdot 6\times 0\cdot 8+0\cdot 4\times 0\cdot 3=0.60$

An urn contains five balls alike in every respect save colour. If three of these balls are white and two are black and we draw two balls at random from this urn without replacing them. If A is the event that the first ball drawn is white and B the event that the second ball drawn is black, are A and B independent? If A and B are independent then enter 1, else enter 0.

Here

$\displaystyle P\left ( A \right )= \dfrac{3}{5},$

$\displaystyle P\left ( B \right )= P\left ( B/A \right )+P\left ( B/B \right )= \dfrac{2}{4}+\dfrac{1}{4}= \dfrac{3}{4}.$

Also

$\displaystyle P\left ( A\cap B \right )= \dfrac{3}{5}.\dfrac{2}{4}= \dfrac{3}{10}$

and

$\displaystyle P\left ( A \right )P\left ( B \right )= \dfrac{3}{5}.\dfrac{3}{4}= \dfrac{9}{20}.$

Hence

$\displaystyle P\left ( A\cap B \right )\neq P\left ( A \right )P\left ( B \right ).$ Therefore A and B are not independent events.

Suppose there are three urns containing 2 white and 3 black balls: 3 white and 2 black balls, and 4 white and one black ball respectively. There is equal probability of each urn being chosen. One ball is drawn from an urn chosen at random. The probability that a white ball is drawn is $\displaystyle \frac{k}{15}$ . Find the value of $k$ .

Let

$\displaystyle A_{i}\left ( i=1,2,3 \right )$ be the event that ith urn is chosen and

$B$ the event that a white ball is drawn.
Since all the urns are equally likey to be selected, we have

$\displaystyle P\left ( A_{1} \right )=P\left ( A_{2} \right )=P\left ( A_{3} \right )=\frac{1}{3}$
and

$\displaystyle P\left ( B/A_{1} \right ),P\left ( B/A_{2} \right )=\frac{3}{5},P\left ( B/A_{3} \right )=\frac{4}{5}$
Hence

$\displaystyle P\left ( B \right )=P\left ( A_{1} \right )P\left ( B/A_{2} \right )P\left ( B/A_{2} \right )P\left ( A_{3} \right )P\left ( B/A_{3} \right )$

$\displaystyle =\frac{1}{3}.\frac{2}{5}+\frac{1}{3}.\frac{3}{5}+\frac{1}{3}.\frac{4}{5}=\frac{3}{5}$

$\displaystyle P\left ( A\cup B \right )=P\left ( A\cap B \right )$ if and only if the ralation between $P(A)$ and $P(B)$ is $P(A)+P(B)=2P(AB)$ . If this is true enter 1, else enter 0.

$\displaystyle P\left ( A\cup B \right )=P\left ( A\cap B \right )$ iff

$\displaystyle P\left ( A \right )+P\left ( B \right )-P\left ( A\cap B \right )=P\left ( A\cap B \right )$
iff

$\displaystyle P\left ( A \right )+P\left ( B \right )=2P\left ( A\cap B \right )$ iff

$\displaystyle P\left ( A \right )+P\left ( B \right )=2P\left ( A \right )P\left ( B/A \right )=2 P(AB)$
which is the required relation
Note that we may write

$\displaystyle P\left ( B \right )P\left ( A /B\right )$ for

$P\left ( A\cap B \right )$ instead of

$\displaystyle P\left ( A \right )P\left ( B /A\right ).$

In a test an examine either gusses or copies or knows the answer to a multiple choice question with $m$ choices out of which exactly one is correct. The probability that he makes a guess is 1/3 and the probability that he copies the answer is 1/The probability that his answer is correct given that the copied it, is 1/If the probability that he knew the answer to the question given that he correctly answered it is 120/141, find $m$

Let

$G$ denote the event the examine guesses,

$C$ the event that the examinee copies and

$K$ the event that the examinee knows. Let

$R$ denote the event that the answer is right. We have

$P(G)=\displaystyle \frac {1}{3}, P(C)=\frac {1}{6}$
and

$P(K)=1-\displaystyle \frac {1}{3}-\frac {1}{6}=\frac {1}{2}$
Also,

$P(R|G)=\displaystyle \frac {1}{m}, P(R|C)=\frac {1}{8}, P(R|K)=1$ .
By Bayes' rule

$P(K|R)=\cfrac { 24m }{ 16+25m } =\cfrac { 120 }{ 141 } \Rightarrow m=5$

One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 98,If X and Y denote the sum and the product of the digits on the tickets, What is the value of $57P(X=9|Y=0)$ ?

$(Y=0)$ is

$\left \{00, 01, ...09, 10, 20, ..., 90\right \}$ .
Also,

$(X=9)\cap (Y=0)=\left \{09, 90\right \}$ .
We have

$P(Y=0)=\dfrac{19}{100}$
and

$P[(X=9)\cap (Y=0)]=\dfrac{2}{100}$

$\therefore \displaystyle P(X=9|Y=0)=\frac {P[(X=9)\cap (Y=0)]}{P(Y=0)}=\frac {2}{19}$

$\Rightarrow 57P(X=9|Y=0)=57\times \dfrac{2}{19}=6$

A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. The probability that the second selected ball happens to be red if the first selected ball is red, is $\dfrac{2}{k}.$ Find k.

A bag contains

$3$ red and

$7$ black balls.

Therefore, total number of balls in the bag is

$10$ . That is

$n(S)=10$ .

Two balls are selected at random one-by-one without replacement.

The probability that the second selected ball happens to be red if the first selected ball is red, is

$\dfrac{2}{k}$ . We have to find the value of

$k$ .

Let

$A$ be the event of selecting first ball as red.

Therefore,

$P(A)=\dfrac{3}{10}$ .

Without replacement of the ball, let

$B$ be the event of selecting second ball to be red.

Therefore,

$P(B)=\dfrac{2}{9}$

The probability that the second selected ball happens to be red if the first selected ball is red

$=\dfrac{2}{k}$

$\Rightarrow P(B/A)=\dfrac{2}{k}$

$\Rightarrow \dfrac{P(B\cap A)}{P(A)}=\dfrac{2}{k}$

$\Rightarrow \dfrac{P(B) \cdot P(A)}{P(A)}=\dfrac{2}{k}$

$\Rightarrow \dfrac{\dfrac{2}{9} \cdot \dfrac{3}{10}}{\dfrac{3}{10}}=\dfrac{2}{k}$

$\Rightarrow \dfrac{2}{9}=\dfrac{2}{k}$

$\Rightarrow k=9$

A bag contains n+1 coins. It is known that one of these coins has heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that the toss results in heads is 7/12, find n

Let

${ E }_{ 1 }:$ a coin with two heads is selected

${ E }_{ 2 }:$ a fair coin is selected

$A:$ the toss results in heads
Then,

$\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ n+1 } ,\quad P\left( { E }_{ 2 } \right) =\frac { n }{ n+1 } ,P\left( \frac { A }{ { E }_{ 1 } } \right) =1$ and

$\displaystyle P\left( \frac { A }{ { E }_{ 2 } } \right) \quad =\frac { 1 }{ 2 }$

$\displaystyle \therefore P\left( A \right) =P\left( { E }_{ 1 } \right) P\left( \frac { A }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) P\left( \frac { A }{ { E }_{ 2 } } \right)$

$\displaystyle \Rightarrow \frac { 7 }{ 12 } =\frac { 1 }{ n+1 } .1+\frac { n }{ n+1 } .\frac { 1 }{ 2 } \\ \Rightarrow 12+6n=7n+7\Rightarrow n=5$

A speaks truth $3$ out of $4$ times. He reported that Mohan Bagan has won the match. The probability that his report was correct is $\displaystyle \frac { k }{ 5 }$ , then $k=$

Let

$T:$ A speaks the truth and

$B:$ Mohan Bagan won the match

Given:

$\displaystyle P\left( T \right) =\frac { 3 }{ 4 } \therefore P\left( \overline { T } \right) =1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 }$

A match can be won, drawn or lost

$\displaystyle \therefore P\left( \frac { B }{ T } \right) =\frac { 1 }{ 3 } ,P\left( \frac { B }{ \overline { T } } \right) =\frac { 2 }{ 3 }$

Using Baye's theorem we get

$\displaystyle P\left( \frac { T }{ B } \right) =\frac { P\left( T \right) .P\left( \frac { B }{ T } \right) }{ P\left( T \right) .P\left( \frac { B }{ T } \right) +P\left( \overline { T } \right) .P\left( \frac { B }{ \overline { T } } \right) }$

$\displaystyle =\frac { \dfrac { 3 }{ 4 } \times \dfrac { 1 }{ 3 } }{ \dfrac { 3 }{ 4 } \times \dfrac { 1 }{ 3 } +\dfrac { 1 }{ 2 } \times \dfrac { 2 }{ 3 } } =\dfrac { \dfrac { 1 }{ 4 } }{ \dfrac { 5 }{ 12 } } =\dfrac { 3 }{ 5 }$

$\therefore k=3$

he hits the target on first shot is $0.x1$ . Find $x$ .

Given,The probability of an archer to hit the target when it is wind,

$P(H/W)=0.4$

The probability of an archer to hit the target when it is not windy,

$P(H/ \overline W)=0.7$

The probability of gust of wind,

$P(W)=0.3$

The probability that gust of wind doesn't exist,

$P(\overline W)=1-P(W)=1-0.3=0.7$

$\therefore The\;proability\;that\;archer\;will\;hit\;target\;on\;first\;shot,P(H)=P(H/W) \times P(W)+P(H/\overline W) \times P(\overline W)$

$=0.4 \times 0.3+0.7 \times 0.7=0.12+0.49=0.61$

$\therefore x=6$

If three coins are tossed, represent the sample space and the event of getting atleast two heads, then find the number of elements in them.

1339789_310544_ans_33a9a87d5f5643d5a9dfca9b042ffdaf.PNG

Box-I contains $5$ red and $4$ white balls while box-II contains $4$ red and $2$ white balls. A fair die is thrown. If it turns up a multiple of $3,$ a ball is drawn, from box-I else a ball is drawn from box-II. The probability that the ball drawn is white is $\dfrac{a}{27}$ . Find $a$ .

Let

$A$ be the event 'a multiple of

$3$ turns up on the die' and

$R$ be the event 'the ball drawn is white'
then

$P$ (ball drawn is white)

$=P\left ( A \right ).P\left ( R/A \right )+P\left ( \bar{A} \right )P\left ( R/\bar{A} \right )$

$\displaystyle =\frac{2}{6}\times \frac{4}{9}+\left ( 1-\frac{2}{6} \right )\frac{2}{6}=\frac{10}{27}$

$A$ speaks the truth ' $3$ times out of $4$ ' and $B$ speaks the truth ' $2$ times out of $3$ ', $A$ die is thrown. Both assert that the number turned up is $2.$ Find the probability of the truth of their assertion.

Let

$A$ and

$B$ be the events '

$A$ speaks the truth' and '

$B$ speaks the truth' repectively. Let

$C$ be the event 'the number turned up is not

$2$ but both agree to the same conclustion that the die has turned up

$2.$ Then

$\displaystyle P\left ( A \right )=\dfrac {3}{4}, P\left ( B \right )=\dfrac {2}{3}$ and

$\displaystyle P\left ( C \right )=\dfrac {1}{5}\times \dfrac {1}{5}$
There are two hypotheses
(i) the die turns up

$2$
(ii) the die does not turns up

$2$
Let these be the events

$E_{1}$ and

$E_{2}$ respectively, then

$=\left ( E_{1} \right )=\dfrac {1}{6}, P\left ( E_{2} \right )=\dfrac {5}{6}$ (a priori probabilities)
Now let

$E$ be the event the statement made by

$A$ and

$B$ agree to the same conclusion.
then

$\displaystyle P\left ( E/E_{1} \right )=P\left ( A \right ).P\left ( B \right )=\dfrac {3}{4}.\dfrac {2}{3}=\dfrac {1}{2}$

$P\left ( E/E_{2} \right )=P\left ( \bar{A} \right ).P\left ( \bar{B} \right ).P\left ( C \right )=\dfrac {1}{4}.\dfrac {1}{3}.\dfrac {1}{25}=\dfrac {1}{300}$
Thus

$\displaystyle P\left ( E \right )=P\left ( E_{1} \right )P\left ( E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{2} \right )=\dfrac {1}{6}\times \dfrac {1}{2}+\dfrac {5}{6}\times \dfrac {1}{300}=\dfrac {31}{360}$

$\therefore$

$\displaystyle P\left ( E_{1}/E \right )=\dfrac {P\left ( E_{1} \right )P\left ( E/E_{1} \right )}{P\left ( E \right )}=\dfrac {30}{31}$

Cards of an ordinary deck of playing cards are placed into two heaps. Heap-I consists of all the red cards and heap-II consists of all the black cards. A heap is chosen at random and a card is drawn, find the probability that the card drawn is a king.

Let I and II be the events that heap-I and heap-II are choosen respectively. Then

$\displaystyle P\left ( I \right )=P\left ( II \right )=\frac{1}{2}$
Let

$K$ be the event 'the card drawn is a king'

$\therefore$

$\displaystyle P\left ( K/I \right )=\frac{2}{26}$ and

$\displaystyle P\left ( K/II \right )=\frac{2}{26}$

$\therefore$

$\displaystyle P\left ( K \right )=P\left ( I \right )P\left ( K/I \right )+P\left ( II \right )P\left ( K/II \right )=\frac{1}{2}\times \frac{2}{26}+\frac{1}{2}\times \frac{2}{26}=\frac{1}{13}.$

A fair die is rolled. Consider events $E=\left\{ 1,3,5 \right\} ,F=\left\{ 2,3 \right\}$ and $G=\left\{ 2,3,4,5 \right\}$ Find
(i) $P\left( E|F \right)$ and $P\left( F|E \right)$
(ii) $P\left( E|G \right)$ and $P\left( G|E \right)$
(iii) $P\left( \left( E\cup F \right) |G \right)$ and $P\left( \left( E\cap F \right) |G \right)$

When a fair die is rolled, the sample space

$S$ will be

$S = \left\{ 1,2,3,4,5,6 \right\}$

It is given that

$E=\left\{ 1,3,5 \right\} ,F=\left\{ 2,3 \right\}$ and

$G=\left\{ 2,3,4,5 \right\}$

$\therefore P\left( E \right) = \displaystyle\frac { 3 }{ 6 } = \displaystyle\frac { 1 }{ 2 }$

$P\left( F \right) = \displaystyle\frac { 2 }{ 6 } = \displaystyle\frac { 1 }{ 3 }$

$P\left( G \right) = \displaystyle\frac { 4 }{ 6 } = \displaystyle\frac { 2 }{ 3 }$

$(i)$

$E\cap F=\left\{ 3 \right\}$

$\therefore P\left( E\cap F \right) = \displaystyle\frac { 1 }{ 6 }$

$\therefore P\left( E|F \right) = \displaystyle\frac { P\left( E\cap F \right) }{ P\left( F \right) } = \displaystyle\frac { \displaystyle\frac { 1 }{ 6 } }{ \displaystyle\frac { 1 }{ 3 } } = \displaystyle\frac { 1 }{ 2 }$

$P\left( F|E \right) = \displaystyle\frac { P\left( E\cap F \right) }{ \left( E \right) } =\displaystyle\frac { \displaystyle\frac { 1 }{ 6 } }{ \displaystyle\frac { 1 }{ 2 } } = \displaystyle\frac { 1 }{ 3 }$

$(ii)$

$E\cap G=\left\{ 3,5 \right\}$

$\therefore P\left( E\cap G \right) = \displaystyle\frac { 2 }{ 6 } = \displaystyle\frac { 1 }{ 3 }$

$\therefore P\left( E|G \right) = \displaystyle\frac { P\left( E\cap G \right) }{ P\left( G \right) } = \displaystyle\frac { \displaystyle\frac { 1 }{ 3 } }{ \displaystyle\frac { 2 }{ 3 } } = \displaystyle\frac { 1 }{ 2 }$

$P\left( G|E \right) = \displaystyle\frac { P\left( E\cap G \right) }{ P\left( E \right) } = \displaystyle\frac { \displaystyle\frac { 1 }{ 3 } }{ \displaystyle\frac { 1 }{ 2 } } =\displaystyle\frac { 2 }{ 3 }$

$(iii)$

$E\cup F=\left\{ 1,2,3,5 \right\}$

$\left( E\cup F \right) \cap G=\left\{ 1,2,3,5 \right\} \cap \left\{ 2,3,4,5 \right\} =\left\{ 2,3,5 \right\}$

$E\cap F=\left\{ 3 \right\}$

$\left( E\cap F \right) \cap G=\left\{ 3 \right\} \cap \left\{ 2,3,4,5 \right\} =\left\{ 3 \right\}$

$\therefore P\left( E\cup G \right) = \displaystyle\frac { 4 }{ 6 } = \displaystyle\frac { 2 }{ 3 }$

$P\left( \left( E\cup F \right) \cap G \right) = \displaystyle\frac { 3 }{ 6 } = \displaystyle\frac { 1 }{ 2 }$

$P\left( E\cap F \right) =\displaystyle\frac { 1 }{ 6 }$

$P\left( \left( E\cap F \right) \cap G \right) = \displaystyle\frac { 1 }{ 6 }$

$\therefore P\left( \left( E\cup F \right) |G \right) = \displaystyle\frac { P\left( \left( E\cup F \right) \cap G \right) }{ P\left( G \right) } = \displaystyle\frac { \displaystyle\frac { 1 }{ 2 } }{ \displaystyle\frac { 2 }{ 3 } } =\displaystyle\frac { 1 }{ 2 } \times \displaystyle\frac { 3 }{ 2 } =\displaystyle\frac { 3 }{ 4 }$

$P\left( \left( E\cap F \right) |G \right) =\displaystyle\frac { P\left( \left( E\cap G \right) \cap G \right) }{ P\left( G \right) } =\displaystyle\frac { \displaystyle\frac{1}{6}}{ \displaystyle\frac { 2 }{ 3 } } = \displaystyle\frac { 1 }{ 6 } \times \displaystyle\frac { 3 }{ 2 } = \displaystyle\frac { 1 }{ 4 }$

Evaluate $P\left( A \cup B \right)$ , if $2 P\left( A \right) = P\left( B \right) = \displaystyle\frac { 5 }{ 13 }$ and $P\left( A | B \right) = \displaystyle\frac { 2 }{ 5 }$ . Find $P\left( A \cup B \right)\times 100$

It is given that,

$2 P\left( A \right) = P\left( B \right) = \displaystyle\frac { 5 }{ 13 }$

$\Rightarrow P\left( A \right) = \displaystyle\frac { 5 }{ 26 }$ and

$P\left( B \right) = \displaystyle\frac { 5 }{ 13 }$

$P\left( A | B \right) = \displaystyle\frac { 2 }{ 5 }$

$\Rightarrow \displaystyle\frac { P\left( A \cap B \right) }{ P\left( B \right) } = \displaystyle\frac { 2 }{ 5 }$

$\Rightarrow P\left( A \cap B \right) = \displaystyle\frac { 2 }{ 5 } \times P\left( B \right) = \displaystyle\frac { 2 }{ 5 } \times \displaystyle\frac { 5 }{ 13 } = \displaystyle\frac { 2 }{ 13 }$

It is known that,

$P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)$

$\Rightarrow P\left( A \cup B \right) = \displaystyle\frac { 5 }{ 26 } + \displaystyle\frac { 5 }{ 13 } - \displaystyle\frac { 2 }{ 13 }$

$\Rightarrow P\left( A \cup B \right) = \displaystyle\frac { 5 + 10- 4 }{ 26 }$

$\Rightarrow P\left( A \cup B \right) = \displaystyle\frac { 11 }{ 26 }=0.42$

$\Rightarrow P\left( A \cup B \right) \times 100=42$

Mother, father and son line up at random for a family picture
$E :$ son on one end,
$F :$ father in middle.
Determine $P\left( E|F \right)$

If mother

$\left( M \right)$ , father

$\left( F \right)$ and son

$\left( S \right)$ line up for a family picture, then the sample space will be

$S = \left\{ MFS, MSF, FMS, FSM, SMF, SFM \right\}$

$\Rightarrow E = \left\{ MFS, FMS, SMF, SFM \right\}$

$F = \left\{ MFS, SFM \right\}$

$\therefore E \cap F = \left\{ MFS, SFM \right\}$

$P\left( E \cap F \right) = \displaystyle\frac { 2 }{ 6 } = \displaystyle\frac { 1 }{ 3 }$

$P\left( F \right) = \displaystyle\frac { 2 }{ 6 } = \displaystyle\frac { 1 }{ 3 }$

$\therefore P\left( E | F \right) = \displaystyle\frac { P\left( E \cap F \right) }{ P\left( F \right) } = \displaystyle\frac{\displaystyle\frac{1}{3}}{\displaystyle\frac{1}{3}} = 1$

A die is thrown three times,
$E\ :\ 4$ appears on the third toss, $F\ :\ 6$ and $5$ appears respectively on first two tosses.
Determine $36P$ where $P\left( E|F \right)$

If a die is thrown three times, then the number of elements in the sample space will be

$6 \times 6 \times 6 = 216$

$E = \left\{ \left( 1, 1, 4 \right) , \left( 1, 2, 4 \right) , \dots \left( 1, 6, 4 \right) \\ \left( 2, 1, 4 \right) , \left( 2, 2, 4 \right) , \dots \left( 2, 6, 4 \right) \\ \left( 3, 1, 4 \right) , \left( 3, 2, 4 \right) , \dots \left( 3, 6, 4 \right) \\ \left( 4, 1, 4 \right) , \left( 4, 2, 4 \right) , \dots \left( 4, 6, 4 \right) \\ \left( 5, 1, 4 \right) , \left( 5, 2, 4 \right) , \dots \left( 5, 6, 4 \right) \\ \left( 6, 1, 4 \right) , \left( 6, 2, 4 \right) , \dots \left( 6, 6, 4 \right) \right\}$

$F = \left\{ \left( 6, 5, 1 \right) , \left( 6, 5, 2 \right) , \left( 6, 5, 3 \right) , \left( 6, 5, 4 \right) , \left( 6, 5, 5 \right) , \left( 6, 5, 6 \right) \right\}$

$\therefore E \cap F = \left\{ \left( 6, 5, 4 \right) \right\}$

$P\left( F \right) = \displaystyle\frac { 6 }{ 216 }$ and

$P\left( E \cap F \right) = \displaystyle\frac { 1 }{ 216 }$

$\therefore P\left( E | F \right) = \displaystyle\frac { P\left( E \cap F \right) }{ P\left( F \right) } = \displaystyle\frac { \displaystyle\frac { 1 }{ 216 } }{ \displaystyle\frac { 6 }{ 216 } } = \displaystyle\frac { 1 }{ 6 }=0.16$

If $P\left(A\right) = 0.8, P\left(B\right) = 0.5$ and $P\left(B | A\right) = 0.4$ , find
(i) $P\left(A \cap B\right)$
(ii) $P\left(A | B\right)$
(iii) $P\left(A \cup B\right)$

It is given that

$P\left( A \right) = 0.8, P\left( B \right) = 0.5$ and

$P\left( B | A \right) = 0.4$
(i)

$P\left( B | A \right) = 0.4$

$\therefore \displaystyle\frac { P\left( A \cap B \right) }{ P\left( A \right) } = 0.4$

$\Rightarrow \displaystyle\frac { P\left( A \cap B \right) }{ 0.8 } = 0.4$

$\Rightarrow P\left( A \cap B \right) = 0.32$
(ii)

$P\left( A | B \right) = \displaystyle\frac { P\left( A \cap B \right) }{ P\left( B \right) }$

$\Rightarrow P\left( A | B \right) = \displaystyle\frac { 0.32 }{ 0.5 } = 0.64$
(iii)

$P\left( A \cup B \right) = P\left( A \right) + P\left( B \right) - P\left( A \cap B \right)$

$\Rightarrow P\left( A \cup B \right) = 0.8 + 0.5 - 0.32 = 0.98$

Of the students in a college, it is known that $60\%$ reside in hostel and $40\%$ are day scholars (not residing in hostel). Previous year results report that $30\%$ of all students who reside in hostel attain $A$ grade and $20\%$ of day scholars attain $A$ grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an $A$ grade, what is the probability that the student is a hosteller?

Let

${ E }_{ 1 }$ and

${ E }_{ 2 }$ be the events that the student is a hostler and a day scholar respectively and

$A$ be the event that the chosen student gets grade

$A$ .

$\therefore P\left( { E }_{ 1 } \right) = 60$ %

$= \displaystyle\frac { 60 }{ 100 } = 0.6$

$P\left( { E }_{ 2 } \right) = 40$ %

$= \displaystyle\frac { 40 }{ 100 } = 0.4$

$P\left( A|{ E }_{ 1 } \right) = P\left( student \ getting \ an \ A \ grade \ is \ a \ hostler \right) = 30$ %

$= 0.3$

$P\left( A|{ E }_{ 2 } \right) = P\left( student \ getting \ an \ A \ grade \ is \ a \ day \ scholar \right) = 20$ %

$= 0.2$

The probability that a randomly chosen student is a hostler, given that he has an

$A$ grade, is given by

$P\left( { E }_{ 1 }|A \right)$ .
By using Baye's theorem, we obtain

$P\left( { E }_{ 1 }|A \right) = \displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( A|{ E }_{ 2 } \right) }$

$= \displaystyle\frac { 0.6\times 0.3 }{ 0.6\times 0.3+0.4\times 0.2 }$

$= \displaystyle\frac { 0.18 }{ 0.26 }$

$= \displaystyle\frac { 18 }{ 26 }$

$= \displaystyle\frac { 9 }{ 13 }=0.69$

A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Let

${E}_{1}$ and

${E}_{2}$ be the events of selecting first bag and second bag respectively.

$\therefore P\left({E}_{1}\right) = P\left({E}_{2}\right) = \displaystyle\frac{1}{2}$
Let

$A$ be the event of getting a red ball.

$\Rightarrow P\left(A|{E}_{1}\right) = P\left(\text{drawing a red ball from first bag}\right) = \displaystyle\frac{4}{8} = \displaystyle\frac{1}{2}$

$\Rightarrow P\left(A|{E}_{2}\right) = P\left(\text{drawing a red ball from second bag}\right) = \displaystyle\frac{2}{8} = \displaystyle\frac{1}{4}$
The probability of drawing a ball from the first bag, given that it is red, is given by

$P\left({E}_{2}|A\right)$ .
By using Baye's theorem, we obtain

$P\left( { E }_{ 1 }|A \right) = \displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) + P\left( { E }_{ 2 } \right) \cdot P\left( A|{ E }_{ 2 } \right) }$

$= \displaystyle\frac { \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 1 }{ 2 } }{ \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 1 }{ 2 } + \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 1 }{ 4 } }$

$= \displaystyle\frac { \displaystyle\frac { 1 }{ 4 } }{ \displaystyle\frac { 1 }{ 4 } + \displaystyle\frac { 1 }{ 8 } }$

$= \displaystyle\frac { 4 }{ \displaystyle\frac { 3 }{ 8 } }$

$= \displaystyle\frac { 2 }{ 3 }=0.66$

Consider the experiment of throwing a die, if a multiple of $3$ comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a $3$ '.

The outcomes of the given experiment can be represented by the following tree diagram.
The sample space of the experiment is,

$S=\left\{ \left( 1,H \right) ,\left( 1,T \right) ,\left( 2,H \right) ,\left( 2,T \right) ,\left( 3,1 \right) ,\left( 3,2 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) ,\\ \left( 4,H \right) ,\left( 4,T \right) ,\left( 5,H \right) ,\left( 5,T \right) ,\left( 6,1 \right) ,\left( 6,2 \right) ,\left( 6,3 \right) ,\left( 6,4 \right) ,\left( 6,5 \right) ,\left( 6,6 \right) \right\}$
Let

$A$ be the event that the coin shows a tail and

$B$ be the event that at least one die shows

$3$ .

$\therefore A=\left\{ \left( 1,T \right) ,\left( 2,T \right) ,\left( 4,T \right) ,\left( 5,T \right) \right\}$

$B=\left\{ \left( 3,1 \right) ,\left( 3,2 \right) ,\left( 3,3 \right) ,\left( 3,4 \right) ,\left( 3,5 \right) ,\left( 3,6 \right) ,\left( 6,3 \right) \right\}$

$\Rightarrow A\cap B=\phi$

$\therefore P\left( A\cap B \right) =0$
Then,

$P\left( B \right) =P\left( \left\{ 3,1 \right\} \right) +P\left( \left\{ 3,2 \right\} \right) +P\left( \left\{ 3,3 \right\} \right) +P\left( \left\{ 3,4 \right\} \right) +P\left( \left\{ 3,5 \right\} \right) +P\left( \left\{ 3,6 \right\} \right) +P\left( \left\{ 6,3 \right\} \right)$

$=\displaystyle\frac { 1 }{ 36 } + \displaystyle\frac { 1 }{ 36 } + \displaystyle\frac { 1 }{ 36 } + \displaystyle\frac { 1 }{ 36 } + \displaystyle\frac { 1 }{ 36 } + \displaystyle\frac { 1 }{ 36 } + \displaystyle\frac { 1 }{ 36 }$

$= \displaystyle\frac { 7 }{ 36 }$
Probability of the event that the coin shows a tail, given that at least one die shows

$3$ , is given by

$P\left( A|B \right)$ .
Therefore,

$P\left( A|B \right) = \displaystyle\frac { P\left( A\cap B \right) }{ P\left( B \right) } =\displaystyle\frac { 0 }{ \displaystyle\frac { 7 }{ 36 } } =0$

Two cards are drawn at random and without replacement from a pack of $52$ playing cards. Find the probability that both the cards are black.

There are

$26$ black cards in a deck of

$52$ cards.

Let

$P\left( A \right)$ be the probability of getting a black card in the first draw.

$\therefore P\left( A \right) = \displaystyle\frac { 26 }{ 52 } = \displaystyle\frac { 1 }{ 2 }$

Let

$P\left( B \right)$ be the probability of getting a black card on the second draw.

Since the card is not replaced,

$\therefore P\left( B \right) = \displaystyle\frac { 25 }{ 51 }$
Thus, probability of getting both the cards black

$= \displaystyle\frac { 1 }{ 2 } \times \displaystyle\frac { 25 }{ 51 } = \displaystyle\frac { 25 }{ 102 }=0.24$ .

An instructor has a question bank consisting of $300$ easy True / False questions, $200$ difficult True / False questions, $500$ easy multiple choice questions and $400$ difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question ?

$\textbf{Step 1: Finding P(B)}$

$\text{Let A be the event that the question is easy}$

$\text{Let B be the event that the question is a multiple choice question}$

$\text{We have to find }P(A|B)$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$\text{There are }500+400=900\text{ multiple choice questions}$

$\text{Total number of questions}=300+200+500+400=1400$

$\therefore P(B)=\dfrac{900}{1400}=\dfrac{9}{14}$

$\textbf{Step 2: Finding }\mathbf{P(A\cap B)}$

$A\cap B\text{ is the event that the question is an easy multiple choice question}$

$\text{Number of such questions}=500$

$\text{Total number of questions}=300+200+500+400=1400$

$\therefore P(A\cap B)=\dfrac{500}{1400}=\dfrac{5}{14}$

$\textbf{Step 2: Finding }\mathbf{P(A|B)}$

$P(A|B)=\dfrac{P(A\cap B)}{P(B)}$

$\implies P(A|B)=\dfrac{\dfrac{5}{14}}{\dfrac{9}{14}}=\dfrac{5}{9}\approx0.556$

$\textbf{Hence, probability that the question is easy given that the question is a multiple choice}$

$\textbf{question is }\mathbf{\dfrac{5}{9}}$

A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing $15$ oranges out of which $12$ are good and $3$ are bad ones will be approved for sale.

$\textbf{Step 1: Find Probability of Three Oranges Drawn}$

$\text{Given: Total Number of Oranges=15 }$

$\text{ Good Oranges=12}$

$\text{ Bad Oranges= 3}$

$\text{Let A,B and C be the respective Events that the First, Second, and Third drawn Orange are Good.}$

$\therefore \text{Probability that First Drawn Orange is Good=}$

$P\left( A \right) = \displaystyle\frac { 12 }{ 15 }$

$\therefore\text{Probability of getting Second Orange is Good=}$

$P\left( B \right) = \displaystyle\frac { 11 }{ 14 }$

$\textbf{[Since,the Oranges are not replaced]}$

$\text{Similarly, Probability of getting Third Orange Good=}$

$P\left( C \right) = \displaystyle\frac { 10 }{ 13 }$

$\textbf{Step 2: Find Probability if All Drawn Oranges are Good}$

$\text{The Box is Approved for Sale if all the Three Oranges are Good.}$

$\text{Thus, Probability of getting all the oranges Good}$

$= \displaystyle\frac { 12 }{ 15 } \times \displaystyle\frac { 11 }{ 14 } \times \displaystyle\frac { 10 }{ 13 }$

$= \displaystyle\frac { 44 }{ 91 }$

$=0.48$

$\textbf{Therefore, the Probability that the Box is Approved for Sale is}$

$0.48$

In answering a question on a multiple-choice test, a student either knows the answer or guesses. Let ${ 3 }/{ 4 }$ be the probability that he knows the answer and ${ 1 }/{ 4 }$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability ${ 1 }/{ 4 }$ . What is the probability that the student knows the answer given that he answered it correctly?

Let

${ E }_{ 1 }$ and

${ E }_{ 2 }$ be the respective events that the student knows the answer and he guesses the answer.
Let

$A$ be the event that the answer is correct.

$\therefore P\left( { E }_{ 1 } \right) = \displaystyle\frac { 3 }{ 4 }$

$P\left( { E }_{ 2 } \right) = \displaystyle\frac { 1 }{ 4 }$

The probability that the student answered correctly, given that he knows the answer, is 1.

$\therefore P\left( A|{ E }_{ 1 } \right) =1$
Probability that the student answered correctly, given that he guessed, is

$\displaystyle\frac { 1 }{ 4 }$

$\therefore P\left( A|{ E }_{ 2 } \right) = \displaystyle\frac { 1 }{ 4 }$
The probability that the student knows the answer, given that he answered it correctly, is given by

$P\left( { E }_{ 1 }|A \right)$

By using Baye's theorem, we obtain

$P\left( { E }_{ 1 }|A \right) = \displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( A|{ E }_{ 2 } \right) }$

$= \displaystyle\frac { \displaystyle\frac { 3 }{ 4 } \cdot 1 }{ \displaystyle\frac { 3 }{ 4 } \cdot 1 + \displaystyle\frac { 1 }{ 4 } \cdot \displaystyle\frac { 1 }{ 4 } }$

$= \displaystyle\frac { \displaystyle\frac { 3 }{ 4 } }{ \displaystyle\frac { 3 }{ 4 } + \displaystyle\frac { 1 }{ 16 } }$

$= \displaystyle\frac { \displaystyle\frac { 3 }{ 4 } }{ \displaystyle\frac { 13 }{ 16 } }$

$= \displaystyle\frac { 12 }{ 13 }=0.92$ .

Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are $0.6$ and $0.4$ respectively. Further, if the first group wins, the probability of introducing a new product is $0.7$ and the corresponding probability is $0.3$ if the second group wins. Find the probability that the new product introduced was by the second group.

Let

${ E }_{ 1 }$ and

${ E }_{ 2 }$ be the respective events that the first group and the second group win the competition. Let

$A$ be the event of introducing a new product.

$P\left( { E }_{ 1 } \right) =$ Probability that the first group wins the competition

$= 0.6$

$P\left( { E }_{ 2 } \right) =$ Probability that the second group wins the competition

$= 0.4$

$P\left( A|{ E }_{ 1 } \right) =$ Probability of introducing a new product if the first group wins

$= 0.7$

$P\left( A|{ E }_{ 2 } \right) =$ Probability of introducing a new product if the second group wins

$= 0.3$
The probability that the new product is introduced by the second group is given by

$P\left( { E }_{ 2 }|A \right)$ .
By using Baye's theorem, we obtain

$P\left( { E }_{ 2 }|A \right) =\displaystyle\frac { P\left( { E }_{ 2 } \right) \cdot P\left( A|{ E }_{ 2 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( A|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( A|{ E }_{ 2 } \right) }$

$= \displaystyle\frac { 0.4\times 0.3 }{ 0.6\times 0.7+0.4\times 0.3 }$

$= \displaystyle\frac { 0.12 }{ 0.42+0.12 }$

$= \displaystyle\frac { 0.12 }{ 0.54 }$

$= \displaystyle\frac { 12 }{ 54 }$

$= \displaystyle\frac { 2 }{ 9 }=0.22$

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

It is known that the sum of all the probabilities in a probability distribution is one.
(i) Sum of the probabilities

$= 0.4 + 0.4 + 0.2 = 1$
Therefore, the given table is a probability distribution of random variables.
(ii) It can be seen that for

$X = 3, P\left(X\right) =-0.1$
It is known that probability of any observation is not negative. Therefore, the given table is not a probability distribution of random variables.
(iii) Sum of the probabilities

$= 0.6 + 0.1 + 0.2 = 0.9 \neq 1$
Therefore, the given table is not a probability distribution of random variables.
(iv) Sum of the probabilities

$= 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05 \neq 1$
Therefore, the given table is not a probability distribution of random variables.

A manufacturer has three machine operators $A, B$ and $C$ . The first operator $A$ produces $1\%$ defective items, whereas the other two operators $B$ and $C$ produce $5\%$ and $7\%$ defective items respectively. $A$ is on the job for $50\%$ of the time, $B$ is on the job for $30\%$ of the time and $C$ is on the job for $20\%$ of the time. A defective item is produced, what is the probability that it was produced by $A$ ?

Let

${ E }_{ 1 }, { E }_{ 2 }$ and

${ E }_{ 3 }$ be the respective events of the time consumed by machines

$A, B$ and

$C$ for the job.

$P\left( { E }_{ 1 } \right) =50$ %

$=\displaystyle\frac { 50 }{ 100 } =\displaystyle\frac { 1 }{ 2 }$

$P\left( { E }_{ 2 } \right) =30$ %

$=\displaystyle\frac { 30 }{ 100 } =\displaystyle\frac { 3 }{ 10 }$

$P\left( { E }_{ 3 } \right) =20$ %

$=\displaystyle\frac { 20 }{ 100 } =\displaystyle\frac { 1 }{ 5 }$

Let

$X$ be the event of producing defective items.

$P\left( X|{ E }_{ 1 } \right) =1$ %

$=\displaystyle\frac { 1 }{ 100 }$

$P\left( X|{ E }_{ 2 } \right) =5$ %

$=\displaystyle\frac { 5 }{ 100 }$

$P\left( X|{ E }_{ 3 } \right) =7$ %

$=\displaystyle\frac { 7 }{ 100 }$

The probability that the defective item was produced by

$A$ is given by

$P\left( { E }_{ 1 }|A \right)$ .
By using Baye's theorem, we obtain

$P\left( { E }_{ 1 }|X \right) =\displaystyle\frac { P\left( { E }_{ 1 } \right) \cdot P\left( X|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) \cdot P\left( X|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) \cdot P\left( X|{ E }_{ 2 } \right) +P\left( { E }_{ 3 } \right) \cdot P\left( X|{ E }_{ 3 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 1 }{ 100 } }{ \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 1 }{ 100 } +\displaystyle\frac { 3 }{ 10 } \cdot \displaystyle\frac { 5 }{ 100 } +\displaystyle\frac { 1 }{ 5 } \cdot \displaystyle\frac { 7 }{ 100 } }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 100 } \cdot \displaystyle\frac { 1 }{ 2 } }{ \displaystyle\frac { 1 }{ 100 } \left( \displaystyle\frac { 1 }{ 2 } +\displaystyle\frac { 3 }{ 2 } +\displaystyle\frac { 7 }{ 5 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 2 } }{ \displaystyle\frac { 17 }{ 5 } }$

$=\displaystyle\frac { 5 }{ 34 }=0.14$

Suppose a girl throws a die. If she gets a $5$ or $6$ , she tosses a coin three times and notes the number of heads. If she gets $1, 2, 3$ or $4$ , she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw $1, 2, 3$ or $4$ with the die ?

$\textbf{Step - 1: Define Formula}$

$\text{Probability, that exactly one head is obtained, when throwing} 1, 2, 3, 4, P(A|C) \text{ is }$

$P(A|C) = \dfrac{{ P(A).P(C|A)}}{{P(A).P(C|A)+P(B).P(C|B)}}$

$\text{ A is Getting } 1, 2, 3, 4 \text{ appear on the die}$

$\text{ B is Getting } 5, 6 \text{ appear on the die}$

$\text{ C is Getting exactly one head}$

$\textbf{Step - 2: Finding the components of formula}$

$P(A) \text{ is probability that } 1,2,3,4 \text{ appears on die}$

$P(A) = \dfrac{{4}}{{6}} = \dfrac{{2}}{{3}}$

$P(C|A) \text{ is probability that exactlly one head appears, when getting } 1, 2, 3, 4$

$P(C|A) = \dfrac{{1}}{{2}}$

$P(B) \text{ is probability that } 5,6 \text{ appears on die}$

$P(B) = \dfrac{{2}}{{6}} = \dfrac{{1}}{{3}}$

$P(C|B) \text{ is probability that exactly one head appears, when getting } 5, 6$

$P(C|B) = \dfrac{{3}}{{8}}$

$\textbf{Step - 3: Putting everything in the formula}$

$P(A|C) = \dfrac{{ \dfrac{2}{3}\times \dfrac{1}{2}}}{{\dfrac{2}{3} \times \dfrac{1}{2}+\dfrac{1}{3} \times \dfrac{3}{8}}}$

$P(A|C) = \dfrac{{ \dfrac{1}{3}}}{{\dfrac{1}{3} + \dfrac{1}{8}}}$

$= \dfrac{{1}}{{3}} \times \dfrac{{24}}{{11}}$

$= \dfrac{{8}}{{11}}$

$\textbf{Hence, the required probability is } \boldsymbol{\dfrac{{8}}{{11}}}$

Find the probability distribution of
(i) number of heads in two tosses of a coin.

(i) When one coin is tosses twice, the sample space is

$\left\{ HH,HT,TH,TT \right\}$
Let

$X$ represent the number of heads.

$\therefore X\left( HH \right) =2,X\left( HT \right) =1,X\left( TH \right) =1,X\left( TT \right) =0$
Therefore,

$X$ can take the value of

$0, 1$ or

$2$ .
It is known that,

$P\left( HH \right) -P\left( HT \right) -P\left( TH \right) -P\left( TT \right) =\displaystyle\frac { 1 }{ 4 }$

$P\left( X=0 \right) =P\left( TT \right) =\displaystyle\frac { 1 }{ 4 }$

$P\left( X=1 \right) =P\left( HT \right) +P\left( TH \right) =\displaystyle\frac { 1 }{ 4 } +\displaystyle\frac { 1 }{ 4 } =\displaystyle\frac { 1 }{ 2 }$

$P\left( X=2 \right) =P\left( HH \right) =\displaystyle\frac { 1 }{ 4 }$
Thus, the required probability distribution is as follows.
(ii) When three coins are tossed simultaneously, the sample space is

$\left\{ HHH,HHT,HTH,HTT,THH,THT,TTH,TTT \right\}$
Let

$X$ represents the number of tails.
It can be seen that

$X$ can take the value of

$0, 1, 2$ or

$3$ .

$P\left( X=0 \right) =P\left( HHH \right) =\displaystyle\frac { 1 }{ 8 }$

$P\left( X=1 \right) =P\left( HHT \right) +P\left( HTH \right) +P\left( THH \right) =\displaystyle\frac { 1 }{ 8 } +\displaystyle\frac { 1 }{ 8 } +\displaystyle\frac { 1 }{ 8 } =\displaystyle\frac { 3 }{ 8 }$

$P\left( X=2 \right) =P\left( HTT \right) +P\left( THT \right) +P\left( TTH \right) =\displaystyle\frac { 1 }{ 8 } +\displaystyle\frac { 1 }{ 8 } +\displaystyle\frac { 1 }{ 8 } =\displaystyle\frac { 3 }{ 8 }$

$P\left( X=3 \right) =P\left( TTT \right) =\displaystyle\frac { 1 }{ 8 }$
Thus, the probability distribution is as follows.
(iii) When a coin is tossed four times, the sample space is

$S=\left\{ HHHH,HHHT,HHTH,HHTT,HTHT,HTHH,HTTH,HTTT,\\ THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT \right\}$
Let

$X$ be the random variable, which represents the number of heads.
It can be seen that

$X$ can take the value of

$0, 1, 2, 3$ or

$4$ .

$P\left( X=0 \right) =P\left( TTTT \right) =\displaystyle\frac { 1 }{ 16 }$

$P\left( X=1 \right) =P\left( TTTH \right) +P\left( TTHT \right) +P\left( THTT \right) +P\left( HTTT \right) =\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } =\displaystyle\frac { 4 }{ 16 } =\displaystyle\frac { 1 }{ 4 }$

$P\left( X=2 \right) =P\left( HHTT \right) +P\left( THHT \right) +P\left( TTHH \right) +P\left( HTTH \right) +P\left( HTHT \right) +P\left( THTH \right) =\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } =\displaystyle\frac { 6 }{ 16 } =\displaystyle\frac { 3 }{ 8 }$

$P\left( X=3 \right) =P\left( HHHT \right) +P\left( HHTH \right) +P\left( HTHH \right) +P\left( THHH \right) =\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } +\displaystyle\frac { 1 }{ 16 } =\displaystyle\frac { 4 }{ 16 } =\displaystyle\frac { 1 }{ 4 }$

$P\left( X=4 \right) =P\left( HHHH \right) =\displaystyle\frac { 1 }{ 16 }$
Thus, the probability distribution is as follows.

There are three coins. One is a two-headed coin another is a biased coin that comes up heads $75$ % of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?

${\textbf{Step-1: Let}}$ ${{\text{E}}_{\text{1}}}$ ${{\text{E}}_2}$ ${{\text{E}}_3}$ ${\textbf{be the resp. events of choosing a two headed coin, a biased}}$

${\textbf{coin, and an unbiased coin.}}$

${\text{P}}$ ( ${{\text{E}}_{\text{1}}}$ ${{\text{E}}_2}$ ${{\text{E}}_3}$ $\dfrac{{\text{1}}}{{\text{3}}}$

${\text{Let A be the event that the coin shows heads. A two-headed coin will always show}}$

${\text{heads.}}$

${\text{P ( A |}}$ ${{\text{E}}_{\text{1}}}$ ) ${\text{= P (coin showing heads, given that it is a two - heads) = 1}}$

${\text{Probability of heads coming up, given that it is a biased coin = 75 %}}$

${\text{P ( A |}}$ ${{\text{E}}_2}$ $\dfrac{{{\text{75}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{\text{3}}}{{\text{4}}}$

${\text{Since the third coin is unbiased, the probability that it shows heads is always}}$ $\dfrac{{\text{1}}}{{\text{2}}}$

${\textbf{Step-2: The probability that the coin is two-headed, given that it shows heads, is given by}}$

${\text{P}}$ ( ${{\text{E}}_{\text{1}}}$ ${\text{| A )}}$

${\text{By using Baye’s theore, we obtain}}$

${\text{P}}$ ( ${{\text{E}}_{\text{1}}}$ $\dfrac{{{\text{P(}}{{\text{E}}_{\text{1}}}{\text{)}}{\text{.P(A|}}{{\text{E}}_{\text{1}}}{\text{)}}}}{{{\text{P(}}{{\text{E}}_{\text{1}}}{\text{)}}{\text{.P(A|}}{{\text{E}}_{\text{1}}}{\text{) + P(}}{{\text{E}}_{\text{2}}}{\text{)}}{\text{.P(A|}}{{\text{E}}_{\text{2}}}{\text{) + P(}}{{\text{E}}_{\text{3}}}{\text{)}}{\text{.P(A|}}{{\text{E}}_{\text{3}}}{\text{)}}}}$

${\text{=}}$ $\dfrac{{\dfrac{{\text{1}}}{{\text{3}}}{\text{.1}}}}{{\dfrac{{\text{1}}}{{\text{3}}}{\text{.1 + }}\dfrac{{\text{1}}}{{\text{3}}}{\text{.}}\dfrac{{\text{3}}}{{\text{4}}}{\text{ + }}\dfrac{{\text{1}}}{{\text{3}}}{\text{.}}\dfrac{{\text{1}}}{{\text{2}}}}}$

${\text{=}}$ $\dfrac{{\text{1}}}{{\dfrac{{\text{9}}}{{\text{4}}}}}$

${\text{=}}$ $\dfrac{{\text{4}}}{{\text{9}}}$

${\text{= 0.44}}$

${\textbf{Hence, Probability is 0.44}}$

An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents are $0.01, 0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ?

Let

${ E }_{ 1 }, { E }_{ 2 }$ , and

${ E }_{ 3 }$ be the respective events that the driver is a scooter driver, a car driver, and a truck driver.
Let

$A$ be the event that the person meets with an accident.
There are

$2000$ scooter drivers,

$4000$ car drivers, and

$6000$ truck drivers.
Total number of drivers

$= 2000 + 4000 + 6000 = 12000$

$P\left( { E }_{ 1 } \right) =P\left( driver is a scooter driver \right) =\displaystyle\frac { 2000 }{ 12000 } = \displaystyle\frac { 1 }{ 6 }$

$P\left( { E }_{ 2 } \right) =P\left( driver is a car driver \right) =\displaystyle\frac { 4000 }{ 12000 } =\displaystyle\frac { 1 }{ 3 }$

$P\left( { E }_{ 3 } \right) =P\left( driver is a truck driver \right) =\displaystyle\frac { 6000 }{ 12000 } =\displaystyle\frac { 1 }{ 2 }$

$P\left( A|{ E }_{ 1 } \right) =P\left( scooter driver met with an accident \right) =0.01=\displaystyle\frac { 1 }{ 100 }$

$P\left( A|{ E }_{ 2 } \right) =P\left( car driver met with an accident \right) =0.03=\displaystyle\frac { 3 }{ 100 }$

$P\left( A|{ E }_{ 3 } \right) =P\left( truck driver met with an accident \right) =0.15=\displaystyle\frac { 15 }{ 100 }$
The probability that the driver is a scooter driver, given that he met with an accident, is given by

$P\left( { E }_{ 1 }|A \right)$ .
By using Baye's theorem, we obtain

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 6 } \cdot \displaystyle\frac { 1 }{ 100 } }{ \displaystyle\frac { 1 }{ 6 } \cdot \displaystyle\frac { 1 }{ 100 } +\displaystyle\frac { 1 }{ 3 } \cdot \displaystyle\frac { 3 }{ 100 } +\displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 15 }{ 100 } }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 6 } \cdot \displaystyle\frac { 1 }{ 100 } }{ \displaystyle\frac { 1 }{ 100 } \left( \displaystyle\frac { 1 }{ 6 } +1+\displaystyle\frac { 15 }{ 2 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 6 } }{ \displaystyle\frac { 52 }{ 6 } }$

$=\displaystyle\frac { 1 }{ 6 } \times \displaystyle\frac { 12 }{ 104 }$

$=\displaystyle\frac { 1 }{ 52 }=0.019$

A factory has two machines $A$ and $B$ . Past record shows that machine $A$ produced $60\%$ of the items of output and machine $B$ produced $40\%$ of the items. Further, $2\%$ of the items produced by machine $A$ were defective and $1\%$ produced by machine $B$ were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine $B$ ?

$\textbf{Step 1: Computing the individual probabilities}$

$\text{Let the probability that the product was made by Machine A be }E_1$

$\text{Let the probability that the product was made by Machine B be }E_2$

$\text{Let the probability that the product was defective be }A$

$P(E_1)=\dfrac{60}{100}=0.6$

$P(E_2)=\dfrac{40}{100}=0.4$

$2\%\text{ of items produced by A was defective}$

$P(A|E_1)=\dfrac{2}{100}=0.02$

$1\%\text{ of items produced by B was defective}$

$P(A|E_2)=\dfrac{1}{100}=0.01$

$\textbf{Step 2: Using Bayes' Theorem to compute the required probability}$

$\text{The probability that the product was produced by machine B given it was defective is }P(E_2|A)$

$\text{Using Bayes' Theorem}$

$P(E_2|A)=\dfrac{P(E_2)\times P(A|E_2)}{P(E_1)\times P(A|E_1)+P(E_2)\times P(A|E_2)}$

$=\dfrac{0.4\times0.01}{0.6\times0.02+0.4\times0.01}$

$=0.25$

$\textbf{Hence ,The probability that the product was produced by machine B given it was defective}$

$\textbf{is }\mathbf{0.25}$

A laboratory blood test is $99\%$ effective in detecting a certain disease when it is present. However, the test also yields a false-positive result for $0.5\%$ of the healthy person tested (i.e., if a healthy person is tested, then, with probability $0.005$ , the test will imply he has the disease). If $0.1$ percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Let

${ E }_{ 1 }$ and

${ E }_{ 2 }$ be the respective events that a person has a disease and a person has no disease

.
Since

${ E }_{ 1 }$ and

${ E }_{ 2 }$ are events complimentary to each other.

$\therefore P\left( { E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) =1$

$\Rightarrow P\left( { E }_{ 2 } \right) =1-P\left( { E }_{ 1 } \right) =1-0.001=0.999$

Let

$A$ be the event that the blood test result is positive.

$P\left( { E }_{ 1 } \right) =0.1$ %

$=\displaystyle\frac { 0.1 }{ 100 } =0.001$

$P\left( A|{ E }_{ 1 } \right) =P\left( result \ is \ positive \ given \ the \ person \ has \ disease \right) =99$ %

$=0.99$

$P\left( A|{ E }_{ 2 } \right) =P\left( result \ is \ positive \ given \ that \ the \ person \ has \ no \ disease \right) =0.5$ %

$=0.005$

Probability that a person has a disease, given that his test result is positive, is given by

$P\left( { E }_{ 1 }|A \right)$ .
By using Baye's theorem, we obtain

$=\displaystyle\frac { 0.001\times 0.99 }{ 0.001\times 0.99+0.999\times 0.005 }$

$=\displaystyle\frac { 0.00099 }{ 0.00099+0.004995 }$

$=\displaystyle\frac { 0.00099 }{ 0.005985 }$

$=\displaystyle\frac { 990 }{ 5985 }$

$=\displaystyle\frac { 110 }{ 665 }$

$=\displaystyle\frac { 22 }{ 133 }=0.165$

Two numbers are selected at random (without replacement) the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E\left(X\right)$ .

The two positive integers can be selected from the first six positive integers without replacement in

$6 \times 5 = 30$ ways.

$X$ represent the larger of the two numbers obtained. Therefore,

$X$ can take the value of 2, 3, 4, 5, or 6.
For

$X = 2$ , the possible observations are

$\left(1, 2\right)$ and

$\left(2, 1\right)$ .

$\therefore P\left(X = 2\right) = \displaystyle\frac{2}{30} = \displaystyle\frac{1}{15}$
For

$X = 3$ , the possible observations are

$\left(1, 3\right), \left(2, 3\right), \left(3, 1\right), and \left(3, 2\right)$ .

$\therefore P\left(X = 3\right) = \displaystyle\frac{4}{30} = \displaystyle\frac{2}{15}$
For

$X = 4$ , the possible observations are

$\left(1, 4\right), \left(2, 4\right), \left(3, 4\right), \left(4, 3\right), \left(4, 2\right)$ , and

$\left(4, 1\right)$ .

$\therefore P\left(X = 4\right) = \displaystyle\frac{6}{30} = \displaystyle\frac{1}{5}$
For

$X = 5$ , the possible observations are

$\left(1, 5\right), \left(2, 5\right), \left(3, 5\right), \left(4, 5\right), \left(5, 4\right), \left(5, 3\right), \left(5, 2\right)$ , and

$\left(5, 1\right)$ .

$\therefore P\left(X = 5\right) = \displaystyle\frac{8}{30} = \displaystyle\frac{4}{15}$
For

$X = 6$ , the possible observations are

$\left(1, 6\right), \left(2, 6\right), \left(3, 6\right), \left(4, 6\right), \left(5, 6\right), \left(6, 4\right), \left(6, 3\right), \left(6, 2\right)$ , and

$\left(6, 1\right)$ .

$\therefore P\left(X = 6\right) = \displaystyle\frac{10}{30} = \displaystyle\frac{1}{3}$ .
Therefore, the required probability distribution is as follows.
Then,

$E\left( X \right) =\sum { { X }_{ i }P\left( { X }_{ i } \right) }$

$=2\cdot \displaystyle\frac { 1 }{ 15 } +3\cdot \displaystyle\frac { 2 }{ 15 } +4\cdot \displaystyle\frac { 1 }{ 5 } +5\cdot \displaystyle\frac { 4 }{ 15 } +6\cdot \displaystyle\frac { 1 }{ 3 }$

$=\displaystyle\frac { 2 }{ 15 } +\displaystyle\frac { 2 }{ 5 } +\displaystyle\frac { 4 }{ 5 } +\displaystyle\frac { 4 }{ 3 } +2$

$=\displaystyle\frac { 70 }{ 15 }$

$=\displaystyle\frac { 14 }{ 3 }$

An urn contains 25 balls of which 10 balls bear a mark $'X'$ and the remaining 15 bear a mark $'Y'$ . A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that
(i) all will bear $'X'$ mark
(ii) not more than 2 will bear $'Y'$ mark
(iii) at least one ball will bear $'Y'$ mark
(iv) the number of balls with $'X'$ mark and $'Y'$ mark will be equal.

Total number of balls in the urn

$= 25$
Balls bearing mark

$'X' = 10$
Balls bearing mark

$'Y' = 15$

$p=P\left( ball bearing mark 'X' \right) =\displaystyle\frac { 10 }{ 25 } =\displaystyle\frac { 2 }{ 5 }$

$q=P\left( ball bearing mark 'Y' \right) =\displaystyle\frac { 15 }{ 25 } =\displaystyle\frac { 3 }{ 5 }$
Six balls are drawn with replacement. Therefore, the number of trials are Bernoulli trials.
Let

$Z$ be the random variable that represents the number of balls with

$'Y'$ mark on them in the trials.
Clearly,

$Z$ has a binomial distribution with

$n = 6$ and

$p=\displaystyle\frac { 2 }{ 5 }$

$\therefore P\left( Z=z \right) =_{ }^{ n }{ { C }_{ z } }{ p }^{ n-z }{ q }^{ z }$
(i)

$P\left( all will bear 'X' mark \right) =P\left( Z=0 \right) =_{ }^{ 6 }{ { C }_{ 0 } }{ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 6 }={ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 6 }$
(ii)

$P\left( not more than 2 bear 'Y' mark \right) =P\left( Z\le 2 \right)$

$=P\left( Z=0 \right) +P\left( Z=1 \right) +P\left( Z=2 \right)$

$=_{ }^{ 6 }{ { C }_{ 0 } }{ \left( p \right) }^{ 6 }{ \left( q \right) }^{ 0 }+_{ }^{ 6 }{ { C }_{ 1 } }{ \left( p \right) }^{ 5 }{ \left( q \right) }^{ 1 }+_{ }^{ 6 }{ { C }_{ 2 } }{ \left( p \right) }^{ 4 }{ \left( q \right) }^{ 2 }$

$={ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 6 }+6{ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 5 }\left( \displaystyle\frac { 3 }{ 5 } \right) +15{ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 4 }{ \left( \displaystyle\frac { 3 }{ 5 } \right) }^{ 2 }$

$={ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 4 }\left[ { \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 2 }+6\left( \displaystyle\frac { 2 }{ 5 } \right) \left( \displaystyle\frac { 3 }{ 5 } \right) +15{ \left( \displaystyle\frac { 3 }{ 5 } \right) }^{ 2 } \right]$

$={ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 4 }\left[ \displaystyle\frac { 4 }{ 25 } +\displaystyle\frac { 36 }{ 25 } +\displaystyle\frac { 135 }{ 25 } \right]$

$={ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 4 }\left[ \displaystyle\frac { 175 }{ 25 } \right]$

$=7{ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 4 }$
(iii)

$P\left( at least one ball bears 'Y' mark \right) =P\left( Z\ge 1 \right) =1-P\left( Z=0 \right)$

$=1-{ \left( \displaystyle\frac { 2 }{ 5 } \right) }^{ 6 }$
(iv)

$P\left( equal number of balls with 'X' mark and 'Y' mark \right) =P\left( Z=3 \right)$

$=_{ }^{ 6 }{ { C }_{ 3 } }{ \left( \displaystyle\frac { 2 }{ 54 } \right) }^{ 3 }{ \left( \displaystyle\frac { 3 }{ 5 } \right) }^{ 3 }$

$=\displaystyle\frac { 20\times 8\times 27 }{ 15625 }$

$=\displaystyle\frac { 864 }{ 3125 }$

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

The probability of success is twice the probability of failure.
Let the probability of failure be

$x$ .

$\therefore$ Probability of success

$= 2x$

$x + 2x = 1$

$\Rightarrow 3x=1$

$\Rightarrow x=\displaystyle\frac { 1 }{ 3 }$

$\therefore 2x=\displaystyle\frac { 2 }{ 3 }$
Let

$p=\displaystyle\frac { 1 }{ 3 }$ and

$q=\displaystyle\frac { 2 }{ 3 }$
Let

$X$ be the random variable that represents the number of successes in six trials.
By binomial distribution, we obtain

$P\left( X=x \right) =_{ }^{ n }{ { C }_{ x } }{ p }^{ n-x }{ q }^{ x }$
Probability of at least 4 successes

$=P\left( X\ge 4 \right)$

$=P\left( X=4 \right) +P\left( X=5 \right) +P\left( X=6 \right)$

$=_{ }^{ 6 }{ { C }_{ 4 } }{ \left( \displaystyle\frac { 2 }{ 3 } \right) }^{ 4 }{ \left( \displaystyle\frac { 1 }{ 3 } \right) }^{ 2 }+_{ }^{ 6 }{ { C }_{ 5 } }{ \left( \displaystyle\frac { 2 }{ 3 } \right) }^{ 5 }\left( \displaystyle\frac { 1 }{ 3 } \right) +_{ }^{ 6 }{ { C }_{ 6 } }{ \left( \displaystyle\frac { 2 }{ 3 } \right) }^{ 6 }$

$=\displaystyle\frac { 15{ \left( 2 \right) }^{ 4 } }{ { 3 }^{ 6 } } +\displaystyle\frac { 6{ \left( 2 \right) }^{ 5 } }{ { 3 }^{ 6 } } +\displaystyle\frac { { \left( 2 \right) }^{ 6 } }{ { 3 }^{ 6 } }$

$=\displaystyle\frac { { \left( 2 \right) }^{ 4 } }{ { \left( 3 \right) }^{ 6 } } \left[ 15+12+4 \right]$

$=\displaystyle\frac { 31\times { 2 }^{ 4 } }{ { \left( 3 \right) }^{ 6 } }$

$=\displaystyle\frac { 31 }{ 9 } { \left( \displaystyle\frac { 2 }{ 3 } \right) }^{ 4 }$

Two dice are thrown simultaneously. If $X$ denotes the number of sixes, find the expectation of $X$ .

Here,

$X$ represents the number of sixes obtained when two dice are thrown simultaneously. Therefore,

$X$ can take the value of 0, 1, or 2.

$\therefore P\left(X = 0\right) = P\left(not getting six on any of the dice\right) = \displaystyle\frac{25}{36}$

$P\left(X = 1\right) = P\left(six on first die and no six on second die\right) + P\left(no six on first die and six on second die\right)$

$= 2\left(\displaystyle\frac{1}{6} \times \displaystyle\frac{5}{6}\right) = \displaystyle\frac{10}{36}$

$P\left(X = 2\right) = P\left(six on both the dice\right) = \displaystyle\frac{1}{36}$
Therefore, the required probability distribution is as follows.
Then, expectation of

$X=E\left( X \right) =\sum { { X }_{ i }P\left( { X }_{ i } \right) }$

$=0\times \displaystyle\frac { 25 }{ 36 } +1\times \displaystyle\frac { 10 }{ 36 } +2\times \displaystyle\frac { 1 }{ 36 }$

$=\displaystyle\frac { 1 }{ 3 }=0.33$

The random variable $X$ has a probability distribution $P\left(X\right)$ of the following form, where $k$ is some number :
$P\left(X\right) = \left\{k, if x = 0 \\ 2k, if x = 1 \\ 3k, if x = 2 \\ 0, otherwise\right\}$
(a) Determine the value of $k$
(b) Find $P\left(X < 2\right), P\left(X \le 2\right), P\left(X \ge 2\right)$ .

It is known that the sum of probabilities of a probability distribution of random variables is one.

$\therefore k + 2k + 3k + 0 = 1$

$\Rightarrow 6k = 1$

$\Rightarrow k = \displaystyle\frac{1}{6}$
(b)

$P\left(X < 2\right) = P\left(X = 0\right) + P\left(X = 1\right)$

$= k + 2k$

$= 3k$

$= \displaystyle\frac{3}{6}$

$= \displaystyle\frac{1}{2}$

$P\left(X \le 2\right) = P\left(X = 0\right) + P\left(X = 1\right) + P\left(X = 2\right)$

$= k + 2k + 3k$

$= 6k$

$= \displaystyle\frac{6}{6}$

$= 1$

$P\left(X \ge 2\right) = P\left(X = 2\right) + P\left(X > 2\right)$

$= 3k + 0$

$= 3k$

$= \displaystyle\frac{3}{6}$

$= \displaystyle\frac{1}{2}$

What is the minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than $90\%$ ?

Let the man toss the coin

$n$ times. The

$n$ tosses are

$n$ Bernoulli trials.
Probability

$\left(p\right)$ of getting a head at the toss of a coin is

$\displaystyle\frac { 1 }{ 2 }$

$p=\displaystyle\frac { 1 }{ 2 } ,q=\displaystyle\frac { 1 }{ 2 }$

$\therefore P\left( X=x \right) =_{ }^{ n }{ { C }_{ x } }{ p }^{ n-x }{ q }^{ x }=_{ }^{ n }{ { C }_{ x } }{ \left( \displaystyle\frac { 1 }{ 2 } \right) }^{ n-x }{ \left( \displaystyle\frac { 1 }{ 2 } \right) }^{ x }=_{ }^{ n }{ { C }_{ x } }{ \left( \displaystyle\frac { 1 }{ 2 } \right) }^{ n }$
It is given that,

$P\left(\text{getting at least one head} \right) > \displaystyle\frac { 90 }{ 100 }$

$P\left( X\ge 1 \right) > 0.9$

$\Rightarrow 1-P\left( X=0 \right) > 0.9$

$1-_{ }^{ n }{ { C }_{ 0 } }\cdot \displaystyle\frac { 1 }{ { 2 }^{ n } } > 0.9$

$_{ }^{ n }{ { C }_{ 0 } }\cdot \displaystyle\frac { 1 }{ { 2 }^{ n } } < 0.1$

$\displaystyle\frac { 1 }{ { 2 }^{ n } } < 0.1$

${ 2 }^{ n } > \displaystyle\frac { 1 }{ 0.1 }$

${ 2 }^{ n } > 10$ ....(1)
The minimum value of

$n$ that satisfies the given inequality is

$4$ .
Thus, the man should toss the coin

$4$ or more than

$4$ times.

Bag $I$ contains $3$ red and $4$ black balls and Bag $II$ contains $4$ red and $5$ black balls. One ball is transferred from Bag $I$ and Bag $II$ and then a ball is drawn from Bag $II$ . The ball so drawn is found to be red in colour. Find $310$ times the probability that the transferred ball is black.

Let

${ E }_{ 1 }$ and

${ E }_{ 2 }$ respectively denote the events that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.

$P\left( { E }_{ 1 } \right) =\displaystyle\frac { 3 }{ 7 }$ and

$P\left( { E }_{ 2 } \right) =\displaystyle\frac { 4 }{ 7 }$
Let

$A$ be the event that the ball drawn is red.
When a red ball is transferred from bag I to II,

$P\left( A|{ E }_{ 1 } \right) =\displaystyle\frac { 5 }{ 10 } =\displaystyle\frac { 1 }{ 2 }$
When a black ball is transferred from bag I to II,

$P\left( A|{ E }_{ 2 } \right) =\displaystyle\frac { 4 }{ 10 } =\displaystyle\frac { 2 }{ 5 }$

$\therefore P\left( { E }_{ 2 }|A \right) =\displaystyle\frac { P\left( { E }_{ 2 } \right) P\left( A|{ E }_{ 2 } \right) }{ P\left( { E }_{ 1 } \right) P\left( A|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) P\left( A|{ E }_{ 2 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 4 }{ 7 } \times \displaystyle\frac { 2 }{ 5 } }{ \displaystyle\frac { 3 }{ 7 } \times \displaystyle\frac { 1 }{ 2 } +\displaystyle\frac { 4 }{ 7 } \times \displaystyle\frac { 2 }{ 5 } }$

$=\displaystyle\frac { 16 }{ 31 }=0.51$

In a hurdle race, a player has to cross $10$ hurdles. The probability that he will clear each hurdle is $\displaystyle\frac { 5 }{ 6 }$ . What is the probability that he will knock down fewer than $2$ hurdles ?

Let

$p$ and

$q$ respectively be the probabilities that the player will clear and knock down the hurdle.

$\therefore p=\displaystyle\frac { 5 }{ 6 }$

$\Rightarrow q=1-p=1-\displaystyle\frac { 5 }{ 6 } =\displaystyle\frac { 1 }{ 6 }$
Let

$X$ be the random variable that represents the number of times the player will knock down the hurdle.
Therefore, by binomial distribution, we obtain

$P\left( X=x \right) =_{ }^{ n }{ { C }_{ x } }{ p }^{ n-x }{ q }^{ x }$

$P\left( player \, knocking \, down \, less \, than \, 2 \, hurdles \right) =P\left( X<2 \right)$

$=P\left( X=0 \right) +P\left( X=1 \right)$

$=_{ }^{ 10 }{ { C }_{ 0 } }{ \left( q \right) }^{ 0 }{ \left( p \right) }^{ 10 }+_{ }^{ 10 }{ { C }_{ 1 } }\left( q \right) { \left( p \right) }^{ 9 }$

$={ \left( \displaystyle\frac { 5 }{ 6 } \right) }^{ 10 }+10\cdot \displaystyle\frac { 1 }{ 6 } \cdot { \left( \displaystyle\frac { 5 }{ 6 } \right) }^{ 9 }$

$={ \left( \displaystyle\frac { 5 }{ 6 } \right) }^{ 9 }\left[ \displaystyle\frac { 5 }{ 6 } +\displaystyle\frac { 10 }{ 6 } \right]$

$=\displaystyle\frac { 5 }{ 2 } { \left( \displaystyle\frac { 5 }{ 6 } \right) }^{ 9 }$

$=\displaystyle\frac { { \left( 5 \right) }^{ 10 } }{ 2\times { \left( 6 \right) }^{ 9 } }$

In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he win / loses.

In a throw of a die, the probability of getting a six is

$\displaystyle\frac { 1 }{ 6 }$ and the probability of not getting a 6 is

$\displaystyle\frac { 5 }{ 6 }$
Three cases can be obtain.
i. If he gets a six in the first throw, then the required probability is

$\displaystyle\frac { 1 }{ 6 }$ .
Amount he will receive

$=1$ Rs
ii. If he does not get a six in the first throw and gets a six in the second throw, then probability

$=\left( \displaystyle\frac { 5 }{ 6 } \times \displaystyle\frac { 1 }{ 6 } \right) =\displaystyle\frac { 5 }{ 36 }$
Amount he will receive

$=-1+1=0$ Rs.
iii. If he does not get a six in the first two throws and gets a six in the third throw, then probability

$=\left( \displaystyle\frac { 5 }{ 6 } \times \displaystyle\frac { 5 }{ 6 } \times \displaystyle\frac { 1 }{ 6 } \right) =\displaystyle\frac { 25 }{ 216 }$
Amount he will receive = -Re 1 - Re 1 + Re 1 = -1
Expected value he can win

$=\displaystyle\frac { 1 }{ 6 } \left( 1 \right) +\left( \displaystyle\frac { 5 }{ 6 } \times \displaystyle\frac { 1 }{ 6 } \right) \left( 0 \right) +\left[ { \left( \displaystyle\frac { 5 }{ 6 } \right) }^{ 2 }\times \displaystyle\frac { 1 }{ 6 } \right] \left( -1 \right)$

$=\displaystyle\frac { 36-25 }{ 216 } =\displaystyle\frac { 11 }{ 216 }=0.05$

Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is $6$ .

Favourable outcomes to get product of

$6$ are

$\left \{ (1,6),(6,1),(2,3)(3,2) \right \}$

And total no of outcomes are

$36$

So probability

$=\dfrac{4}{36}=\dfrac{1}{9}$

Three cards are drawn successively with replacement from a well shuffled pack of $52$ cards. Find the probability distribution of the number of spades. Hence find the mean of the distribution.

$0$ spades	$1$ spades	$2$ spades	$3$ spades	Total
$\dfrac { 3 }{ 4 } \times \dfrac { 3 }{ 4 } \times \dfrac { 3 }{ 4 } \\= \dfrac { 27 }{ 64 }$	$3\times \dfrac { 1 }{ 4 } \times \dfrac { 3 }{ 4 } \times \dfrac { 3 }{ 4 } \\=\dfrac { 27 }{ 64 }$	$3\times \dfrac { 1 }{ 4 } \times \dfrac { 1 }{ 4 } \times \dfrac { 3 }{ 4 } \\=\dfrac { 9 }{ 64 }$	$\dfrac { 1 }{ 4 } \times \dfrac { 1 }{ 4 } \times \dfrac { 1 }{ 4 } \\=\dfrac { 1 }{ 64 }$	$\dfrac { 27 }{ 64 } +\dfrac { 27 }{ 64 } +\dfrac { 9 }{ 64 } +\dfrac { 1 }{ 64 } \\=\dfrac { 64 }{ 64 } =1$

The probability of getting a spade remains the same in every pick because the cards are replaced after each pick thus making the picks independent events.

The mean of the distribution is

$\sum x_i P(X=x_i) = \dfrac{3}{4}$ .

An article manufactured by a company consists of two parts $X$ and $Y$ . In the process of manufacture of the part $X$ . $9$ out of $100$ parts may be defective. Similarly $5$ out of $100$ are likely to be defective in part $Y$ . Calculate the probability that the assembled product will not be defective.

Let

$A=$ Part

$X$ is not defective

Probability of

$A$ is

$P(A)=\dfrac {91}{100}$

$B=$ Part

$Y$ is not defective.

Probability of

$B$ is

$P(B)=\dfrac {95}{100}$

Required probability

$=P(A\cap B)=P(A)P(B)=\dfrac {91}{100}\times \dfrac {95}{100}=\dfrac{8645}{10000}$

$8p(\bar A \cup \bar B)$

Given :

$p(A)=\dfrac{3}{8} ; p(B)=\dfrac{1}{2} ; p(A\cap B) = \dfrac{1}{4}$

To find :

$8p(A'\cup B')$

Solution :

$p(A'\cup B')=p(A')+p(B')-p(A'\cap B')$

$p(A'\cup B')=1-p(A)+1-p(B)-(1-p(A\cup B))$

$p(A' \cup B')=1-p(A)-p(B)+p(A\cup B)$ ......(i)

now,

$p(A\cup B)=p(A)+p(B)-p(A\cap B)$

$p(A\cup B)=\dfrac{3}{8}+\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{5}{8}$

putting values in eq. (i), we get

$p(A'\cup B')=1-\dfrac{3}{8}-\dfrac{1}{2}+\dfrac{5}{8}=\dfrac{6}{8}$

so,

$8p(A'\cup B')=6$

$8p(\bar A \cap B)$

Given:

$p(A)=\dfrac 38, p(B)=\dfrac 12, p(A\cap B)=\dfrac 14$

To find:

$8p(\overline A\cap B)$

We know,

$p(\overline A\cap B)=p(B)-p(A\cap B)=\dfrac 12-\dfrac 14=\dfrac {2-1}4=\dfrac {1}4$

Hence,

$8p(\overline A\cap B)=8\times \dfrac 14=2$

If A and B are both random events.
$p(A)=\dfrac{3}{8}$ ; $p(B)=\dfrac{1}{2}$ ; $p(A\ \cap\ B)=\dfrac{1}{4}$
Calculate:
$8p(\bar A \cap \bar B)$

Given:

$p(A)=\cfrac 38, p(B)=\cfrac 12, p(A\cap B)=\cfrac 14$

$p(\overline { A } \cup \overline { B } )=p(\overline { A } )+p(\overline { B } )-p(\overline { A } \cap \overline { B } )=5/8\\ 8p(\bar { A } \cup \bar { B } )=3$

$4p(\bar {B})$

Given:

$p(A)=\dfrac 38, p(B)=\dfrac 12, p(A\cap B)=\dfrac 14$

To find:

$4p(\overline B)$

We know,

$p(\overline B)=1-p(B)=1-\dfrac 12=\dfrac 12$

Hence,

$4p(\overline B) = 4\times \dfrac 12=2$

$8p( A \cap \bar B)$

Given:

$p(A)=\dfrac 38, p(B)=\dfrac 12, p(A\cap B)=\dfrac 14$

To find:

$8p( A\cap\overline B)$

We know,

$p( A\cap \overline B)=p(A)-p(A\cap B)=\dfrac 38-\dfrac 14=\dfrac {3-2}8=\dfrac {1}8$

Hence,

$8p(\overline A\cap B)=8\times \dfrac 18=1$

From a pack of $52$ cards, two cards are drawn randomly one by one without replacement. Find the probability that both of them are red.

Probabilty of getting first card as red is

$\cfrac { 26 }{ 52 }$ as we have

$26$ red cards in a deck.

Now probability of getting second card as red is

$\cfrac { 25 }{ 51 }$ , as we have only

$25$ of red cards left and the total number of cards are

$51$ so total probabilty is:

$\cfrac { 26\times 25 }{ 52\times 51 } =\cfrac { 650 }{ 2652 } = \cfrac{25}{102}$

A die is thrown twice and the sum of the numbers appearing is observed to be $7$ . Find the conditional probability that the number $3$ has appeared at least once.

$\textbf{Step -1: Calculating probability}$

$\text{Let A be the set that the sum of the numbers appearing on top be 7}$

$\implies A=\text{{(6,1)(1,6)(5,2)(2,5)(3,4)(4,3)}}$

$\text{Let B be the set that the number 3 has appeared at least once.}$

$\implies B=\text{{(1,3)(2,3)(3,3)(4,3)(5,3)(6,3(3,1)(3,2)(3,4)(3,5)(3,6)}}$

$\implies A\cap B=\text{{(3,4)(4,3)}}$

$\implies P(B/A)=\dfrac{n(A\cap B)}{n(A)}$

$\implies \dfrac26=\dfrac13$

$\textbf{Thus, the required probability is }\mathbf{\dfrac13 .}$

A bag contains 8 red and 5 white balls. Two successive draws of 3 balls are made at random from the bag without replacements. Find the probability that the first draw yields 3 white balls and the second draw 3 red balls.

Let A : Event that 3 balls in the first draw are all white
B : Event that 3 balls in the second draw are all red.
3 balls can be drawn out 13 in

$^{13}C_3$ and 3 white ball can be drawn out of 5 in

$^5C_3$ ways.

$P(A) = \dfrac{^5C_3}{^{13}C_3} = \dfrac{5}{143}$
Since 3 balls are not replaced before the second draw, we are left with 8 red and 2 white balls.
Now, 3 balls can be drawn in

$^{10}C_3$ ways and 3 red balls can be drawn in

$^8C_3$ ways.

$P(B/A) = \dfrac{^8C_3}{^{10}C_3} = \dfrac{7}{15}$

$\therefore P(A \cap B) = P(A). P(B/A) = \dfrac{5}{143} . \dfrac{7}{15} = \dfrac{7}{429}$

Bag A contains 2 red and 3 black balls while another bag B contains 3 red and 4 black balls. One ball is drawn at random from one of the bag and it is found to be red. Find the probability that it was drawn from bag B.

Probability of drawing a red ball from bag A =

$\cfrac{2}{2 + 3} = \cfrac{2}{5}$

Probability of drawing a red ball from bag B =

$\cfrac{3}{3 + 4} = \cfrac{3}{7}$

therefore probability that the drawn red ball belongs to bag B =

$\cfrac{\cfrac{3}{7}}{\cfrac{2}{5} + \cfrac{3}{7}}$

$= \cfrac{3}{7} \times \cfrac{5 \times 7}{14 + 15}$

$= \cfrac{15}{29}$

Three boxes ${B}_{1},{B}_{2},{B}_{3}$ contain balls with different colours as follows:
White Black Red
${B}_{1}$ 2 1 2
${B}_{2}$ 3 2 4
${B}_{3}$ 4 3 2
A dice is thrown. If $1$ or $2$ turns up on the dice, box ${B}_{1}$ is selected; if $3$ or $4$ turns up, ${B}_{2}$ is selected if $5$ or $6$ turns up, then ${B}_{3}$ is selected. If a box is selected like this, a ball is drawn from that box. If the ball is red, then find the probability that it was drawn from ${B}_{2}$ .

Probability of

${ B }_{ 2 }$ if it is red

$=\dfrac {\text{ Probability of red from} \ { B }_{ 2 } }{\text{ Probability of red from}\ { B }_{ 1 }+\text{Probability of red from}\ { B }_{ 2 }+\text{Probability of red from}\ { B }_{ 3 } }$

$=\dfrac { \dfrac { 2 }{ 6 } \times \dfrac { 4 }{ 9 } }{ \dfrac { 2 }{ 6 } \times \dfrac { 2 }{ 5 } +\dfrac { 2 }{ 6 } \times \dfrac { 4 }{ 9 } +\dfrac { 2 }{ 6 } \times \dfrac { 2 }{ 9 } }$

$=\dfrac { \dfrac { 4 }{ 27 } }{ \dfrac { 2 }{ 15 } +\dfrac { 4 }{ 27 } +\dfrac { 2 }{ 27 } } =\dfrac { \dfrac { 4 }{ 27 } }{ \dfrac { 2 }{ 15 } +\dfrac { 2 }{ 9 } }$

[Probability of

${ B }_{ n }'s$ selection

$\times$ Probability than red is selected from

${ B }_{ n }$ .]

$=\dfrac { \dfrac { 4 }{ 27 } }{ \dfrac { 6+10 }{ 45 } } =\dfrac { 45 }{ 27\times 16 }$

$=\dfrac { 5 }{ 48 }$

Box-I contains 2 gold coins, while another Box-II contains 1 gold and 1 silver coin. A person choose a box at random and takes out a coin. If the coin is of gold, what is the probability that the order coin in the box is also of gold?

Let

$E_1$ and

$E_2$ be the events that the boxes

$I$ and

$II$ are chosen respectively.

$\implies P(E_1)=\dfrac{1}{2}=P(E_2)$

Let

$A$ be the event that the coin drawn is of gold.

$\implies P(A|E_1)=P(\text{a gold coin from box I})=\dfrac{2}{2}=1$

$P(A|E_2)=P(\text{a gold coin from box II})=\dfrac{1}{2}$

Probability that the other coin in the box is of gold

$=$ The probability that the gold coin is drawn from box

$\text{I}=P(E_1|A)$

By using Bayes' theorem,

$P(E_1|A)=\dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}$

$=\dfrac{\cfrac{1}{2}\times 1}{\cfrac{1}{2}\times 1 + \cfrac{1}{2}\times \cfrac{1}{2}}$

$=\dfrac{2}{3}$

Hence,

$P(A|E_1)=\dfrac{2}{3}$

State and prove Bayes' theorem

The probability of two events

$A$ and

$B$ happening,

$P(A\cap B)$ , is the probability of

$A$

$(P(A))$ ,times the probability of

$B$ given that

$A$ has occurred

$(P(B|A))$ .

$P(A\cap B)=P(A)P(B|A)$ ....

$(1)$

On the other hand, the probability of

$A$ and

$B$ is also equal to the probability of

$B$ times the probability of

$A$ given

$B$ .

$P(A\cap B)=P(B)P(A|B)$ ....

$(2)$

Equating the two yields:

$P(B)P(A|B)=P(A)P(B|A)$ .....

$(3)$

and thus

$P(A|B)=P(A)\dfrac {P(B|A)}{P(B)}$ ....

$(4)$

This equation, known as Bayes' theorem is the basis of statistical inference.

If $P(A)=\dfrac{4}{5}$ and $P(B/A)=\dfrac{2}{5}$ , find $P(A\cap B).$

Given that

$P(A)=\dfrac{4}{5}$ and

$P(B/A)=\dfrac{2}{5}$

We have to find the value of

$P(A\cap B)$

From Conditional probability theorem, we get

$P(B/A)=\dfrac{P(A\cap B)}{P(A)}$

$\implies P(A\cap B)=P(B/A)P(A)=\dfrac{2}{5} \times \dfrac{4}{5}=\dfrac{8}{25}$

Hence,

$P(A\cap B)=\dfrac{8}{25}$

If $P(A)=0.8$ and $P(B|A)=0.4$ , then find $P(A\cap B)$ .

Given :

$P(A)=0.8$ ,

$P(B|A)$

We have the formula

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}$

$P(A\cap B)=P(B|A) P(A)=0.4\times 0.8=0.32$

Hence,

$P(A \cap B)=0.32$

Three boxes numbered I, II, III contains the balls as follows:
White Black Red
I $1$ $2$ $3$
II $2$ $1$ $1$
III $4$ $5$ $3$
One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that it is from box II

Probability of a red ball being drawn from box I is

$\cfrac{3}{1 + 2 + 3} = \cfrac{1}{2}$ .

Probability of a red ball being drawn from box II is

$\cfrac{1}{1 + 2 + 1} = \cfrac{1}{4}$ .

Probability of a red ball being drawn from box III is

$\cfrac{3}{4 + 5 + 3} = \cfrac{1}{4}$ .

Now, given that we have the red ball, the probability that it has been drawn from box II becomes

$\cfrac{\cfrac{1}{4}}{\cfrac{1}{2} + \cfrac{1}{4} + \cfrac{1}{4}}= \cfrac{1}{4}$ .

A person buys a lottery ticket in 50 lotteries in each of which his chance of winning a price is $\dfrac{1}{100}$ . What is the probability that he will win a price exactly once?

Out of

$(n=50)$ lotteries, let

$X$ denote the number of times the person wins a prize, with

$p=\dfrac{1}{100}$ and

$q=\dfrac{99}{100}$ .

It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii)independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.

Since X has a binomial distribution, the probability of

$x$ success in

$n$ -Bernoulli trials,

$P(X=x)={ }^{n}C_{x} p^x q^{n-x}$

$\therefore\ P(X=1)={ }^{50}C_{1}\left(\dfrac{1}{100}\right)^1 \left(\dfrac{99}{100}^{(50-1)}\right)=50\left(\dfrac{1}{100}\right)\left(\dfrac{99}{100}\right)^{49}=\dfrac{1}{2}\left(\dfrac{99}{100}\right)^{49}$

Hence,

$P(X=1)=\dfrac{1}{2}\left(\dfrac{99}{100}\right)^{49}$

In a factory, there are two machines $A$ and $B$ producing bulbs. Their contributions are respectively $60\%$ and $40\%$ of the total production. It is found that $1\%$ and $3\%$ of the items produced by these machines are defective. An item randomly chosen from a day's production was found to be defective. Find the probability that the defective item was produced by machine $B$ .

Let

$E_{1}$ and

$E_{2}$ be the events that the items produced by machine

$A$ and

$B$ respectively.

Then

$P(E_1)=\dfrac{60}{100}=\dfrac{3}{5}$ and

$P(E_2)=\dfrac{40}{100}=\dfrac{2}{5}$

Let

$C$ be the event that we choose a defective item at random.

Then,

$P(\text{The defective item came from machine } A)=P(C|E_1)=1\%=\dfrac{1}{100}$

$P(\text{The defective item came from machine } B)=P(C|E_2)=3\%=\dfrac{3}{100}$

To find the probability that the defective item was produced by machine

$B$ , we need to find

$P(E_2|C)$

So, we use Baye's formula and then we have

$P(E_2|C)=\dfrac{P(E_2)P(C|E_2)}{P(E_1)P(C|E_1)+P(E_2)P(C|E_2)}$

$\displaystyle =\displaystyle \dfrac{\cfrac{2}{5}\times \cfrac{3}{100}}{\cfrac{3}{5}\times \cfrac{1}{100}+\cfrac{2}{5}\times \cfrac{3}{100}}$

$=\dfrac{6}{3+6}=\dfrac{6}{9}=\dfrac{2}{3}$

Hence, the probability that the defective item was produced by machine

$B$ is

$\dfrac{2}{3}$ .

There are two small boxes $A$ and $B$ . In $A$ there are $9$ white beads and $8$ black beads.
In $B$ there are $7$ white and $8$ black beads. We want to take a bead from a box.
(a) What is the probability of getting a white bead from each box?
(b) A white bead and a black bead are added to box $B$ an then a bead is taken from it.
What is probability of getting a white bead from it.

Total number of beads in box

$A = 9W + 8B = 17$
Total number of beads in box

$B = 7W + 8B = 15$
(a) P(getting a white bead from box A) =

$\dfrac{9}{17}$
P(getting a white bead from box B) =

$\dfrac{7}{15}$
(b) When a white bead and a black bead are added to box B,
Number of white beads in box

$B = 7W + 1 W = 8W$
Number of black beads in box

$B = 8B + 1B = 9B$
Thus, total number of beads in box

$B = 8W + 9B = 17$
Hence, P(getting a white bead from box B) =

$\dfrac{8}{17}$ .

Find the probability distribution of number of heads in two tosses of a coin.

Given that a single coin is tossed twice, therefore the sample space

$S=\left\{HH,HT,TH,TT\right\}$

Since the

$4$ events are equally likely, we have

$P(HH)=P(HT)=P(TH)=P(TT)=\cfrac{1}{4}$

Let

$X$ be the number of heads in

$S$ such that

$X(HH)=2 , X(HT)=1 , X(TH)=1 , X(TT)=0$ , It follows that

$P(X=0)=P(TT)=\cfrac{1}{4}$

$P(X=1)=P(TH,HT)=\cfrac{1}{2}$

$P(X=2)=P(HH)=\cfrac{1}{4}$

Therefore the probability distribution table is shown below:

$X$

$0$

$1$

$2$

$P(X)$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

$\dfrac{1}{4}$

A coin is tossed twice. Find the probability distribution of the number of heads.

To get head on a coin comes

$p=\displaystyle\frac{1}{2}$
not comes head

$q=1-\displaystyle\frac{1}{2}=\frac{1}{2}$

$H$	$P_1$
$0$	$qq$	$\displaystyle\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$
$1$	$2(pq)$	$2\left(\displaystyle\frac{1}{2}\cdot\frac{1}{2}\right)=\displaystyle \frac{1}{2}$
$2$	$pp$	$\displaystyle\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

If $P(A)=0.6, P(B)=0.7$ and $P(A\cup B)=0.9$ , then find $P\left(\dfrac{A}{B}\right)$ and $P\left(\dfrac{B}{A}\right)$ .

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\Longrightarrow 0.9=0.7+0.6-P(A\cap B)\\ \Longrightarrow P(A\cap B)=0.4$

$P\left(\dfrac { A }{ B }\right)=\dfrac{ P(A\cap B) }{ P(B)}$

$\Longrightarrow P\left(\dfrac { A }{ B }\right)=\dfrac { 0.4 }{ 0.7 } =\dfrac { 4 }{ 7 }$

$P\left(\dfrac { B }{ A }\right)=\dfrac { P(A\cap B) }{ P(A) } =\dfrac { 0.4 }{ 0.6 } =\dfrac { 2 }{ 3 }$

If $mn$ coins have been distributed into na purses, n into each, find $(1)$ the chance that two specified coins will be found in the same purse; and $(2)$ what the chance becomes when $r$ purses have been examined and found not to contain either of the specified coins.

Represent the two coins by A and B. then the chance that B is with A is

$\cfrac { n-1 }{ mn-1 }$ for whenever A is placed, there remain

$mn-1$ possible position for B,

$n-1$ of which are favourable.

Hence the chance that A and B are not together is

$\cfrac { n(m-1) }{ mn-1 }$

Now consider the

$m-r$ purses which have not been examined.

If A and B are together, the chance that they occur in these purses is

$\cfrac { m-r }{ m }$ .

If A and B are apart the chance that they occur in these purses is

$\cfrac { \left( m-r \right) \left( m-r-1 \right) }{ m\left( m-1 \right) }$ ;

For

$m(m-1)$ is the total no. of ways in which they can occur separately in any

$2$ purses whatever and

$(m-r)(m-r-1)$ is the no. of ways in which they can occur separately in any two of the purses we are considering

Hence the required chance,

$=\cfrac { n-1 }{ mn-1 } .\cfrac { m-r }{ m } \div \left\{ \cfrac { n-1 }{ mn-1 } \cfrac { m-r }{ m } +\cfrac { n(m-1) }{ mn-1 } \cfrac { (m-r)(m-r-1) }{ m(m-1) } \right\}$

$=\cfrac { n-1 }{ mn-nr-1 }$

A certain stake is to be won by the first person who throws an ace with an octahedral die, if there are $4$ persons what is the chance of the last to win the stake?

Person

$\longrightarrow A,B,C,D$
Win probability

$=\cfrac { 1 }{ 8 } \quad A^{ ' }=\cfrac { 1 }{ 8 }$
Loss probability

$=\cfrac { 7 }{ 8 } \quad { B }^{ ' }=\cfrac { 7 }{ 8 }$

${ C }^{ ' }=\cfrac { 7 }{ 8 }$

$D=\cfrac { 1 }{ 8 }$ chance of D

$\cfrac { 7 }{ 8 } \times \cfrac { 7 }{ 8 } \times \cfrac { 7 }{ 8 } \times \cfrac { 1 }{ 8 }$
Required probability

$={ \left( \cfrac { 7 }{ 8 } \right) }^{ 3 }\times \cfrac { 1 }{ 8 }$

A bag contain 6 Red and 4 White balls. 4 balls are drawn one by one without replacement and were found to be atleast 2 white. Then the probability that next draw of a ball from this bag will give a white ball.

Given:

Total number of balls = 6+4=10

4 balls drawn, possible scenarios are,

(i) 2 white, 2 red ball - 2 white balls left and 4 red balls left, hence next draw to be white in this scenario=

$\dfrac 26$

(ii) 3 white balls, 1 red balls - 1 white ball left and 5 red balls left, hence next draw to be white in this scenario =

$\dfrac 16$

(iii) 4 white balls, 0 red balls - 0 white balls left and 6 red balls left, hence next draw to be white in this scenario =

$\dfrac 06$

Hence, the probability that next draw of a ball from this bag will give a white ball.= case (i) or case (ii) or case (iii) =

$\dfrac 26+\dfrac 16+\dfrac 06=\dfrac 36=\dfrac 12$

$10$ witnesses each of whose probability of speaking the truth is $\cfrac { 5 }{ 6 }$ . Asserts that the certain event took place, if the probability of this event before their statement is $\cfrac { 1 }{ 1+{ 5 }^{ 9 } }$ , find the probability of the truth of their assertion.

Probability that all 10 would say the truth =

$\left(\dfrac56\right)^{10}$

Probability that all 10 would lie =

$\left(\dfrac16\right)^{10}$

Hence, probability that the event occurred

$=\dfrac{(\dfrac56)^{10}}{(\dfrac56)^{10}+(\dfrac16)^{10}}=\dfrac{5^{10}}{1+5^{10}}$

A biased coin which comes up heads three times as often as tails is tossed. If it shows heads, a chip is drawn from urn-I which contains $2$ white chips and $5$ red chips. If the coin comes up tail, a chip is drawn from urn-II which contains $7$ white and $4$ red chips. Given that a red chip was drawn, what is the probability that the coin came up heads?

Let

$E_1$ be the event of getting a head, and

$E_2$ be the event of getting a tail in the biased coin.

$P(X)$ is a probability function on event

$X$ ,

$P(E_1)=3P(E_2)=3(1-P(E_1))$

$\Rightarrow P(E_1)=\dfrac 34$ and

$P(E_2)=\dfrac 14$

Let

$A$ be the event of getting a red chip.

So,

$P(A/E_1)=\dfrac 57$ (

$5$ red chips in a total of

$7$ chips)

$P(A/E_2)=\dfrac 4{11}$ (

$4$ red chips in a total of

$11$ chips)

So, according to Baye's theorem,

$P(E_1/A)=\dfrac {P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}$

Where

$P(E_1/A)$ is the probability of getting a heads given a red chip was drawn.

ie,

$P(E_1/A)=\dfrac {\dfrac 34\dfrac 57}{\dfrac 34\dfrac 57+\dfrac 14\dfrac 4{11}}=\dfrac {15\times 11}{15\times 11+28}=\dfrac {165}{193}$

Suppose a girl throws a die. If she gets $1$ or $2$ , she tosses a coin three times and notes the number of tails. If she gets $3, 4, 5$ or $6$ , she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', what is the probability that she threw $3, 4, 5$ or $6$ with the die?

Let

$E_1$ be the event that the girl gets

$1$ or

$2$

Thus,

$P(E_1) = \dfrac{2}{6} = \dfrac{1}{3}$

Let

$E_2$ be the event that the girl gets

$3,4,5$ or

$6$

$P(E_1) = \dfrac{4}{6} = \dfrac{2}{3}$

Let

$A$ be the event that she obtained exactly one tail.

If she tossed a coin

$3$ times and exactly

$1$ tail showed up, then the total number of favorable outcomes

$=\{(THH), (HTH), (HHT)\}=3$ .

Thus

$P(A|E_1) = \dfrac{3}{8}$

If she tossed the coin only once and exactly

$1$ tail showed up, the total number of favorable outcomes

$= 1$ .

Thus

$P(A|E_2) = \dfrac{1}{2}$

The probability that she threw

$3,4,5$ or

$6$ with the die, given that she got exactly one tail can be found as follows:

Or in other words, we need to find

$P(E_2|A)$

We can use the Baye's theorem, according to which

$P(E_2|A) = \dfrac{P(E_2)(P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}$

$P(E_2|A)= \dfrac{\dfrac{1}{2}\cdot \dfrac{2}{3}}{\dfrac{1}{2}\cdot \dfrac{2}{3}+\dfrac{3}{8}\cdot \dfrac{1}{3}} = \dfrac{\dfrac{1}{3}}{\dfrac{1}{3}+\dfrac{1}{8}} = \dfrac{8}{11}$

A is one of the $6$ horses entered for a race and is to be ridden by one of two jockeys B or C. It is $2:1$ that B rides A, in which case all the horses are equally likely to win; if C rides A, his chance is gets tripled. If the odds against his winning is $\dfrac{p}{q}$ , where p and q are co-primes then find the value of $p+q$ .

To find the odd against

$A$ 's winning, first we are to find the probability of

$A$ 's winning. And this can happen in one of the two mutual exclusive ways:

$B$ rides and

$A$ wins.

ii)

$C$ rides and

$A$ wins.

The chance for

$B$ to ride the horse

$A$ is

$2$ to

$1$ , i.e.

$P(B)=\dfrac{2}{3}\Rightarrow P(C)=\dfrac{1}{3}$ .

The chance for

$A$ to win is equally likely if

$B$ rides, i.e.

$\dfrac{1}{6}$ .

The probability for

$A$ to win if

$C$ rides, is trippled i.e.

$\dfrac{3}{6}$ .

The probability of

$A$ 's win

$=\dfrac{2}{3}\times\dfrac{1}{6}+\dfrac{1}{3}\times\dfrac{3}{6}=\dfrac{5}{18}$ .

The probability of

$A$ 's losing is

$\dfrac{13}{18}$ .

The odds against

$A$ 's winning are

$13:5$ .

According to the problem

$p=13, q=5$ .

$\therefore p+q=18$ .

There is a probability that 4 out of 10 students appearing in CA CPT exams will qualify for CA course after passing CA CPT exam. The probability that of 5 atudents all will qualify CA CPT exam.

Probability that the students will qualify,

$p=\dfrac{4}{10}$

Probability that they can't qualify,

$q=\dfrac{6}{10}$

$\Rightarrow n=5$

By Bernoulli's equation,

$\Rightarrow p={^nC-rp^rq{n-r}}$

p ( at least one will get selected )

$={^5C_5}\left( \dfrac { 4 }{ 10 } \right) ^{ 5}\left( \dfrac { 6 }{ 10 } \right) ^{ 0}$

$=0.0024.$

Hence, the answer is

$0.0024.$

From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find the probability distribution of X.

We have a lot of

$6$ items, 2 are defective

$\rightarrow 6-2=4$ are non-defective.

P (drawing a defective item) =

$\dfrac{2}{6}=\dfrac{1}{3}$

P (drawing a non-defective item) = 1 - P (drawing a defective item)

$=1-\dfrac{1}{3}=\dfrac{2}{3}$

Note

$: 4$ items can be drawn from a lot of 6 items in

$^{6}C_4$ ways.

Let

$X$ be the random variable of drawing 2 defective items from the lot.

$P (X = 0) =$ P (no defective item in the sample) =

$\dfrac{^{4}C_4}{^{6}C_4}=\dfrac{1}{15}$

$P (X = 1) =$ P (one defective item from the lot) = P(one defective, 3 non-defective items)

$=(\large\frac{^{2}C_1\;^{4}C_3}{^{6}C_4}) = \dfrac{8}{15}$

$P (X = 2) =$ P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =

$\large\frac{^{2}C_2\;^{4}C_2}{^{6}C_4} = \dfrac{6}{15}$

Therefore the required probability distribution is as follows:

$X: \space\space\space\space\space\space\space\space\space 0\space\space\space\space 1\space\space\space\space 2$

$P(X) :\dfrac{1}{15}\space\dfrac{8}{15}\space\dfrac{6}{15}$

If A and B are events such that $P(A)=\displaystyle\frac{1}{2}$ , $P(B)=\displaystyle\frac{1}{3}$ and $P(A\cap B)=\displaystyle\frac{1}{4}$ , then find $P(B/A)$ .

This is conditional probability,

$P(A\bigcap B)=P(A)P(\dfrac{B}{A})$

$\Rightarrow P(\dfrac{B}{A})=\dfrac{P(A\bigcap B)}{P(A)}=\dfrac{\dfrac{1}{4}}{\dfrac{1}{2}}=\dfrac{2}{4}=\dfrac{1}{2}$

If A and B are events such that $P(A)=\displaystyle\frac{1}{2}$ , $P(B)=\displaystyle\frac{1}{3}$ and $P(A\cap B)=\displaystyle\frac{1}{4}$ , then find $P(A/B)$ .

Given

$P(A)=\displaystyle\frac{1}{2}$ ,

$P(B)=\displaystyle\frac{1}{3}$ and

$P(A\cap B)=\displaystyle\frac{1}{4}$

This is conditional probability,

$P(A\bigcap B)=P(B)P(\dfrac{A}{B})$

$\Rightarrow P(\dfrac{A}{B})=\dfrac{P(A\bigcap B)}{P(B)}=\dfrac{\dfrac{1}{4}}{\dfrac{1}{3}}=\dfrac{3}{4}$

Assume that the chances of a patient having a heart attack are $40\%$ . It is also assumed that a meditation and yoga course reduce the risk of heart attack by $30\%$ and the prescription of certain drugs reduces its chances by $25\%$ . At a time a patient can choose any one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Probability of patient having heart attack

$=0.4$

Probability that patient won't have heart attack

$=0.6$

Let

$E_1$ be the event of the patient following a course of meditation and yoga.

Let

$E_2$ be the event of the patient following a course of prescription of drugs.

$\Rightarrow P\left( \dfrac { A }{ { E }_{ 1 } } \right) =P\left( \dfrac { A }{ { E }_{ 2 } } \right) =0.4$

$\Rightarrow P\left( { E }_{ 1 } \right) =0.3$

$P\left( { E }_{ 2 } \right) =0.25$

$\therefore P\left( \dfrac { E }{ { E }_{ 1 } } \right) =\dfrac { P\left( { E }_{ 1 } \right) P\left( \dfrac { A }{ { E }_{ 1 } } \right) }{ P\left( { E }_{ 1 } \right) P\left( \dfrac { A }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) P\left( \dfrac { A }{ { E }_{ 1 } } \right) }$

$=\dfrac { 0.3\times 0.4 }{ 0.3\times 0.4+0.25\times 0.4 }$

$=0.55$

Hence, the answer is

$0.55.$

A family has two children. What is the probability that both the children are boys, given that atleast one of them is a boy?

$E:$ Both the children are boys

$E=\{(b,b)\}$

$P(E)=\frac{1}{4}$

$F:$ At least one of the children is a boy

$F=\{(g,b),(b,g),(b,b)\}$

$P(F)=\dfrac{3}{4}$

Also,

$E\cap F=\{(b,b)\}$

$P(E\cap F)=\dfrac{1}{4}$

Now,

$P(E|F)=\dfrac{P(E\cap F)}{P(F)}$

$=\dfrac{\frac{1}{4}}{\dfrac{3}{4}}=\dfrac{1}{3}$

If $P(A) = \dfrac {7}{13}, P(B) = \dfrac {9}{13}$ and $P(A\cap B) = \dfrac {4}{13}$ , find $P(A/ B)$ .

$P(A/ B) = \dfrac {P(A\cap B)}{P(B)} = \dfrac {4/13}{9/13} = \dfrac {4}{9}$ .

Three fair coins are tossed simultaneously. If X denotes the number of heads, find the probability distribution of X.

Since three coins are tossed simultaneously.

Therefore sample space

$S$ will be:

$\left\{ HHH, HHT, HTH, THH, HTT, THT, TTH, TTT \right\}$

Given that

$X$ denotes the no. of heads such that

$X{\left( 0 \right)} = TTT = 1$

$X{\left( 1 \right)} = HTT, THT, TTH = 3$

$X{\left( 2 \right)} = HHT, HTH, THH = 3$

$X{\left( 3 \right)} = HHH = 1$

Total no. of outcomes

$= {2}^{n}$

whereas,

$n =$ no. of coins tossed

$\therefore$ Total no. of outcomes

$= {2}^{3} = 8$

Therefore,

$P{\left( X = 0 \right)} = P{\left( TTT \right)} = \cfrac{1}{8}$

$P{\left( X = 1 \right)} = P{\left( HTT, THT, TTH \right)} = \cfrac{3}{8}$

$P{\left( X = 2 \right)} = P{\left( HHT, HTH, THH \right)} = \cfrac{3}{8}$

$P{\left( X = 3 \right)} = P{\left( HHH \right)} = \cfrac{1}{8}$

Therefore probability distribution of

$X$ -

$X$	$P{\left( X \right)}$
$0$	$\cfrac{1}{8}$
$1$	$\cfrac{3}{8}$
$2$	$\cfrac{3}{8}$
$3$	$\cfrac{1}{8}$

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$
$P(X)$ $0.1$ $k$ $2k$ $2k$ $k$
Determine:
(i) $k$
(ii) $P(X \geq 2)$ .

(i) Since

$\sum P(x) = 1$

$\Rightarrow 0.1 + k + 2k + 2k + k = 1$
or

$6k = 1 - 0.1$

$k = \dfrac {0.9}{6}\Rightarrow k = 0.15$

(ii)

$P(x\geq 2) = P(x = 2) + P(x = 3) + P(x = 4)$

$= 2k + 2k + k$

$= 5k = 5(0.15)$

$P(x\geq 2) = 0.75$ .

Five coins are tossed simultaneously. Find the probability of getting atleast onehead.

$P(E)=\dfrac{Favorable Outcomes}{Total Outcomes}$

So the to count the favorable outcomes: There are two possible outcomes for each toss.

And there are 5 total tosses.

So for each outcome there will be another outcome, so we'll multiply

$2∗2∗2∗2∗2=2^5$

Sample Space or

$n(S)=2^5$

Now think of getting one or more heads. We know that sum all probabilities is 1.

So we think of the other side.

Why don't we calculate the number of possibilities for no head?

And that would just be

$\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2}*\dfrac{1}{2} = \dfrac{1}{32}$

Probability of getting no head

$=\dfrac{1}{32}$

$P(E)=1−\dfrac{1}{32} = \dfrac{31}{32}$

So this probability will be the required one.

$P(A)=\dfrac {1}{Z}$
$P(B)=\dfrac {1}{4}$
$P(A \cap B)=\dfrac {1}{5}$
$P(A^{}/B^{})=?$

$P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 1/5 }{ 1/4 } =4/5$

A is known to speak truth $5$ out of $7$ times. What is the probability that $A$ reports that it is a $7$ when a die is thrown?

Probability that it speaks truth=5/7

When a die is thrown A reports that it is seven is only possible when he is lying, and he lies with probability=2/7

A purse contains n coins of unknown value, a coin drawn at random is found to be a rupee, what is the chance that it is the only rupee in the purse? Assume all numbers of rupee coins in the purse to be equally likely.

$P = \frac{{\frac{1}{n} + \frac{2}{n} + \frac{3}{n} + \frac{4}{n} + ..... + \frac{n}{n}}}{{{}^n{C_1}}}$

$= \frac{1}{n}\left( {\frac{{1 + 2 + 3 + 4 + .... + n}}{{{}^n{C_1}}}} \right)$

$= \frac{1}{n} \times \frac{{n\left( {n + 1} \right)}}{2} \times \frac{1}{n}$

$= \frac{{n + 1}}{2} \times \frac{1}{n} = \frac{{n + 1}}{{2n}}$

Consider the experiment of throwing a die, if a multiple of $3$ comes up, thrown the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a $3$ '.

On throwing a die multiple of

$3$ comes

$P(of\,coming\,multiple):\,,\dfrac{3}{6}=\dfrac{1}{2}$

$P(of\,coming\,other\,no):\,,\dfrac{3}{6}=\dfrac{1}{2}$

Now coins is tosed

$P(tail\,comes)=\dfrac{1}{2}$

$Conditional\,probability: P(tail\,comes)\times P(other\,no\,comes)$

$=\dfrac{1}{2}\times \dfrac{1}{2}$

$=\dfrac{1}{4}$

The probability is

$\dfrac{1}{4}$

If $A$ ad $B$ are two events such that $P(A)=\dfrac{1}{4}, P(B)=\dfrac{1}{3}$ and $P(A\cup B)=\dfrac{1}{2}$ , show that $A$ and $B$ are independent events

1379426_1065075_ans_b1015cdf123f410ca7849ec2618c773f.png

Find the value of K.

Given

$P(X,x)=\begin{cases} Kx\quad ,\quad if\quad x=0\quad or\quad 1 \\ 2Kx\quad ,\quad if\quad x=2 \\ K(5-x)\quad ,\quad if\quad x=3\quad or\quad 4 \end{cases}$

We know

$\sum _{ x=0 }^{ 4 }{ P(X=x)=1 }$

$\Rightarrow 1=Kx+Kx+2Kx+K(5-x)+K(5-x)\\ =0+K+4K+K(2)+K(1)\\ \Rightarrow 1=8K\\ \Rightarrow K=\cfrac { 1 }{ 8 }$

Consider rolling of die.
Find probability that a multiple of $3$ appear in third roll.

Multiple of 3 are 3,6

First and second roll can have any outcomes

Probability=(6/6)*(6/6)*2/6=1/3

Two coins are tossed once. Let $A=$ Tail appears on one coin $B=$ coin show Head. Find $P(A|B)$ .

Given that two coins are tossed once. Let head be denoted by

$h$ and tails by

$t$

Possible outcomes are

$:-$

$S = (h,h) , (h,t) , (t,h) , (t,t)$

We need to find the probability that tails appears on one coin, if we know that tails appears on one coin, if we know that one coin shows head.

Let

$A :$ One coin shows head

$B :$ Tail appears on one coin

We need to find

$P(A|B)$

$A = (h,t),(t,h)$

$B = (h,t),(t,h)$

$\Rightarrow P(A) = P(B) = \dfrac{2}{4} = \dfrac{1}{2}$

Now,

$P(A|B) = \dfrac{P(A\cap B)}{P(B)}$

$= \dfrac{\dfrac{1}{2}}{\dfrac{1}{2}}$

$= 1$

In how many ways can $10$ people be made to sit around a circular table?

$'n'$ people can be arrange around a circular table in

$(n-1)!$ ways

$\therefore 10$ people to arrange around a circular table it takes

$(10-1)!$ ways

$=9!$

What is the probability that you will get admission in atleast two colleges?

Given

$P(X,x)=\begin{cases} Kx\quad ,\quad if\quad x=0\quad or\quad 1 \\ 2Kx\quad ,\quad if\quad x=2 \\ K[5-x]\quad ,\quad if\quad x=3\quad or\quad 4 \end{cases}$

Possibility for atleast

$2$ colleges

$=P(X=2)+P(X=3)+P(X=4)\\ =2Kx+K[5-x]+K[5-x]\\ =2Kx+5K-Kx+5K-Kx\\ =10K$

A fair die is rolled. Consider the events $E = \{1, 3, 5\}$ and $F = \{2, 3\}$ , find $P(E| F)$ .

$E = \left \{ 1,3,5 \right \} ; F = \left \{ 2, 3 \right \}$

$E\cap F = \left \{ 3 \right \}$

Total no. of possible outcomes = 6

$P(E\cap F) = \dfrac{1}{6}$

$P(F) = \dfrac{2}{6} = \dfrac{1}{3}$

$P(E/F) = \dfrac{P(E\cap F)}{P(F)}$

$= \dfrac{1/6}{1/3}$

$= \dfrac{1}{2}$

$\boxed{\therefore P(E/F) = \dfrac{1}{2}}$

$10\%$ of the tools produced by a machine are defective. Find the probability distribution of the number of defective tools when $3$ tools are drawn one by one with replacement.

$X$ denote the no of defective machines.

$X = 0$

$P(X) =(0.9)^3$

$X = 1$

$P(X) = (0.9)^2 \times (0.1) \times \cfrac{3!}{2!}$

$X = 2$

$P(X) = (0.9) \times (0.1)^2 \times \cfrac{3!}{1! \times 2!}$

$X = 3$

$P(X) = (0.1)^3$

$20\%$ of the bulbs produced by a machine are defective. Find the probability distribution of the number of defective bulbs when $4$ bulbs are drawn one by one with replacement.

$X$ denote the no of defective bulbs.

$X = 0$

$P(X) =(0.8)^4$

$X = 1$

$P(X) = (0.8)^3 \times (0.2) \times \cfrac{4!}{3!}$

$X = 2$

$P(X) = (0.8)^2 \times (0.2)^2 \times \cfrac{4!}{2! \times 2!}$

$X = 3$

$P(X) = (0.8) \times (0.2)^3 \times \cfrac{4!}{3!}$

$X = 4$

$P(X) = (0.2)^4$

The probability that a bulb produced by a factory will fuse after 160 days of use is 0.Find the probability that out of 5 such bulbs, at the most one bulb will fuse after 160 days of use.

1373627_1124660_ans_dc414625715344239411c8fc0116ab76.jpg

In a certain college, 65% of the students are fully time students 55% of the students are female, 35% of the students are male full times students, Find the probability that
i) a student chosen at random from all the students in the college is a part time student
ii) a student chosen at random from all the students in the college is female and a part time student
iii) a student chosen at random from all the female students in the college is a part time.

1078883_1122835_ans_bfa748e295264c7faa0134dc57f5156c.png

Find P(AUB) when 2P(A)=P(B)= $\frac { 5 }{ 13 }$ and P(A|B)= $\frac { 2 }{ 5 }$ .

$P(A|B)=\frac { P(A\cap B) }{ P(B) } \\ \frac { 2 }{ 5 } =\frac { P(A\cap B) }{ 5/13 } \\ P(A\cap B)=2/13\\ P(AUB)=P(A)+P(B)-P(A\cap B)\\ \quad \quad \quad \quad \quad \quad =3P(A)-\frac { 2 }{ 13 } \\ \quad \quad \quad \quad \quad \quad =\frac { 15 }{ 26 } -\frac { 2 }{ 13 } \\ \quad \quad \quad \quad \quad \quad =11/26$

There are two bags 1 and 2, containing $3$ red and $4$ white balls and $2$ red and $3$ white balls respectively. A bag is selected at random and a ball is drawn from it. If it is found to be a red ball, Find the probability that it is drawn from the first bag.

1156534_1114024_ans_d6e45e18693b4740a91fb90402a34b01.jpg

A family has two children. What is the probability that both the children are boys given that of them is a boy.

$P\left(\dfrac{B}{A}\right)=\dfrac{P(A\cap B}{P(A)}$

Sample space

$S=\left\{MM, MF, FF, FM\right\}$

Both will be boys

$A:$ Event both children are boys.

$B:$ Event that at least one is big

$P(A)=\dfrac{1}{4}, P(B)=\dfrac{3}{4}, P(A\cap B)=\dfrac{1}{4}$

$P\left(\dfrac{B}{A}\right)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{\dfrac{1}{4}}{\dfrac{3}{7}}=\dfrac{1}{3}$

If $P(A)=\dfrac{1}{4}, P(B)=\dfrac{1}{5}$ and $P(A\cap B)=\dfrac{1}{7}$ , find $P(\overline{A}/\overline{B})$ .

Given,

$P(A)=\dfrac{1}{4}, P(B)=\dfrac{1}{5}$ and

$P(A\cap B)=\dfrac{1}{7}$ .

Therefore,

$P(\dfrac{\overline{A}}{\overline{B}})=\dfrac{P(\overline{A}\cap \overline{B})}{P(\overline{B})}=\dfrac{1-P(A \cup B)}{1-P(B)}$

$=\dfrac{1-[P(A)+P(B)-P(A\cap B)]}{1-\dfrac{1}{5}}$

$=\dfrac{1-[\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{7}]}{\dfrac{4}{5}}$

$=\dfrac{(140-43)}{140}\times \dfrac{5}{4}=\dfrac{97}{112}$

A person goes to work by car, train or bus. The probabilities of person travelling by there modes are $3/10,1/2$ and $1/5$ respectively. It is found that chance of person being late for work are $3/7,1/7$ and $5/7$ by respective modes. Given that on particular day person reaches late, find the probability that on that day he had travelled by train.

$Using\quad Baye's\quad Rule\\ Probability\quad that\quad he\quad travelled\quad by\quad train\quad given\quad he\quad is\quad late=\frac { Probability\quad he\quad is\quad late\quad and\quad used\quad train }{ Probability\quad that\quad \quad he\quad is\quad late\quad by\quad other\quad mode\quad of\quad transport } \\ =\frac { (1/2)\times (1/7) }{ (3/10)\times (3/7)+(1/2)\times (1/7)+(1/5)\times (5/7) } =\frac { 5 }{ 24 }$

Two cards are drawn from a well shuffled pack of $52$ acrds. Find the probability that one of them is a red card & the other is a queen.

Let

$P{\left( {R}_{1} \right)}$ be the probability of getting a red card in the

$I$ draw

$= \cfrac{26}{52} = \cfrac{1}{2}$

Let

$P{\left( {Q}_{1} \right)}$ be the probability of getting a queen in the

$II$ draw

$= \cfrac{4}{51} + \cfrac{3}{51} = \cfrac{7}{51}$

Let

$P{\left( {Q}_{2} \right)}$ be the probability of getting a red card in the

$I$ draw

$= \cfrac{4}{52} = \cfrac{1}{13}$

Let

$P{\left( {R}_{2} \right)}$ be the probability of getting a queen in the

$II$ draw

$= \cfrac{26}{51} + \cfrac{25}{51} = \cfrac{51}{51} = 1$

$\therefore$ Required probability

$= \cfrac{1}{2} \times \cfrac{7}{51} + \cfrac{1}{13} \times 1 = \cfrac{7}{102} + \cfrac{1}{13} = \cfrac{193}{1326}$

Hence the correct answer is

$\cfrac{193}{1326}$ .

Three bags contain $2$ silver, $5$ copper coins, and $3$ silver, $4$ copper coins and $5$ silver, $2$ copper coins respectively. A bag is chosen at random and a coin is drawn from it which happens to be silver. What is the probability that it has come from third bag?

$Using\quad Baye's\quad Rule,\\ Selection\quad of\quad each\quad bag\quad is\quad equally\quad likely\quad with\quad probability\quad 1/3\\ Probability\quad that\quad its\quad coming\quad from\quad the\quad third\quad bag\quad given\quad its\quad silver\\ =\frac { Probability\quad that\quad its\quad coming\quad from\quad the\quad third\quad bag\quad given\quad its\quad silver }{ Probability\quad that\quad \quad the\quad coin\quad is\quad silver\quad from\quad any\quad bag } \\ =\frac { (1/3)\times (5/7) }{ (1/3)\times (2/7)+(1/3)\times (3/7)+(1/3)\times (5/7) } =\frac { 5 }{ 10 } =1/2$

What is the probability that the total of two dice will be greater than 9, given that the first die is a 5 ?

Let A = first die is 5

Let B = total of two dice is greater than 9

$P(A)=\dfrac{1}{6}$

Possible outcomes for A and B : {(5,5), (5,6)}

P(A and B)=

$\dfrac{2}{36}=\dfrac{1}{18}$

$P(B |A)=\dfrac{P(A \cap B)}{P(A)}$

$=\dfrac{1}{18} \times 6=\dfrac{1}{3}$

Assume that a factory has two machines $A$ and $B$ . Past records shows that machine $A$ produces 60% of the items of output and machine $B$ produces 40% of the items. Further, 2% of the items produced by machine $A$ were defective and only 1% produced by machine $B$ were defective. If a detective item is drawn at random, what is the probability that it was produced by machine $A$ ?

Consider the problem

Let,

${E_1}\,and\,{E_2}$ be the respective events of items produced by machine

$A$ and

$B$ .

And let

$x$ be the event that the produced item was found to be defective.

Therefore,

Probability of items produced by machine

$A,P$

$\left( {{E_1}} \right)$

$= 60\% = \frac{3}{5}$

Probability of items produced by machine

$B,P$

$\left( {{E_2}} \right)$

$= 40\% = \frac{2}{5}$

And,

Probability that machine

$A$ produced defective items,

$P\left( {\frac{x}{{{E_1}}}} \right)$

$= 2\% = \frac{2}{100}$

Probability that machine

$B$ produced defective items,

$P\left( {\frac{x}{{{E_2}}}} \right)$

$= 1\% = \frac{1}{100}$

So, the probability that randomly selected items was from machine

$A$ is given by

$P\left( {\frac{{{E_1}}}{x}} \right)$

Now, Apply Bayes' Theorem

$\begin{array}{l} P\left( { \frac { { { E_{ 1 } } } }{ x } } \right) =\frac { { P\left( { { E_{ 1 } } } \right) P\left( { \frac { x }{ { { E_{ 1 } } } } } \right) } }{ { P\left( { { E_{ 1 } } } \right) P\left( { \frac { x }{ { { E_{ 1 } } } } } \right) +P\left( { { E_{ 2 } } } \right) P\left( { \frac { x }{ { { E_{ 2 } } } } } \right) } } \\ =\frac { { \frac { 3 }{ 5 } \times \frac { 2 }{ { 100 } } } }{ { \frac { 2 }{ 5 } \times \frac { 1 }{ { 100 } } +\frac { 3 }{ 5 } \times \frac { 2 }{ { 100 } } } } \\ =\frac { 6 }{ { 2+5 } } \\ =\frac { 6 }{ { 11 } } \end{array}$

Hence, the required probability of machine

$A$ is

$\frac{6}{{11}}$

A factory has three machines A,B and C, which produce $100,200\ and\ 300 \$ items of a particular type daily. The machines produce $2\%, 3\%\ and \ 5\%$ defective items respectively. One day when the production was over, an item was picked up randomly and it was found to be defective. Find the probability that it was produced by machine A.

Let

${E_{1,}}{E_2},{E_3}$ is machine

$A,B,C$

$D=$ defective items

total production

$100+200+300=600$

$\begin{array}{l} p\left( { { E_{ 1 } } } \right) =\dfrac { { 100 } }{ { 600 } } \\ p\left( { { E_{ 1 } } } \right) =\dfrac { 1 }{ 6 } \\ p\left( { { E_{ 2 } } } \right) =\dfrac { { 200 } }{ { 600 } } =\dfrac { 1 }{ 3 } \\ p\left( { { E_{ 3 } } } \right) =\dfrac { { 300 } }{ { 600 } } =\dfrac { 1 }{ 2 } \end{array}$

now

$\begin{array}{l} p\left( { \frac { D }{ { { E_{ 1 } } } } } \right) =0.02 \\ p\left( { \frac { D }{ { { E_{ 2 } } } } } \right) =0.03 \\ p\left( { \frac { D }{ { { E_{ 3 } } } } } \right) =0.05 \end{array}$

probability

$\begin{array}{l} P\left( { \dfrac { { { E_{ 1 } } } }{ D } } \right) =\dfrac { { p\left( { { E_{ 1 } } } \right) p\left( { \dfrac { D }{ { { E_{ 1 } } } } } \right) } }{ { p\left( { { E_{ 1 } } } \right) p\left( { \dfrac { D }{ { { E_{ 1 } } } } } \right) +p\left( { { E_{ 2 } } } \right) p\left( { \frac { D }{ { { E_{ 2 } } } } } \right) +p\left( { { E_{ 3 } } } \right) p\left( { \dfrac { D }{ { { E_{ 3 } } } } } \right) } } \\ =\dfrac { { \frac { 1 }{ 6 } \times 0.02 } }{ { \dfrac { 1 }{ 6 } \times 0.02+\dfrac { 1 }{ 3 } \times 0.03+\dfrac { 1 }{ 2 } \times 0.05 } } \\ =\dfrac { 2 }{ { 23 } } \end{array}$

In a bolt factory, three machines A,B, and C manufacture 25,35 and 40 percent of the total bolts respectively. Out of the total bolts manufactured by the machines 5,4 and 2 percent are defective from machine A, B & C respectively. A bolt is drawn at random and is found to be defective. Find the probability that it was manufactured by (i) Machine A or C (ii) Machine B.

Consider the problem

Let

$A:$ bolt manufactured from machine

$A$

$B:$ bolt manufactured from machine

$B$

$C:$ bolt manufactured from machine

$C$

$D:$ bolt is defective

WE need find the probability that the bolt is manufactured by machine

$A\;or\;C$ and machine

$B$ , if it is defective.

That is

$P\left( {\dfrac{B}{D}} \right)$

So,

$P\left( {\dfrac{B}{D}} \right) = \frac{{P\left( B \right).P\left( {\dfrac{D}{B}} \right)}}{{P\left( A \right).P\left( {\dfrac{D}{A}} \right) + P\left( B \right).P\left( {\dfrac{D}{B}} \right) + P\left( C \right).P\left( {\dfrac{D}{C}} \right)}}$

$P(A)=$ probability that the bolt is made by machine

$A$

$\begin{array}{l} =25\%=\dfrac { { 25 } }{ { 100 } } \\ =0.25 \end{array}$

$P(B)=$ probability that the bolt is made by machine

$B$

$\begin{array}{l} =35\%=\dfrac { { 35 } }{ { 100 } } \\ =0.35 \end{array}$

$P(C)=$ probability that the bolt is made by machine

$C$

$\begin{array}{l} =40\%=\dfrac { { 40 } }{ { 100 } } \\ =0.40 \end{array}$

$P(A\; or \;C)=$ probability that the bolt is made by machine

$A\; or\;C$

$\begin{array}{l} =65\%=\dfrac { { 65 } }{ { 100 } } \\ =0.65 \end{array}$

And,

$P\left( {\dfrac{D}{A}} \right)=$ probability of a defective bolt from machine

$A$

$\begin{array}{l} =5\%=\dfrac { { 5 } }{ { 100 } } \\ =0.05 \end{array}$

$P\left( {\dfrac{D}{B}} \right)=$ probability of a defective bolt from machine

$B$

$\begin{array}{l} =4\%=\dfrac { { 4 } }{ { 100 } } \\ =0.04 \end{array}$

$P\left( {\dfrac{D}{C}} \right)=$ probability of a defective bolt from machine

$C$

$\begin{array}{l} =2\%=\dfrac { { 2 } }{ { 100 } } \\ =0.02 \end{array}$

$P\left( {\dfrac{D}{A\;or\;C}} \right)=$ probability of a defective bolt from machine

$A\;or\;C$

$=0.07$

Now apply Bayes' Theorem

$\begin{array}{l} P\left( { \dfrac { B }{ D } } \right) =\dfrac { { 0.35\times 0.04 } }{ { 0.25\times 0.05+0.35\times 0.04+0.04\times 0.02 } } \\ =\dfrac { { 0.014 } }{ { 0.0125+0.014+0.008 } } \\ =\dfrac { { 0.014 } }{ { 0.0345 } } \\ =\dfrac { { 140 } }{ { 345 } } \\ =\dfrac { { 28 } }{ { 69 } } \end{array}$

Therefore, required probability by the machine

$B$ is

$\dfrac{{28}}{{69}}$

And Machine

$A \;or\; C$ is

Apply Bayes' Theorem

$\begin{array}{l} P\left( { \dfrac { { A\, or\, C } }{ D } } \right) =\dfrac { { 0.65\times 0.07 } }{ { 0.65\times 0.07+0.35\times 0.04 } } \\ =\dfrac { { 13 } }{ { 17 } } \end{array}$

There are 4 boys and 2 girls in Room no. 1 and 5 boys and 3 girls in room no 2 Find the probability that a girl from one of the two rooms laughed loudly was from Room no. 2?

Consider the problem,

The probability that girl laugh from room 1 is:

$\begin{array}{l} P\left( { { E_{ 1 } } } \right) =\frac { 1 }{ 2 } \\ P\left( { { E_{ 2 } } } \right) =\frac { 1 }{ 2 } \\ P\left( { { E_{ 1 } }/G } \right) =\frac { 2 }{ 6 } \\ P\left( { { E_{ 2 } }/G } \right) =\frac { 3 }{ 8 } \\ Apply\, the\, Baye's\, theorem \\ P\left( { G/{ E_{ 2 } } } \right) =\frac { { \frac { 1 }{ 2 } \left( { \frac { 3 }{ 8 } } \right) } }{ { \frac { 1 }{ 2 } \left( { \frac { 2 }{ 6 } } \right) +\frac { 1 }{ 2 } \left( { \frac { 3 }{ 8 } } \right) } } =\frac { 9 }{ { 17 } } \end{array}$

State which of the following variables are continuous and which are discrete :
a)number of children in your class
b) distance traveled by a car
c) sizes of shoes
d) time
e) number of patients in a hospital

$a\rightarrow$ Discrete

$b\rightarrow$ Continous

$c\rightarrow$ Discrete

$d\rightarrow$ Continous

$d\rightarrow$ Discrete

Obtain the probability distribution of the number of sixes in two tosses of a fair die .

$n(s)=6\times 6=36$

$x=$ no of since in the toss

$P(x=0)=P(no\,\sin)=\dfrac{25}{36}$

Sample space=

$\begin{Bmatrix} 11 & 12 & 13 & 14 & 15 & 16 \\ 21 & 22 & 23 & 24 & 25 & 26 \\ 31 & 32 & 33 & 34 & 35 & 36 \\ 41 & 42 & 43 & 44 & 45 & 46 \\ 51 & 52 & 53 & 54 & 55 & 56 \\ 61 & 62 & 63 & 64 & 65 & 66 \end{Bmatrix}$

$P(x=1)=P(one\,\sin)$

$=\dfrac{10}{36}$

$P(x=2)=P(two\,sin)$

$=\dfrac{1}{36}$

Probability distribution

$\begin{matrix} x& 0& 1& 2 \\ p(x)& \dfrac{25}{36}& \dfrac{10}{36}& \dfrac{1}{36}\end{matrix}$

Amit and Nisha apear for an interview for two vacancies in a company. The probability of amit's selection is $1/5$ and that Nisha's selection is $1/6.$ What is the probability that both of then selected?

Probability that both of them are success=

$=\frac { 1 }{ 5 } \times \frac { 1 }{ 6 } =1/30$

In a certain university , the percentage of a Hindus, Muslim and Christians among student are 50, and 25 and 25 respectively. if 50% of Hindus, 90% of Muslim and 80% of Christians are smokers, Find the probability that a randomly selected smoker student in Muslim.

Consider the problem

Let

${E_1},{E_2},{E_3}$ be the event of choosing a hindu, Muslim and christian respectively.

And Let

$E$ be the event that the selected student smoke.

Then,

$\begin{array}{l} p\left( { { E_{ 1 } } } \right) =\frac { { 50 } }{ { 100 } } =\frac { 1 }{ 2 } \\ p\left( { { E_{ 2 } } } \right) =\frac { { 25 } }{ { 100 } } =\frac { 1 }{ 4 } \\ p\left( { { E_{ 3 } } } \right) =\frac { { 25 } }{ { 100 } } =\frac { 1 }{ 4 } \end{array}$

And

$\begin{array}{l} p\left( { \frac { E }{ { { E_{ 1 } } } } } \right) =\frac { { 50 } }{ { 100 } } \\ p\left( { \frac { E }{ { { E_{ 2 } } } } } \right) =\frac { { 90 } }{ { 100 } } \\ p\left( { \frac { E }{ { { E_{ 3 } } } } } \right) =\frac { { 80 } }{ { 100 } } \end{array}$

$\begin{array}{l} p\left( { \frac { { { E_{ 2 } } } }{ E } } \right) =\frac { { p\left( { { E_{ 2 } } } \right) \times p\left( { \frac { E }{ { { E_{ 2 } } } } } \right) } }{ { p\left( { { E_{ 1 } } } \right) p\left( { \frac { E }{ { { E_{ 1 } } } } } \right) +p\left( { { E_{ 2 } } } \right) p\left( { \frac { E }{ { { E_{ 2 } } } } } \right) +p\left( { { E_{ 3 } } } \right) p\left( { \frac { E }{ { { E_{ 3 } } } } } \right) } } \\ =\frac { { \frac { { 25 } }{ { 100 } } \times \frac { { 90 } }{ { 100 } } } }{ { \frac { { 50 } }{ { 100 } } \times \frac { { 50 } }{ { 100 } } +\frac { { 25 } }{ { 100 } } \times \frac { { 90 } }{ { 100 } } +\frac { { 25 } }{ { 100 } } \times \frac { { 80 } }{ { 100 } } } } \\ =\frac { { 25\times 90 } }{ { 50\times 50+25\times 90+25\times 80 } } \\ =\frac { { 90 } }{ { 2\times 50+90+80 } } \\ =\frac { { 45 } }{ { 50+45+40 } } \\ =\frac { { 45 } }{ { 135 } } \\ =\frac { 1 }{ 3 } \end{array}$

If the chance that any of $5$ telephone lines is busy at any instance is $0.01$ . What is the probability that all the lines are busy? What is the probability that more than $3$ lines are busy?

We have:

P(telephone line is busy ) = 0.01

Q(telephone lines are not busy) = 1 - 0.01 = 0.99

N =5

So it is problem of binomial distribution

1) The probability for all lines are busy (which is equivalent to no lines are free) = ${ P\left( X = 0 \right) } =\quad { { ^5 }{ C }_{ 0 } }{ p }^{ 5 }{ q }^{ 0 }\quad =\quad { 0.01 }^{ 5 }={ 10 }^{ -10 }$

2) The probability for more than 3 lines are busy = ${ P\left( X=0 \right) }+{ P\left( X=1 \right) }={ ^{ 5 }{ C }_{ 0 } }{ p }^{ 5 }{ q }^{ 0 }+{ ^{ 5 }{ C }_{ 1 } }{ p }^{ 4 }{ q }^{ 1 }\\ =({ .01 })^{ 5 }+{ 5 }({ .01) }^{ 4 }({ 0.99 }^{ 1 })\\ ={ 10 }^{ -10 }+{ 495 }\times { 10 }^{ -10 }\\ ={ 496 }\times { 10 }^{ -10 }\\ ={ 4.96 }\times { 10 }^{ -10 }$

A lot of 20 bulbs contain 4 defective ones. One bulb is selected at random from the lot. What is the probability that this bulb is defective? Suppose the bulb selected in the previous case is not defective and is not replaced. Now one bulb is selected at random from the rest. What is the probability that this bulb is not defective?

${\textbf{Step1: Finding probability of getting defective bulb}}$

${\text{Total number of bulbs}}$ $=20$

${\text{Number of defective bulbs}}$ $=4$

${\text{Therefore,}}$

${\text{P (getting a defective bulbs) =}}$ $\dfrac{{{\text{No}}{\text{. of defective bulbs}}}}{{{\text{Total No}}{\text{.of bulbs}}}}{\text{ = }}\dfrac{{\text{4}}}{{{\text{20}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{5}}}$

${\textbf{Step2: Finding Probability of not getting defective bulb}}$

${\text{One non defective bulb is drawn then the total number of bulb left is}}$ $19$

${\text{Total number of bulbs}}$ $=19$

${\text{Number of defective bulbs}}$ $=4$

${\text{Number of non-defective bulbs}}$ $=19-4$

$=15$

${\text{P(getting a not defective bulbs) =}}$ $\dfrac{{{\text{No}}{\text{. of non-defective bulbs}}}}{{{\text{Total no}}{\text{. of bulbs}}}}{\text{ = }}\dfrac{{{\text{15}}}}{{{\text{19}}}}$

${\textbf{Hence, Probability that the bulb is not defective is 15/19}}$

Find the probability success when a die is thrown , where a success is defined as
(i) number greater than $4$ (ii) six appears on the die

$i)$ The favourable outcomes are

$5,6= 2$

Total no. of outcomes =

$6$

Probability is

$\cfrac{2}{6}=\cfrac{1}{3}$

$ii)$ Thee favourable outcome is

$6 = 1$

Total no.of outcomes is

$6$

Probability is

$\cfrac{1}{6}$

Three unbaised coins are tossed find the probability distribution of the number of heads occurring on the topmost faces.

Sample space of tossing of three unbiased coins = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Total out comes =

$8$

No. of outcomes in which

$3$ heads are there =

$1$

Probability =

$\dfrac{1}{8}$

No. of outcomes in which

$2$ heads are there =

$3$

Probability =

$\dfrac{3}{8}$

No. of outcomes in which

$1$ head is there =

$3$

Probability =

$\dfrac{3}{8}$

No. of outcomes in which no heads are there =

$1$

Probability =

$\dfrac{1}{8}$

Let X denotes no. of heads coming when three unbiased coins are tossed probability distribution :-

$X$

$0$

$1$

$2$

$3$

$(x)$

$1/8$

$3/8$

$1/8$

A card from a pack 52 cards is lost. From the remaining card of the pack, one card is drawn and is found be heart, Find the probability that the lost cards were both hearts.

Consider the problem

We have

$52$ cards initially,

$13$ hearts and

$39$ hearts

Consider the events

${E_1}$ be both lost card are hearts

${E_2}$ be both lost cards are non-hearts

${E_3}$ One lost card is non-heart and One is heart

And,

$A$ is the probability that heart is picked from the remaining

$50$ cards.

So,

$\begin{array}{l} P\left( { { E_{ 1 } } } \right) =\dfrac { { ^{ 13 }{ C_{ 2 } } } }{ { ^{ 52 }{ C_{ 2 } } } } \\ =\dfrac { { 13\times 12 } }{ { 52\times 51 } } \\ =\dfrac { 1 }{ { 17 } } \end{array}$

$P\left( {\dfrac{A}{{{E_1}}}} \right) = \dfrac{{11}}{{50}}$

And

$\begin{array}{l} P\left( { { E_{ 2 } } } \right) =\dfrac { { ^{ 39 }{ C_{ 2 } } } }{ { ^{ 52 }{ C_{ 2 } } } } \\ =\dfrac { { 39\times 38 } }{ { 52\times 51 } } \\ =\dfrac { { 19 } }{ { 34 } } \end{array}$

$P\left( {\dfrac{A}{{{E_2}}}} \right) = \dfrac{{13}}{{50}}$

And,

$\begin{array}{l} P\left( { { E_{ 3 } } } \right) =\dfrac { { \left( { One\, heart } \right) and\left( { One\, non\, heart } \right) } }{ { ^{ 52 }{ C_{ 2 } } } } \\ =\dfrac { { 13\times 39 } }{ { ^{ 52 }{ C_{ 2 } } } } \\ =\dfrac { { 13\times 39\times 2 } }{ { 52\times 51 } } \\ =\dfrac { { 39 } }{ { 102 } } \end{array}$

$P\left( {\dfrac{A}{{{E_3}}}} \right) = \dfrac{{12}}{{50}}$

Now Required probability is

$\begin{array}{l} =\dfrac { { P\left( { { E_{ 1 } } } \right) \times P\left( { \dfrac { A }{ { { E_{ 1 } } } } } \right) } }{ { P\left( { { E_{ 1 } } } \right) \times P\left( { \dfrac { A }{ { { E_{ 1 } } } } } \right) +P\left( { { E_{ 2 } } } \right) \times P\left( { \dfrac { A }{ { { E_{ 2 } } } } } \right) +P\left( { { E_{ 3 } } } \right) \times P\left( { \dfrac { A }{ { { E_{ 3 } } } } } \right) } } \\ =\dfrac { { \dfrac { 1 }{ { 17 } } \times \dfrac { { 11 } }{ { 50 } } } }{ { \dfrac { 1 }{ { 17 } } \times \dfrac { { 11 } }{ { 50 } } +\dfrac { { 19 } }{ { 34 } } \times \dfrac { { 13 } }{ { 50 } } +\dfrac { { 39 } }{ { 102 } } \times \dfrac { { 12 } }{ { 50 } } } } \\ =\dfrac { { \dfrac { 1 }{ { 17 } } \times \dfrac { { 11 } }{ { 50 } } } }{ { \dfrac { 1 }{ { 17 } } \times \dfrac { { 11 } }{ { 50 } } +\dfrac { { 19 } }{ { 34 } } \times \dfrac { { 13 } }{ { 50 } } +\dfrac { { 39 } }{ { 17 } } \times \dfrac { 2 }{ { 50 } } } } \\ =\dfrac { { 11 } }{ { 11+\dfrac { { 247 } }{ 2 } +78 } } \, \, \, \, \, \, \left( { Multiplying\, each\, term\, by\, 17\times 50 } \right) \\ =\dfrac { { 11\times 2 } }{ { 22+247+156 } } \\ =\dfrac { { 22 } }{ { 425 } } \end{array}$

An integer is chosen at random from the first 200 numbers.What is the probability that the integer chosen is divisible by 6 or 8.

One integer can be chosen out of 200 integers in

$^{200}C_1=200\ ways$

Let A be the event that an integer selected is divisible by 6 and

B be the event that the integer is divisible by 8.

i.e

$A={6,12,18,.....198}$

$B={8,16,24,......200}$

$A\cap B={24,48,.....192}$

$\Rightarrow n(A)=33,\ n(B)=25,\ and\ n(A\cap B)=8$

Then

$P(A)=\dfrac{33}{200}$

$P(B)=\dfrac{25}{200}$ and

$P(A\cap B)=\dfrac{8}{200}$

$P(getting\ number\ divisible\ by\ 6\ or\ 8)=P(A\cup B)$

We know

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=\dfrac{33}{200}+\dfrac{25}{200}-\dfrac{8}{200}$

$=\dfrac{1}{4}$

A card from a pack of 52 cards is lost. from the remaining cards of pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Let

$E_{1}$ be the event that the lost card is Diamond

Given that there are 13 diamonds in the deck

$P(E_{1}) =\dfrac {13}{52} =\dfrac 14$

let

$E_{2}$ be the even that the lost card is not a diamond

$P(E_{2})=\dfrac 34$

Let A be the event that the two cards drawn are both diamonds

$P(A/E_{1}) = \dfrac{^{12}C_{2}}{^{51}C_{1}} = \dfrac{No.of\,ways\,of\,drawing\, 2\, diamonds \,cards}{Total\, no. \,of \,ways\, of\, drawing\,2\,cards}$

$= \dfrac{66}{1275}$

$P(A/E_{2}) = \dfrac{^{13}C_{2}}{^{51}C_{2}} = \dfrac{78}{1275}$

Applying Bayes Theorem, where

$P(E_{1}/A) = \dfrac{P(E_{1})\times (P(A/E_{1}))}{P(E_{1})\times P(A/E_{1})+P(E_{2})\times P(A/E_{2})}$

$\displaystyle =\frac{\dfrac{66}{1275}\times \dfrac{1}{4}}{\dfrac{66}{1275}\times \dfrac{1}{4}+\dfrac{78}{1275}\times \dfrac 34} = \dfrac{11}{50}$

In a school, there are $1000$ student, out of which $430$ are girls. It is known that out of $430, 10$ % of the girls study in class $XII$ . What is the probability that a student chosen randomly studies in class $XII$ given that the chosen student is a girl?.

Total number of students

$= 1000$

Total number of girls

$= 430$

Girls studying in class

$XII = 10 \% \text{ of } 430$

$\\ = \cfrac{10}{100} \times 430$

$\\ = 43$

We need to find the probability that a student chosen randomly studies in class

$XII$ , given that the chosen student is a girl.

$A$ : Student is in class

$XII$

$B$ : Studenet is a girl

Therefore,

$P{\left( A | B \right)}$

$= \cfrac{P{\left( A \cap B \right)}}{P{\left( B \right)}} \\ = \cfrac{\text{No. of girls studying in class XII}}{\text{Number of girls}} \\ = \cfrac{43}{430} = 0.1$

In a shop X, $30$ tin pure ghee and $40$ tin adultered ghee are kept for sale while in shop Y, $50$ tin pure ghee and $60$ tin adultered ghee are there. One tin of ghee is purchased from one of the shops randomly and it is found to be adultered. Find the probability that it is purchased from shop B.

Given In shop X pure ghee

$=30$ tin In shop Y pure ghee

$=50$ tin

Adultered ghee

$=40$ Adultered ghee

$=60$

$P(A/B)=\dfrac{P(B/A)(A)}{P(B)}$

$A\rightarrow$ the tin is purchased from shop B

$B\rightarrow$ the tin is adultered

$P(B/A)=$ probability of the tin from shop 'B' being adultered

$P(B/A)=\dfrac{60}{60+50}=\dfrac{60}{110}=\dfrac{6}{11}$

$P(A)=$ Probability that the tin is from shop B

$P(A)=\dfrac{(50+60)}{(50+60)+(30+40)}=\dfrac{110}{180}=\dfrac{11}{18}$

$P(B)=$ probability that the tin is adultered

$P(B)=\dfrac{(60+40)}{(50+60)+(30+40)}=\dfrac{100}{180}=\dfrac{10}{18}$

$P(A/B)=$ probability that the tin is from shop B given it is adultered

$=\dfrac{P(B/A)(P(A))}{P(B)}=\dfrac{\dfrac{6}{11}\times \dfrac{11}{18}}{\left(\dfrac{10}{18}\right)}=\dfrac{6}{10}=\dfrac{3}{5}$

$\therefore$ probability that the adultered tin is from shop B is

$\dfrac{3}{5}$ .

A bag contains $3$ red and $7$ black balls. Two balls are drawn one by one at a time at random without replacement. If second drawn ball is red then what is the probability the first drawn ball is also red?

Number of red balls

$=3$

Number of black balls

$=7$

Total number of balls

$=10$

Let ,

$P(A)=$ Probability that first is red

$P(B)=$ Probability that second is red

If second is red, there are 2 possible ways

Either first is red and second is red.

Or first is black and second is red.

So,

$P(B)=\dfrac{^3C_1}{10}\times\dfrac{^2C_1}{9}+\dfrac{^7C_1}{10}\times\dfrac{^3C_1}{9}$

$=\dfrac{6}{90}+\dfrac{21}{90}$

$=\dfrac{27}{90}$

$=\dfrac{3}{10}$

Probability that first is red, given second is red

$P(A/B)=\dfrac{P(A\cap B)}{P(B)}$

$=\dfrac{\dfrac{^3C_1}{10}\times\dfrac{^2C_1}{9}}{\dfrac{3}{10}}$

$=\dfrac{\dfrac{6}{90}}{\dfrac{3}{10}}$

$=\dfrac{6}{90}\times\dfrac{10}{3}$

$=\dfrac{6}{27}$

$=\dfrac{2}{9}$

From three men and two women, environment committee of two person to be formed.
Condition for event $A :$ There must be at least one woman member.
Condition for event $B :$ One man, one woman committee to be formed.
Condition for event $C :$ There should not be a woman member.

Given

$3\ Men$

$2\ Women$

$P(A)=P(1M1W)+P(2W)$

$=\cfrac { { _{ }^{ 3 }{ C } }_{ 1 }\,{ _{ }^{ 2 }{ C } }_{ 1 } }{ { _{ }^{ 5 }{ C } }_{ 2 } } +\cfrac { { _{ }^{ 2 }{ C } }_{ 2 } }{ { _{ }^{ 5 }{ C } }_{ 2 } }$

$=\cfrac { 3.2 }{ 10 } +\cfrac { 1 }{ 10 } =\cfrac { 7 }{ 10 }$

$P(B)=P(1M1W)$

$=\cfrac { { _{ }^{ 3 }{ C } }_{ 1 } \,{ _{ }^{ 2 }{ C } }_{ 1 } }{ { _{ }^{ 5 }{ C } }_{ 2 } } =\cfrac { 3.2 }{ 10 } =\cfrac { 3 }{ 5 }$

$P(C)=P(0W2M)$

$=\cfrac { { _{ }^{ 3 }{ C } }_{ 2 } }{ { _{ }^{ 5 }{ C } }_{ 2 } } =\cfrac { 3 }{ 10 }$

A dice is manufactured in such a way that probability of getting an even number is twice likely to occur as an odd number. If the dice is tossed twice, find the probability distribution of the random variable representing the perfect square in the both tosses.

If dice is tossed twice, Then, probability of perfect square on both tossed are

Let X be denote the perfect square number

$(1,1),(4,4)$

$P\left( {1,1} \right) = \frac{1}{{36}},P\left( {4,4} \right) = \frac{1}{{36}}$

Probability distribution

1221628_1211559_ans_fd5fe8477f684d4dba1543dc897bba12.JPG

Find the probability distribution of number of doublets in three throws of a pair of dice.

Let X denotes the number of doublets. X can take the value 0,1,2 or 3.

Possible doublets are (1,1), (2,2), (3,3), (4,4), (5,5) and (6,6)

Probability of getting a doublet

$=\dfrac{6}{36}=\dfrac{1}{6}$

Probability of not getting a doublet

$=1-\dfrac{1}{6}=\dfrac{5}{6}$

Now,

$P(X=0)=P(No\ doublet)$

$=\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}$

$=\dfrac{125}{216}$

$P(X=1)=P(One\ doublet\ and\ two\ non-doublet)$

$=\dfrac{1}{6}\times\dfrac{5}{6}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{1}{6}\times\dfrac{5}{6}+\dfrac{5}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}$

$=\dfrac{25}{216}+\dfrac{25}{216}+\dfrac{25}{216}$

$=\dfrac{75}{216}$

$P(X=2)=P(Two\ doublet\ and\ one\ non-doublet)$

$=\dfrac{1}{6}\times\dfrac{1}{6}\times\dfrac{5}{6}+\dfrac{1}{6}\times\dfrac{5}{6}\times\dfrac{1}{6}+\dfrac{5}{6}\times\dfrac{1}{6}\times\dfrac{1}{6}$

$=\dfrac{5}{216}+\dfrac{5}{216}+\dfrac{5}{216}$

$=\dfrac{15}{216}$

$P(X=3)=P(Three\ doublets)$

$=\dfrac{1}{6}\times\dfrac{1}{6}\times\dfrac{1}{6}$

$=\dfrac{1}{216}$

$\therefore$ The required probabilities are shown in the figure.

1115818_1208858_ans_512199c71e1147f485632642f87f03eb.png

A box contains $150$ oranges. If one orange. If one orange is taken out from the box at random the probabilty of its being rotten is $0.06$ , then find the number of good oranges in the box.

Probability of rotten mangoes $= 0.06$

let number of rotten mangoes is $x$ then

$0.06 = x\times150$

and $x = 9$

number of good mangoes $= 150-9 = 141$

Probability of good mangoes $= 141/150 = 0.94$

Bag A contains 3 red and 2 white balls, and Bag B contains 2 red and 5 white balls. A bag selected at random, a ball is drawn and put into the other bag; then a ball is drawn from the second bag. Find the probability that both balls drawn are of the same colour.

Let

$E_{11}$ be the event of transferring a red ball from Bag A to Bag B

$=\dfrac{3}{5}$

Let

$A$ be the event of drawing a red ball from Bag B

$P\left(\dfrac{A}{E_{11}}\right)=\dfrac{3}{8}$

Let

$E_{12}$ be the event of transferring a white ball from Bag A to Bag B

$=\dfrac{2}{5}$

Let

$B$ be the event of drawing a white ball from Bag B

$P\left(\dfrac{B}{E_{12}}\right)=\dfrac{6}{8}$

Let

$E_{21}$ be the event of transferring a red ball from Bag B to Bag A

$=\dfrac{2}{7}$

Let

$C$ be the event of drawing a red ball from Bag A

$P\left(\dfrac{C}{E_{21}}\right)=\dfrac{4}{6}$

Let

$E_{22}$ be the event of transferring a white ball from Bag B to Bag A

$=\dfrac{5}{7}$

Let

$D$ be the event of drawing a white ball from Bag A

$P\left(\dfrac{D}{E_{22}}\right)=\dfrac{3}{6}$

Hence

Required

$P(E_1) =P(E_{11}).P\left(\dfrac{A}{E_{11}}\right)+P(E_{12}).P\left(\dfrac{B}{E_{12}}\right)$

$=\left(\dfrac{3}{5}\right).\left(\dfrac{3}{8}\right)+\left(\dfrac{2}{5}\right)\left(\dfrac{6}{8}\right)$

$=\dfrac{9}{40}+\dfrac{12}{40}$

$=\dfrac{21}{40}$

Required

$P(E_2) =P(E_{21}).P\left(\dfrac{C}{E_{21}}\right)+P(E_{22}).P\left(\dfrac{D}{E_{22}}\right)$

$=\left(\dfrac{2}{7}\right).\left(\dfrac{4}{6}\right)+\left(\dfrac{5}{7}\right)\left(\dfrac{3}{6}\right)$

$=\dfrac{8}{42}+\dfrac{15}{42}$

$=\dfrac{23}{42}$

$\therefore P(E) = 0.5P(E_1)+0.5P(E_2)$ (since a bag was picked randomly with equal probability)

$P(E) = 0.5\left(\dfrac{21}{40}+\dfrac{23}{42}\right) = \dfrac{901}{1680}$

Evaluate $P\left( {A \cup B} \right),$ if $2P\left( A \right) = P\left( B \right) = \dfrac{5}{{13}}$ and $P\left( {A|B} \right) = \dfrac{2}{5}$

To find :

$P(A\cup B)$

Given:

$2P(A)=P(B)=\cfrac { 5 }{ 13 } ,P(A|B)=\cfrac { 2 }{ 5 }$

$\Rightarrow P(A)=\cfrac { 5 }{ 2\times 13 } =\cfrac { 5 }{ 26 }$

$\Rightarrow P(A)=\cfrac { 5 }{ 26 } ;P(B)=\cfrac { 5 }{ 13 }$

We know that

$P(A|B)=\cfrac { P(A\cap B) }{ P(B) }$

$\Rightarrow \cfrac { 2 }{ 5 } =\cfrac { P(A\cap B) }{ \cfrac { 5 }{ 13 } } \Rightarrow P(A\cap B)=\cfrac { 2 }{ 5 } \times \cfrac { 5 }{ 13 } =\cfrac { 2 }{ 13 }$

We know that

$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\cfrac { 5 }{ 26 } +\cfrac { 5 }{ 13 } -\cfrac { 2 }{ 13 } =\cfrac { 5+10-4 }{ 26 } =\cfrac { 11 }{ 26 }$

Find the probability that in a year of $22$ nd century chosen at a random, there will be $53$ sundays.

1339790_1235426_ans_cf68b04d9c2d4eda822509e22b7944a3.PNG

A company has two plants to manufacture televisions. The plant I manufacture $70$ % of televisions and plant II manufacture $30$ %. At plant I, $80$ % of the televisions are rated as of standard quality and at plant II, $90$ % of the televisions are rated as of standard quality. A television is chosen at random and is found to be of standard quality. The probability that it has come from plant II is

${ E }_{ 1 }=$ Plant

$1$ manufactures televisions.

${ E }_{ 1 }=$ Plant

$2$ manufactures televisions.

$P({ E }_{ 1 })=\cfrac { 70 }{ 100 } ,\quad P({ E }_{ 2 })=\cfrac { 30 }{ 100 } \quad ,P(\cfrac { A }{ { E }_{ 1 } } )=\cfrac { 80 }{ 100 } ,\quad P(\cfrac { A }{ { E }_{ 2 } } )=\cfrac { 90 }{ 100 }$

By baye's theorem

$P(\cfrac { { E }_{ 2 } }{ A } )=\cfrac { P({ E }_{ 2 })P(\cfrac { A }{ { E }_{ 2 } } ) }{ P({ E }_{ 2 })P(\cfrac { A }{ { E }_{ 2 } } )+P({ E }_{ 1 })P(\cfrac { A }{ { E }_{ 1 } } ) } \\ =\cfrac { 27 }{ 27+56 } \\ =\cfrac { 27 }{ 83 }$

$\therefore$ Probability it has came from plant

$2$ is

$\cfrac { 27 }{ 83 }$

The probabilities of x, y and z becoming manager are $\frac { 4 }{ 9 } ,\frac { 2 }{ 9 } \& \frac { 1 }{ 3 }$ respectively. The probabilities that the bonus scheme will be introduced if x,y and z become managers are $\frac { 3 }{ 10 } ,\frac { 1 }{ 2 } \& \frac { 2 }{ 5 }$ respectively. Find (i) What is the probatility that the bonus scheme will be introduced? (ii) if the bonus scheme has teen introduced, What is the probability that the manager appointed wax x?

$P(x)=\cfrac{4}{9},P(y)=\cfrac{2}{9},P(z)=\cfrac{1}{3}$

Let event

$A$ be the bonus scheme introduced

so,

$P(A/x)=\cfrac{3}{10}$ and

$P(A/y)=\cfrac{1}{2}$

$(1)$

$P(A)=P(x).P(a/x)+P(y).P(A/y)+P(z).P(A/z)$

$=\cfrac{4}{9}\times \cfrac{3}{10}+|cfrac{2}{9}\times \cfrac{1}{2}+\cfrac{1}{2}\times\cfrac{2}{5}$

$=\cfrac{6+5+6}{45}=\cfrac{17}{45}$

$(2)$ Probability that manager appointed if bonus introduced

$P(z/A)=\cfrac{P(A/z).P(z)}{P(A)}$

$=\cfrac{2\times 45}{15\times 17}=\cfrac{6}{17}$

In a simultaneous throw of a pair of dice, if the probability of getting neither $9$ nor $11$ as the sum of the numbers on the faces is $\dfrac{5}{a}.$ Find $a.$

The outcomes that we gets, in a simultaneous throw of a pair of dice are,

$S= \left\{\begin{matrix} \left( 1,1 \right) \left( 1,2 \right) \left( 1,3 \right) \left( 1,4 \right) \left( 1,5 \right) \left( 1,6 \right) \\ \left( 2,1 \right) \left( 2,2 \right) \left( 2,3 \right) \left( 2,4 \right) \left( 2,5 \right) \left( 2,6 \right) \\ \left( 3,1 \right) \left( 3,2 \right) \left( 3,3 \right) \left( 3,4 \right) \left( 3,5 \right) \left( 3,6 \right) \\ \left( 4,1 \right) \left( 4,2 \right) \left( 4,3 \right) \left( 4,4 \right) \left( 4,5 \right) \left( 4,6 \right) \\ \left( 5,1 \right) \left( 5,2 \right) \left( 5,3 \right) \left( 5,4 \right) \left( 5,5 \right) \left( 5,6 \right) \\ \left( 6,1 \right) \left( 6,2 \right) \left( 6,3 \right) \left( 6,4 \right) \left( 6,5 \right) \left( 6,6 \right) \end{matrix}\right\}$

$n(S)=36$

Now,

Favourable outcomes i.e getting neither

$9$ nor

$11$ as the sum of numbers on the faces are

$E= \left\{\begin{matrix} \left( 1,1 \right) \left( 1,2 \right) \left( 1,3 \right) \left( 1,4 \right) \left( 1,5 \right) \left( 1,6 \right) \left( 2,1 \right) \left( 2,2 \right) \\ \left( 2,3 \right) \left( 2,4 \right) \left( 2,5 \right) \left( 2,6 \right) \left( 3,1 \right) \left( 3,2 \right) \left( 3,3 \right) \\ \left( 3,4 \right) \left( 3,5 \right) \left( 4,1 \right) \left( 4,2 \right) \left( 4,3 \right) \left( 4,4 \right) \left( 4,6 \right) \\ \left( 5,1 \right) \left( 5,2 \right) \left( 5,3 \right) \left( 5,5 \right) \left( 6,1 \right) \left( 6,2 \right) \left( 6,4 \right) \left( 6,6 \right) \end{matrix}\right\}$

$n(E)=30$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{30}{36}$

$P(E)=\dfrac{5}{6}=\dfrac{5}{a}$

$\therefore \boxed {\ a=6\ }$

In a simultaneous throw of a pair of dice, if the probability of getting odd number on the first and $6$ on the second is $\dfrac{1}{a}.$ Find $a.$

In a throw of a pair of dice, total no of possible outcomes

$=36(6\times 6)$ are

$\begin{matrix} \left( 1,1 \right) & \left( 1,2 \right) & \left( 1,3 \right) & \left( 1,4 \right) & \left( 1,5 \right) & \left( 1,6 \right) \\ \left( 2,1 \right) & \left( 2,2 \right) & \left( 2,3 \right) & \left( 2,4 \right) & \left( 2,5 \right) & \left( 2,6 \right) \\ \left( 3,1 \right) & \left( 3,2 \right) & \left( 3,3 \right) & \left( 3,4 \right) & \left( 3,5 \right) & \left( 3,6 \right) \\ \left( 4,1 \right) & \left( 4,2 \right) & \left( 4,3 \right) & \left( 4,4 \right) & \left( 4,5 \right) & \left( 4,6 \right) \\ \left( 5,1 \right) & \left( 5,2 \right) & \left( 5,3 \right) & \left( 5,4 \right) & \left( 5,5 \right) & \left( 5,6 \right) \\ \left( 6,1 \right) & \left( 6,2 \right) & \left( 6,3 \right) & \left( 6,4 \right) & \left( 6,5 \right) & \left( 6,6 \right) \end{matrix}$

Let

$E\rightarrow$ event of getting an odd number on the first and

$6$ on the second

Then

$E=\left\{(1, 6), (3, 6), (5, 6)\right\}$

$n(E)=3$

$\therefore P(E)=\dfrac{n(E)}{n(s)}=\dfrac{3}{36}$

$=\dfrac {1}{12}=\dfrac 1a\Rightarrow a=12$

Hence

$\boxed{\ a=6\ }$

In a simultaneous throw of a pair of dice, if the probability of getting a sum less then $6$ is $\dfrac{5}{a}.$ Find $a.$

Let

$S=$ Total outcomes

$E=$ Favourable outcomes (sum less than

$6$ )

$S=\left\{(1,1)\ (1,2)\ (1,3)\ (1,4)\ (1,5)\ (1,6)\right.$

$(2,1)\ (2,2)\ (2,3)\ (2,4)\ (2,5)\ (2,6)$

$(3,1)\ (3,2)\ (3,3)\ (3,4)\ (3,5)\ (3,6)$

$(4,1)\ (4,2)\ (4,3)\ (4,4)\ (4,5)\ (4,6)$

$(5,1)\ (5,2)\ (5,3)\ (5,4)\ (5,5)\ (5,6)$

$\left.(6,1)\ (6,2)\ (6,3)\ (6,4)\ (6,5)\ (6,6)\right\}$

$n(S)=36$

$E=\left\{(1,1)\ (1,2)\ (1,3)\ (1,4)\ (2,1)\ (2,2)\ (2,3)\ (3,1)\ (3,2)\ (4,1)\right\}$

$n(E)=5$

$\Rightarrow P(E)=\dfrac{n(E)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$

i.e probability of getting a sum less than

$6$ is

$\dfrac{5}{18}=\dfrac{5}{a}$

$\therefore \boxed{\ a=18\ }$

In a simultaneous throw of a pair of dice, if the probability of getting a sum less than $7$ is $\dfrac{a}{b}.$ Find $a+b.$

Let

$S=$ Total outcomes

$=\left\{(1,1)\ (1,2)\ (1,3)\ (1,4)\ (1,5)\ (1,6)\right.$

$(2,1)\ (2,2)\ (2,3)\ (2,4)\ (2,5)\ (2,6)$

$(3,1)\ (3,2)\ (3,3)\ (3,4)\ (3,5)\ (3,6)$

$(4,1)\ (4,2)\ (4,3)\ (4,4)\ (4,5)\ (4,6)$

$(5,1)\ (5,2)\ (5,3)\ (5,4)\ (5,5)\ (5,6)$

$\left.(6,1)\ (6,2)\ (6,3)\ (6,4)\ (6,5)\ (6,6)\right\}$

$E=$ sum less than

$7$

$=\left\{(1,1)\ (1,2)\ (1,3)\ (1,4)\ (1,5)\ (2,1)\ (2,2)\ (2,3)\ (2,4)\ (3,1)\ (3,2)\ (3,3)\ (4,1)\ (4,2)\ (5,1)\right\}$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{15}{36}=\dfrac{5}{12}$

i.e probability of getting sum less than

$7$ is

$\dfrac{5}{12}=\dfrac ab$

$\Rightarrow a=5,b=12$

$\therefore\ a+b=5+12=17$

In a simultaneous throw of a pair of dice, if the probability of getting a number greater than $4$ on each die is $\dfrac{1}{a}.$ Find $a.$

Let

$S=$ Total outcomes

$S=\left\{ \begin{matrix} \left( 1,1 \right) \left( 1,2 \right) \left( 1,3 \right) \left( 1,4 \right) \left( 1,5 \right) \left( 1,6 \right) \\ \left( 2,1 \right) \left( 2,2 \right) \left( 2,3 \right) \left( 2,4 \right) \left( 2,5 \right) \left( 2,6 \right) \\ \left( 3,1 \right) \left( 3,2 \right) \left( 3,3 \right) \left( 3,4 \right) \left( 3,5 \right) \left( 3,6 \right) \\ \left( 4,1 \right) \left( 4,2 \right) \left( 4,3 \right) \left( 4,4 \right) \left( 4,5 \right) \left( 4,6 \right) \\ \left( 5,1 \right) \left( 5,2 \right) \left( 5,3 \right) \left( 5,4 \right) \left( 5,5 \right) \left( 5,6 \right) \\ \left( 6,1 \right) \left( 6,2 \right) \left( 6,3 \right) \left( 6,4 \right) \left( 6,5 \right) \left( 6,6 \right) \end{matrix}\right\}$

$E=$ A number greater than

$4$ on each die.

$\therefore \left\{(5, 5) (5, 6) (6, 5) (6, 6)\right\}$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{4}{36}=\dfrac{1}{9}$

$\therefore$ Probability of getting a number greater than

$4$ on each die is

$\dfrac{1}{9}=\dfrac 1a\Rightarrow a=9$ .

Hence

$boxed {\ a=9\ }$

Cards marked with number $2$ to $101$ are placed in a box and mixed thoroughly. One card is drawn from this box. Find the probability that the number of the card is an even number.

We have,

Cards marked with number $2$ to $101$ are placed in a box.

Total number of cards $=100$

Even cards $=50$

Odd cards $=50$

According to given question,

Probability of even cards $=\dfrac{50}{100}=\dfrac{1}{2}$

Hence, this is the answer.

$3000$ families with $2$ children each were selected randomly and the following date are recorded.
No of girls $2$ $1$ $0$
No of familis $950$ $1628$ $422$
Compute the probability of familes chosen at random, having no $A-2$ girl $B-$ 1 girl $C-N0$ girls.

Total number of Families $=3000\,families$

Children $=2$

(1):- 2 girl

(2):- 1 girl

(3):- No girl

Solve:-

Part(1):-

Total number of families $=3000$

Number of families having 2 girl $=950$

Then,

Probability chosen family has 2 girls $=\dfrac{\text{Number}\,\text{of}\,\text{families}\,\text{having}\,\text{2}\,\text{girls}}{\text{Total}\,\text{number}\,\text{of}\,\text{families}}$

Probability

$\left( P \right)=\dfrac{950}{3000}$

$=\dfrac{95}{300}$

$=\dfrac{19}{60}$

(2):-

Total number of families $=3000$

Number of families having 1 girl $=1628$

Probability of chosen family has 1 girl $=\dfrac{\text{Number}\,\text{of}\,\text{families}\,\text{having}\,\text{1}\,\text{girl}\,}{\text{Total}\,\text{number}\,\text{of}\,\text{families}}$

$P=\dfrac{1628}{3000}$

$P=\dfrac{407}{750}$

Now,

(3):-

Total number of families $=3000$

Number of families having no girl $=422$

Then,

Probability chosen family has no girl $=\dfrac{\text{number}\,\text{of}\,\text{families}\,\text{having}\,\text{no}\,\text{girl}}{\text{Total}\,\text{number}\,\text{of}\,\text{families}}$

$P=\dfrac{422}{3000}$

$P=\dfrac{211}{1500}$

Hence, this is the answer.

Find the value of P for the following distribution whose mean is 10
$x_1$ 5 7 9 11 13 15 20
$f_i$ 4 4 p 7 3 2 1

$x_if_i=20, 28, 9p, 77, 39, 30, 20$

Mean

$=\dfrac{\sum x_if_i}{\sum f_i}=\dfrac{214+9p}{p+21}=10$

$\Rightarrow 9p+214=10p+210$

$\Rightarrow p=4$ .

1181989_1288955_ans_c62579c599c546cab6494be28a57a499.jpg

A card is picked up from a deck of $52$ plying cards.
What is the even that the chosen card id black faces card?

1412273_1279085_ans_7dfcd3cd2f174c27a4bf01ea28e47054.jpg

Ten pairs of shoes are in a closet. Four shoes are selected at random. The probability that there will be at least one pair among the four shoes selected is

1239140_1313009_ans_9fb196cbb8f84bc6bc094d02db3d503f.jpg

In a simultaneous throw of a pair of dice, if the probability of getting a total of $9$ or $11$ is $\dfrac{1}{x}.$ Find $x.$

Let

$S=$ Total outcomes

$S=\left\{\begin{matrix} \left( 1,1 \right) \left( 1,2 \right) \left( 1,3 \right) \left( 1,4 \right) \left( 1,5 \right) \left( 1,6 \right) \\ \left( 2,1 \right) \left( 2,2 \right) \left( 2,3 \right) \left( 2,4 \right) \left( 2,5 \right) \left( 2,6 \right) \\ \left( 3,1 \right) \left( 3,2 \right) \left( 3,3 \right) \left( 3,4 \right) \left( 3,5 \right) \left( 3,6 \right) \\ \left( 4,1 \right) \left( 4,2 \right) \left( 4,3 \right) \left( 4,4 \right) \left( 4,5 \right) \left( 4,6 \right) \\ \left( 5,1 \right) \left( 5,2 \right) \left( 5,3 \right) \left( 5,4 \right) \left( 5,5 \right) \left( 5,6 \right) \\ \left( 6,1 \right) \left( 6,2 \right) \left( 6,3 \right) \left( 6,4 \right) \left( 6,5 \right) \left( 6,6 \right) \end{matrix}\right\}$

$n(S)=6\times 6=36$

Let

$E:$ getting a total of

$9$ or

$11$

$E=\left\{(3, 6) (4, 5) (5, 4) (6, 3) (6, 5) (5, 6)\right\}$

$n(E)=6$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6}$

$\Rightarrow$ Probability of getting a total of

$9$ or

$11$ is

$\dfrac{1}{6}=\dfrac{1}{x}$

$\therefore \boxed {\ x=6\ }$

Find the number of triangles whose vertices are at the vertices of an octagon but none of whose sides happen to come from the octagon .

Required number of triangles

$=$ total number of triangles

$-$ number of triangles having two sides common

$-$ number of triangles having one side common

$= ^{8}{{C}_{3}}-8-8\times 4$

$=\dfrac{8!}{3!5!}-8-32$

$=\dfrac{8\times 7\times 6\times 5!}{3\times 2\times 5!}-40$

$=56-40=16$

A die is thrown twice and the sum of the numbers appearing is observed to beWhat is the conditional probability that the number 4 has appeared at least once?

A dice is thrown twice

$S={ (1,1),(1,2),(1,3),(1,4),(1,5) },(1,6)()2,1,(2,2),(2,3),(2,4),(2,5)(2,6)(3,1),(3,2)(3,3),\\(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2),(5,3),(5,4),\\(5,5),(5,6),(6,1),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)$

Here we need to find the probability that has appeared at least once. Since the sum of number oserved to be

$6$

LEt

$F=$ Sum of numbers is

$6$

$E=4$ has appeared at least once

We need to find

$P(E|F)$

$E={(1,4),(2,4).(3,4),(4,4),(5,4),(6,4),(4,1),(4,2),(4,3),(4,5),(4,6)}$

$P(E)=\cfrac { 11 }{ 16 } \\ F={ (1,5),(5,2),(2,4),(4,2),(3,3) }\\ P(F)=\cfrac { 5 }{ 36 }$

Also

$\quad E\cap F={ (2,4),(4,2) }\\ P(E\cap F)=\cfrac { 2 }{ 36 }$

Now

$\quad P(E/F)=\cfrac { P(E\cap F) }{ P(F) }$

$\quad =\cfrac { \cfrac { 2 }{ 36 } }{ \cfrac { 5 }{ 36 } }$

$\therefore$ required probability

$=\cfrac{2}{5}$

Enter 1 if true else enter 0.
In a simultaneous throw of a pair of dice, the probability of getting a total greater than $8$ is $\dfrac {5}{18}$

1412193_1278570_ans_b14d4e4f8bde4bc69f65f1a038cdadd7.jpg

In a single throw of three dice, if the probability of getting a total of $17$ or $18$ is $\dfrac{1}{a}$ . Find $a.$

The total outcomes we get in a single throw of

$3$ dice are

$6\times 6\times 6=216$

For a sum of

$17$ ,

i.e,

$a+b+c=17$ .

$(6, 6, 5) (6, 5, 6) (5, 6, 6)$ are for which we get a total of

$17$ .

The only outcome for which we get a total of

$18$ is

$(6, 6, 6)$

Hence,

probability of getting a sum of

$17$ or

$18$ is

$\dfrac{4}{216}=\dfrac{1}{54}=\dfrac 1a\Rightarrow a=54$

Hence

$\boxed{\ a=54\ }$

A die is thrown three times, find the probability that $4$ appears on the third toss if it is given that $6$ and $5$ appear respectively on first two tosses.

Let E: 4 appears on the third toss

F: 6 and 5 appear respectively on the first two tosses.

$P(E)=\dfrac{1}{6}$

$P(F)=\dfrac16\times\dfrac16=\dfrac{1}{36}$

Now,

$P(E \cap F)=\dfrac{1}{216}$

Therefore, required probability ,

$P(E|F)=\dfrac{P(E \cap F)}{P(F)}=\dfrac{1}{6}$

How many coins are to be tossed at once to get $64$ outcomes in total?

$2^{m} = 64$

$2^{m} = 2^{6}$

Therefore, m =6

Hence, 6 coins must be tossed at once to get 64 outcomes in total.

If $A$ and $B$ are two events, such that $P(A) = 0.4, P(B) = 0.8$ , and $P(B/A) = 0.6$ , then find $P(A/B)$ .

1272347_1340370_ans_e0519f1f57bb438aa3b77246606bf1cd.jpg

Coin tossed 6 times ,find the probability it shows heads.
i) Exact 5 times. ii) Head in 1st four and tail in rest.

1202525_1379463_ans_199363e5e7b1478e85f92e8346bcf115.JPG

Let $X$ denote the number of hours you study during a randomly selected school day. The probability that $X$ can take the values of $x$ , has the following form, where $k$ is some constant
$P(X=x)=\begin{matrix} \begin{cases} 0,1 & \text{if }x=0 \\ Kx & \text{if }x=1 \text{ or } 2 \\ K(5-x) & \text{if } x = 3 \text{ or } 4 \\ 0 & \text{otherwise} \end{cases} \end{matrix}$
What is the probability that you study.
For at least two hour.

$\sum_{0}^{4}P(X)=1$

$0.1+k+2k+2k+k=1$

$6k=1-0.1$

$\therefore k=0.15$

$P(X\geq 2)=P(X=2)+P(X=3)+P(X=4)$

$=2k+2k+k$

$=5k$

$=5(0.15)=0.75$

Find the number of ways in which 5 boys and 5 girls be seated in a row so that no two girls may sit together.

$5$ boys can be seated in a row in

$^{5}{{P}_{5}}$ ways.

Now, in the

$6$ gaps

$5$ girls can be arranged in

$^{6}{{P}_{5}}$ ways.

Hence, the number of ways in which no two girls sit together

$=5!\times ^{6}{{P}_{5}}=5!\times 6!$

Verify that the following function can be regarded as p.d.f for the random variable $X$
$x$ $-1$ $0$ $1$
$P(x)$ $-0.2$ $1$ $0.2$

A function

$P(x)$ serves as probability distribution function if and only if

$P(x)\ge 0$

But here

$P(-1)= -0.2\implies P(x)\le 0$ which is not possible

So given function is not regarded as p.m.f but regarded as p.d.f.

Three machines $E_{1}$ . $E_{2}, E_{3}$ in a certain factory produce 50 $\%$ . 25 $\%$ and 25 $\%$ respectively of the total daily output of electric tubes. It is known that 4 $\%$ of the tube produced on each of machines $E_{1}$ . and $E_{2},$ are defective and that 5 $\%$ those produced on $E_{3}$ are defective. If one is picked up at random from a day production, calculate the probability that it is defective.

1699111_1433726_ans_a53e5d4766cd4e46914e0691c08cdead.jpeg

Find the number of numbers of $5$ digits that can be formed with the digits $0, 1, 2, 3, 4$ if the digits can be repeated in the same number.

In a five digit number

$0$ cannot be put in ten thousand’s place. So, the number of ways of filling up the ten thousand’s place

$= 4$ .

Since the repetition of digits is allowed, therefore each of the other places can be filled in

$5$ ways.

So, the required number of numbers

$= 4 \times 5 \times 5 \times 5 \times 5 = 2500.$

A die is rolled. If the outcome is an odd number, what is the probability that it is prime?

Sample space,

$S=\left\{1,2,3,4,5,6\right\}$

Let

$A$ be the event of getting an odd number.

Let

$B$ be the event of getting an prime number.

$A=\left\{1,3,5\right\},B=\left\{2,3,5\right\},A\cap B=\left\{3,5\right\}$

Conditional probability of event

$B$ given that

$A$ has already occured,

$P\left(B|A\right)$

$=\dfrac{P\left(A\cap B\right)}{P\left(A\right)}=\dfrac{\dfrac{2}{6}}{\dfrac{3}{6}}=\dfrac{2}{3}$

How many numbers of $3$ digits can be formed with the digits $1, 2, 3, 4, 5$ when digits may be repeated?

The unit’s place can be filled in

$5$ ways. Since, the repetition of digits is allowed, therefore ten’s place can be filled in

$5$ ways and hundred’s place can also be filled in

$5$ ways. Therefore, by the fundamental principle of counting, the required number of three digit numbers

$= 5 \times 5 \times 5 = 125$ .

In a class of $10$ students there are $3$ girls $A, B, C$ . In how many different ways can they be arranged in a row such that no two of the three girls are consecutive.

There are

$7$ boys and

$3$ girls. Seven boys can be arranged in a row in

$^{7}{{P}_{7}}=7!$ ways. Now, we have

$8$ places in which we can arrange

$3$ girls in

$^{8}{{P}_{3}}$ ways.

Hence, by fundamental principle of counting, the number of arrangements

$= 7!\times ^{8}{{P}_{3}} = 7! \times\dfrac{8!}{5!}=7!\times 336$

Find the number of ways in which $5$ boys and $5$ girls be seated in a row so that all the girls sit together and all the boys sit together.

The two groups of girls and boys can be arranged in

$2!$ ways

$5$ girls can be arranged among themselves in

$5!$ Ways. Similarly,

$5$ boys can be arranged among themselves in

$5!$ ways. Hence, by the fundamental principle of counting, the total number of requisite seating arrangements

$= 2!\left(5! \times 5!\right) = 2{\left(5!\right)}^{2}$

In an examination hall there are four rows of chairs. Each row has $8$ chairs one behind the other. There are two classes sitting for the examination with $16$ students in each class. It is desired that in each row, all students belong to the same class and that no two adjacent rows are allotted to the same class. In how many ways can these $32$ students be seated?

Let the two classes be

${C}_{1}$ and

${C}_{2}$ and the four rows be

${R}_{1}, {R}_{2}, {R}_{3}, {R}_{4}$ . There are

$16$ students in each class. So, there are

$32$ students.

According to the given conditions there are two different ways in which

$32$ students can be seated :

	${R}_{1}$	${R}_{2}$	${R}_{3}$	${R}_{4}$
$I$	${C}_{1}$	${C}_{2}$	${C}_{1}$	${C}_{2}$
$II$	${C}_{2}$	${C}_{1}$	${C}_{2}$	${C}_{1}$

Since the seating arrangement can be completed by using any one of these two ways. So, by the fundamental principle of addition, Total no. of seating arrangements

$=$ No. of arrangements in

$I$ case

$+$ No. of arrangements in

$II$ case.

Now,

$16$ students of class

${C}_{1}$ can be seated in

$16$ chairs in

$^{16}{{P}_{16}}=16!$ ways.

And,

$16$ students of class

${C}_{2}$ can be seated in

$16$ chairs in

$^{16}{{P}_{16}}=16!$ ways.

Hence, the total number of seating arrangements

$=\left(16!\times 16!\right)+\left(16!\times 16!\right)=2\left(16!\times 16!\right)$

A factory has two machines A and B. past records shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

$\textbf{Step 1: Computing the individual probabilities}$

$\text{Let the probability that the product was made by Machine A be }E_1$

$\text{Let the probability that the product was made by Machine B be }E_2$

$\text{Let the probability that the product was defective be }A$

$P(E_1)=\dfrac{60}{100}=0.6$

$P(E_2)=\dfrac{40}{100}=0.4$

$2\%\text{ of items produced by A was defective}$

$P(A|E_1)=\dfrac{2}{100}=0.02$

$1\%\text{ of items produced by B was defective}$

$P(A|E_2)=\dfrac{1}{100}=0.01$

$\textbf{Step 2: Using Bayes' Theorem to compute the required probability}$

$\text{The probability that the product was produced by machine B given it was defective is }P(E_2|A)$

$\text{Using Bayes' Theorem}$

$P(E_2|A)=\dfrac{P(E_2)\times P(A|E_2)}{P(E_1)\times P(A|E_1)+P(E_2)\times P(A|E_2)}$

$=\dfrac{0.4\times0.01}{0.6\times0.02+0.4\times0.01}$

$=0.25$

$\textbf{Hence, The probability that the product was produced by machine B given it was defective}$

$\textbf{is }\mathbf{0.25}$

Three urns have the following composition of balls:
Urn I : $1$ white, $2$ black
Urn II : $2$ white, $1$ black
Urn III : $2$ white, $2$ black
One of the urns is selected at random and a ball is drawn. It turns out to be white. Find the probability that it came from urn III.

Let

$P{\left( E \right)}$ be the probability of choosing

$1$ white ball from Urn.

$P{\left( {E}_{i} \right)} =$ Probability of selecting one of three urns .

Therefore,

$P{\left( {E}_{1} \right)} = P{\left( {E}_{2} \right)} = P{\left( {E}_{3} \right)} = \cfrac{1}{3}$

$P{\left( {E}/{{E}_{1}} \right)} = \cfrac{{^{1}{C}_{1}}}{{^{3}{C}_{1}}} = \cfrac{1}{3}$

$P{\left( {E}/{{E}_{1}} \right)} = \cfrac{{^{2}{C}_{1}}}{{^{3}{C}_{1}}} = \cfrac{2}{3}$

$P{\left( {E}/{{E}_{1}} \right)} = \cfrac{{^{2}{C}_{1}}}{{^{4}{C}_{1}}} = \cfrac{1}{2}$

Therefore,

$P{\left( {{E}_{3}}/{E} \right)} = \cfrac{P{\left( {E}_{3} \right)} \cdot P{\left( {E}/{{E}_{3}} \right)}}{P{\left( {E}_{1} \right)} \cdot P{\left( {E}/{{E}_{1}} \right)} + P{\left( {E}_{2} \right)} \cdot P{\left( {E}/{{E}_{2}} \right)} + P{\left( {E}_{3} \right)} \cdot P{\left( {E}/{{E}_{3}} \right)}}$

$\Rightarrow P{\left( {{E}_{3}}/{E} \right)} = \cfrac{\cfrac{1}{3} \cdot \cfrac{1}{2}}{\cfrac{1}{3} \cdot \cfrac{1}{3} + \cfrac{1}{3} \cdot \cfrac{2}{3} + \cfrac{1}{3} \cdot \cfrac{1}{2}} = \cfrac{3}{2 + 4 + 3} = \cfrac{1}{3}$

Hence the probability that the ball came from urn III is

$\cfrac{1}{3}$ .

How many even numbers are there with three digits such that if $5$ is one of the digits, then $7$ is the next digit?

We have to determine the total number of even numbers formed by using the given condition. So, at units place we can use one of the digits

$0, 2, 4, 6, 8$ . If

$5$ is at ten’s place then, as per the given condition,

$7$ should be at unit’s place. In such a case the number will not be an even number. So,

$5$ cannot be at ten’s and one’s places. Hence,

$5$ can be only at hundred’s place. Now two cases arise.

Case

$I:$ When

$5$ is at hundred’s place. If

$5$ is a hundred’s place, then

$7$ will be at ten’s place. So, unit’s place can be filled in

$5$ ways by using the digits

$0, 2, 4, 6, 8$ . So, total number of even numbers

$= 1\times 1\times 5 = 5$ .

Case

$II:$ When

$5$ in not at hundred’s place. Now, hundred’s place can be filled in

$8$ ways (

$0$ and

$5$ cannot be used at hundred’s place). In ten’s place we can use any one of the digits except

$5$ . So, ten’s place can be filled in

$9$ ways. At unit’s place we have to use one of the even digits

$0, 2, 4, 6, 8$ . So, units place can be filled in

$5$ ways. So, total number of even numbers

$= 8\times 9 \times 5 = 360.$

Hence, the total number of required even numbers

$= 360 + 5 = 365$ .

A card is drawn from a pack of 52 cards. The card is drawn at random; find the probability that it is neither club nor queen?

$\Rightarrow$ Total number of cards from pack

$=52$

$\therefore$

$n(S)=52$

$\Rightarrow$ Total number of club and queen cards

$=13+3=16$

$\Rightarrow$ Let

$E$ be the favorable outcomes of getting neither club nor queen.

$\therefore$

$n(E)=52-16=36$

$\therefore$

$n(E)=36$

$\Rightarrow$

$P(E)=\dfrac{n(E)}{n(S)}=\dfrac{36}{52}=\dfrac{9}{13}$

How many different signals can be made by $5$ flags from $8$ flags of different colours?

The total number of signals is the number of arrangements of

$8$ flags by taking

$5$ flags at a time.

Hence, required number of signals

$=^{8}{{P}_{5}}=\dfrac{8!}{\left(8-5\right)!}=\dfrac{8!}{3!}=\dfrac{8\times 7\times 6\times 5\times 4\times 3!}{3!}=6720$

How many times must a man toss a fair coin so that the probability of having at least one head is more than 90% ?

1320814_1480752_ans_ee8f61f79b8a4e39b9819a90167b7c42.jpeg

If $x$ is a random variable with probability distribution $p\left(x = k\right) = \dfrac{\left(k + 1\right)C}{2^k}, k = 0,1,2,3,............$ , then find $C.$

Given:-

$P{\left( x = k \right)} = \cfrac{\left( k + 1 \right) C}{{2}^{k}}$

To find:-

$C = ?$

$P{\left( x = 0 \right)} = \cfrac{\left( 0 + 1 \right) C}{{2}^{0}} = C$

$P{\left( x = 1 \right)} = \cfrac{\left( 1 + 1 \right) C}{{2}^{1}} = \cfrac{2C}{{2}^{1}}$

$P{\left( x = 2 \right)} = \cfrac{\left( 2 + 1 \right) C}{{2}^{2}} = \cfrac{3C}{{2}^{2}}$

$P{\left( x = 0 \right)} = \cfrac{\left( 3 + 1 \right) C}{{2}^{3}} = \cfrac{4C}{{2}^{3}}$

As we know,

$P{\left( x = 0 \right)} + P{\left( x = 1 \right)} + P{\left( x = 2 \right)} + ..... = 1$

$\therefore 1 = \cfrac{C}{{2}^{0}} + \cfrac{2C}{{2}^{1}} + \cfrac{3 C}{{2}^{2} } + ..... \quad \quad ..... \left( 1 \right)$

Dividing the above equation by

$2$ , we get

$\cfrac{1}{2} = \cfrac{C}{{2}^{1}} + \cfrac{2C}{{2}^{2}} + \cfrac{3 C}{{2}^{3} } + ..... \quad \quad ..... \left( 2 \right)$

Subtracting equation

$\left( 2 \right)$ from

$\left( 1 \right)$ , we have

$1 - \cfrac{1}{2} = \left( \cfrac{C}{{2}^{0}} + \cfrac{2C}{{2}^{1}} + \cfrac{3 C}{{2}^{2} } + ..... \right) - \left( \cfrac{C}{{2}^{1}} + \cfrac{2C}{{2}^{2}} + \cfrac{3 C}{{2}^{3}} + ..... \right)$

$\cfrac{1}{2} = C + \left( \cfrac{2C}{{2}^{1}} - \cfrac{C}{{2}^{1}} \right) + \left( \cfrac{2C}{{2}^{2}} - \cfrac{C}{{2}^{2}} \right) + \left( \cfrac{2C}{{2}^{3}} - \cfrac{C}{{2}^{3}} \right) + .....$

$\Rightarrow \cfrac{1}{2} = C + \cfrac{C}{{2}^{1}} + \cfrac{C}{{2}^{2}} + \cfrac{C}{{2}^{3}} + .....$

$\Rightarrow \cfrac{1}{2} = \cfrac{C}{\left( 1 - \cfrac{1}{2} \right)}$

$\Rightarrow 2C = \cfrac{1}{2}$

$\Rightarrow C = \cfrac{1}{4}$

Hence the value of

$C$ is

$\cfrac{1}{4}$ .

A lot consists of $144$ ball pens of which $20$ are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that
She will not buy it ?

Sample space

$=144$ ball pens
Good pens

$144-20=124$
Defective Pens

$=20$
Probability that she will buy

$=\dfrac{good\ pens}{Sample\ Space}$

$=\dfrac{124}{144}$
Probability that she will not buy

$=\dfrac{defective\ pens}{Sample\ Space}$

$=\dfrac{20}{144}=\dfrac{5}{36}$

How many numbers between $400$ and $1000$ can be formed with the digits $0, 2, 3, 4, 5, 6,$ if no digit is repeated in the same number?

Number between

$400$ and

$1000$ consist of three digits with digit at hundred’s place greater than or equal to

$4$ . Hundred’s place can be filled, by using the digits

$4, 5, 6$ in

$3$ ways. Now, ten’s and unit’s places can be filled by the remaining

$5$ digits in

$^{5}{{P}_{2}}$ ways.

Hence, the required number of numbers

$=3\times^{5}{{P}_{2}}=3\times\dfrac{5!}{3!}=3\times\dfrac{5\times 4\times 3!}{3!}=3\times 20=60$

Five defective bulbs are accidently mixed with 20 good ones. It is not possible to just look at a bulb and tell whether or not it is defective. Find the probability distribution from this lot.

1587437_1511112_ans_9e406480093f40f8a03d1414def84284.jpeg

Two cards are drawn (without replacement) from a well-shuffled deck of $52$ cards. Find the probability distribution table and mean.

Total no. of cards in the pack

$= 52$

The number of kings in the pack of cards

$= 4$

Let

$X$ be the discrete random variable denoting the king when two cards are drawn without replacement.

$X$ takes the values

$0, 1, 2$ .

Now,

Case I:- Probability of no king being drawn

$P{\left( X = 0 \right)} = \cfrac{{^{48}{C}_{2}}}{{^{52}{C}_{2}}} = \cfrac{188}{221}$

Case II:- Probability of one king being drawn

$P{\left( X = 1 \right)} = \cfrac{{^{48}{C}_{1}} \times {^{4}{C}_{1}}}{{^{52}{C}_{2}}} = \cfrac{32}{221}$

Case III:- Probability of two king being drawn

$P{\left( X = 2 \right)} = \cfrac{{^{4}{C}_{2}}}{{^{52}{C}_{2}}} = \cfrac{1}{221}$

Therefore, the probability distribution is-

$X$	$0$	$1$	$2$
$P{\left( x \right)}$	$\cfrac{188}{221}$	$\cfrac{32}{221}$	$\cfrac{1}{221}$

Now, mean of probability distribution-

Mean

$= \sum{{x}_{i} {P}_{i}} \\ = 0 \times \cfrac{188}{221} + 1 \times \cfrac{32}{221} + 2 \times \cfrac{1}{221} \\ = \cfrac{0 + 32 + 2}{221} = \cfrac{34}{221}$

Compute $P ( A | B ) ,$ if $P ( B ) = 0.5$ and $P ( A \cap B ) = 0.32$ .

$P ( B ) = 0.5$

$P ( A \cap B ) = 0.32$

$P ( A | B ) =\dfrac{P(A\cap B)}{P(B)}$

$\\=\dfrac{0.32}{0.5}=0.64$

A $4-$ digit number is formed by the digits $0, 1, 2, 5$ and $8$ without repetition. Find the probability that the number is divisible by $5$ .

Total possibility

$= 5!$

Favorable outcomes

$= 2\times 4!$ (to be divisible by

$5$ unit digit can be filled with only

$0$ or

$5$ , so only two possibilities are there, then the remaining can be filled in

$4, 3$ and

$2$ ways respectively)

So probability

$=\dfrac{2\times 4!}{5!}=\dfrac{2\times 4!}{5\times 4!}=\dfrac{2}{5}$

Two persons $A$ and $B$ appear in an interview. The probability of $A$ 's selection is $\dfrac{1}{5}$ and the probability of $B$ 's selection is $\dfrac{2}{7}$ . What is the probability that only one of them is selected?

Given The probability of

$A$ 's selection is

$\dfrac{1}{5}$

the probability of

$B$ 's selection is

$\dfrac{2}{7}$

The probability of

$A$ 's rejection is

$1-\dfrac{1}{5}=\dfrac 45$

The probability of

$B$ 's rejection is

$1-\dfrac27=\dfrac 57$

Probability that one of them is selected

$=A$ selects and

$B$ rejects

$+ B$ selects and

$A$ rejects

$= \dfrac{1}{5}\times\dfrac{5}{7}+\dfrac{4}{5}\times\dfrac{2}{7}=\dfrac{13}{35}$

If $P{\left( A \right)} = \cfrac{7}{13}$ , $P{\left( B \right)} = \cfrac{9}{13}$ , $P{\left( A \cup B \right)} = \cfrac{12}{13}$ , then find (A|B).

Given:-

$P{\left( A \right)} = \cfrac{7}{13}$

$P{\left( B \right)} = \cfrac{9}{13}$

$P{\left( A \cup B \right)} = \cfrac{12}{13}$

As we know that,

$P{\left( A \cap B \right)} = P{\left( A \right)} + P{\left( B \right)} - P{\left( \cup B \right)}$

$P{\left( A \cap B \right)} = \cfrac{7}{13} + \cfrac{9}{13} - \cfrac{12}{13}$

$\Rightarrow P{\left( A \cap B \right)} = \cfrac{4}{13}$

Now,

$P{\left( A | B \right)} = \cfrac{P{\left( B \cap A \right)}}{P{\left( B \right)}}$

$\therefore P{\left( A | B \right)} = \cfrac{\left( \cfrac{4}{13} \right)}{\left( \cfrac{9}{13} \right)}$

$\Rightarrow P{\left( A | B \right)} = \cfrac{4}{9}$

Hence the correct answer is

$\cfrac{4}{9}$ .

If the height of $300$ students are normally distributed with mean $64.5$ inches and standard deviation $3.3$ inches,find the height below which $99\%$ of the student lie.

Let

$X$ be the random variable the height of a student

$X \sim N (64.5, (3.3)^{2})$

To find the height below which

$99\%$ of the students lie

Let

$X_{1}$ be the height

$P( X < X_{1})=0.99$

Let

$Z$ be the standard normal variable

$Z=\dfrac{X-M}{\sigma }=\dfrac{X-64.5}{3.3}$

When

$X=X_{1} , Z=\dfrac{X_{1}-64.5}{3.3}$

$P(Z < Z_{1})= 0.99 \Rightarrow P(Z < Z_{1})=0.5+0.49$

$\Rightarrow P(0 < Z < Z_{1})= 0.49$

$\therefore Z=2.33$ (from table)

Now

$Z=\dfrac{X_{1}-M}{\sigma }$ where

$M=64.5$

$\Rightarrow 2.33 = \dfrac{X_{1}-64.5}{3.3}\Rightarrow X_{1}-64.5=2.33\times 3.3$

$\Rightarrow X_{1}=64.5+(2.33)(3.3)$

$=64.5+7.7$

$=72.2$ inches

Let X denote the number of hours you study during a randomly selected school day. The probability that X can take the value 'x' has the following from, where 'k' is some unknown constant.
p = (X=x) = $\begin{cases} 0.1,\quad \quad \quad \quad \quad \quad if\quad x=0 \\ kx,\quad \quad \quad \quad \quad \quad \quad if\quad x=1\quad or\quad 2\quad \\ k(5-x),\quad \quad \quad \quad if\quad x=3\quad or\quad 4 \\ 0,\quad \quad \quad \quad \quad \quad \quad \quad otherwise \end{cases}$
(a) find the value of 'k'
(b) what is the probability that you study :
(i) at least two hours?
(ii) exactly two hours?
(iii) at most 2 hours?

Bag $I$ contains $2$ blacks and $8$ red balls, bag $II$ contains $7$ black and $3$ red balls and bag $III$ contains $5$ black and $5v$ red balls. One bage is chosen at random and a ball is drawn from it which is found to bered. Find the probability that the ball is drawn from bag $II$

Let

$R$ be the event of drawing a red ball.

Let

${E}_{1}, {E}_{2}$ and

${E}_{3}$ be the event that bag

$I, II$ or

$III$ is chosen.

Therefore,

$P{\left( {E}_{1} \right)} = \cfrac{1}{3}, \quad P{\left( {E}_{2} \right)} = \cfrac{1}{3}, \quad P{\left( {E}_{3} \right)} = \cfrac{1}{3}$

$P{\left( R|{E}_{1} \right)} = \cfrac{8}{10}$

$P{\left( R|{E}_{2} \right)} = \cfrac{3}{10}$

$P{\left( R|{E}_{3} \right)} = \cfrac{5}{10}$

Now,

Probability that the red ball drawn is from bag $$II$-

$P{\left( {E}_{2}|R \right)} = \cfrac{P{\left( {E}_{2} \right)} \cdot P{\left( R|{E}_{2} \right)}}{P{\left( {E}_{1} \right)} \cdot P{\left( R | {E}_{1} \right)} + P{\left( {E}_{2} \right)} \cdot P{\left( R | {E}_{2} \right)} + P{\left( {E}_{3} \right)} \cdot P{\left( R | {E}_{3} \right)}}$

$\Rightarrow P{\left( {E}_{2}|R \right)} = \cfrac{\cfrac{1}{3} \times \cfrac{3}{10}}{\cfrac{1}{3} \times \cfrac{8}{10} + \cfrac{1}{3} \times \cfrac{3}{10} + \cfrac{1}{3} \times \cfrac{5}{10}} = \cfrac{3}{8 + 3 + 5} = \cfrac{3}{16}$

Hence probability that the ball is drawn from bag

$II$ is

$\cfrac{3}{16}$ .

There are two boxes $I$ and $II$ . Box $I$ contains $3$ red and $6$ black balls. Box $II$ contains $5$ red and 'n' black balls. One of the two boxes, box $I$ and box $II$ is selected at random and a ball is drawn at random. The ball drawn is found to be red. If the probability that this red ball comes out from box $II$ is $\dfrac{3}{5}$ , find the value of 'n'.

From Baye`s Theorm

$P(E_k/A)=\dfrac{P(E_k).P(A/E_k)}{\sum P(E_i).P(A/E_i)}$

$P(2/R)=\dfrac{\dfrac{1}{2}\left(\dfrac{5}{5+n}\right)}{\dfrac{1}{2}\left(\dfrac{3}{9}+\dfrac{5}{5+n}\right)}=\dfrac{3}{5}$

$\implies \dfrac{15}{n+20}=\dfrac{3}{5}\implies n=5$

Mother, father and son line up at random for a family photo. If $A$ and $B$ are two events given by $A$ = Son on one end, $B$ = Father in the middle, find $P(A/B)$ and $P(B/A)$ .

The sample space

$S$ is given by

$S=\left\{MFS, MSF, FSM, FMS, SMF, SFM \right\}$

Clearly,

$A=\left\{MFS, FMS, SMF SFM\right\}, B=\left\{MFS, SFM\right\}$ and so

$A\cap B=\left\{MFS, SFM \right\}$

$\therefore \ P(A/B)=\dfrac {P(A\cap B)}{P(B)}=\dfrac {2/6}{2/6}=1$
and,

$P(B/A)=\dfrac {P(A\cap B)}{P(A)}=\dfrac {2/6}{4/6}=\dfrac {1}{2}$

A black and a red dice are rolled.
a) find the conditional probability of obtaining a sum greater then 9, given that the black die resulted in a 5.

$E:$ Sum gr than

$9=\left\{ (4,6),(6,4),(5,5,),(6,5,),(5,6),(6,6,)\right\}$

$F:$ Black resulted in

$5=\left\{ (5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\right\}$

$E \cap F=$ sum gr than

$9$ and black die resulted in

$5$

$P(E)=\dfrac{6}{36}=\dfrac{1}{6}$

$P(F)=\dfrac{6}{36}=\dfrac{1}{6}$

$P(E \cap F)=\dfrac{2}{36}=\dfrac{1}{18}$

$P(E|F)=\dfrac{P(E \cap F)}{P(F)}=\dfrac{\dfrac{2}{36}}{\dfrac{6}{36}}=\dfrac{1}{3}$

Let $X$ be a random variable which assumes values $x_1, x_2, x_3, x_4$ such that $2P(X = x_1) = 3P(X = x_2) = P(X = x_3) = 5P(X = x_4)$ .
Find the probability distribution of $X$ .

Given

$2 P(X=x_1)=3 P(X=x_2)=P(X=x_3)=5 P(X=x_4)=k$

$\displaystyle\sum^{4}_{i=1} P(X=x_i)=1$

$\implies P(X=x_1)+P(X=x_2)+P(X=x_3)+P(X=x_4)=1$

$\implies \dfrac{k}{2}+\dfrac{k}{3}+k+\dfrac{k}{5}=1\implies \dfrac{15{k}+10{k}+30{k}+6{k}}{30}=1\implies k=\dfrac{30}{61}$

$\implies P(X=x_1)=\dfrac{15}{61},P(X=x_2)=\dfrac{10}{61},P(X=x_3)=\dfrac{30}{61},P(X=x_4)=\dfrac{6}{61}$

In a factory there are three types of Machines ${M}_{1}$ , ${M}_{2}$ and ${M}_{3}$ which produces $25\%, 35\%$ and $40\%$ of the total products respectively. ${M}_{1}$ , ${M}_{2}$ and ${M}_{3}$ produces $5\% , 4\%$ and $2\%$ defective products, respectively. A bolt is drawn at random from the product and is found to be defective.What is the probability of that it is manufactured by machine ${M}_{2}$ ?

----------
1256081_1534901_ans_ddad3788f6b24ccdaf58daa70160ee5a.PNG

Two dice are thrown. Find the probability that the numbers appeared has the $8$ , if it is known that the second die always exhibits $4$ .

Second die always shows

$4$

So the favourable outcome for First die

$=4$

The no.of possible outcomes are

$6$

Probabity is given as

$\dfrac{1}{6}$

The probability distribution function of a random variable $X$ is given by
$x_{i}$ : 0 1 2
$p_{i}$ : $3c^{}$ $4c$ $c-1$
where $c > 0$ . Find $c$ .

In a probability Distribution The Sum of all Probabilities is equal to

$1$

$3c+4c+c-1=1$

$\\8c=2$

$\\c=\dfrac 14$

A random variable $X$ has the following probability distribution :
Values of X: -2 -1 0 1 2 3
P(X): 0.1 k 0.2 2k 0.3 k
Find the value of $a$ , if $k=\dfrac1a$

In a probability Distribution, The Sum of all Probabilities is equal to

$1$

$0.1+k+0.2+2k+0.3+k=1$

$\\0.6+4k=1$

$\\4k=0.4$

$k=\dfrac{0.4}4=\dfrac4{40}=\dfrac1{10}$

Now comparing the given answer with

$\dfrac{1}a$

$\therefore a=10$

Let $X$ be a random variable which assumes values $x_{1},x_{2},x_{3},x_{4}$ such that $2P(X=x_{1})=3P(X=x_{2})=P(X=x_{3})=5P(X=x_{4})$ . Find the probability distribution of $X$ .

1503605_1663453_ans_197d924bf1c24a96a0f99e134c0383ff.jpg

A card is thrown from a pack of $52$ cards so that each card equally likely to be selected. Then find whether the events $A$ and $B$ independent?

$A=$ the card drawn is spade, $B=$ the card drawn in an ace.

In a pack of

$52$ cards there are

$13$ spade,

$4$ ace

$A=$ the card drawn is a spade.

$P\left(A\right)=\dfrac{13}{52}=\dfrac{1}{4}$

$B=$ the card drawn is an ace

$P\left(B\right)=\dfrac{4}{52}=\dfrac{1}{13}$

$A\cap B=$ Card is an ace from spade

$P\left(A\cap B\right)=\dfrac{1}{52}$

$P\left(A\right)P\left(B\right)=\dfrac{1}{4}\times\dfrac{1}{13}=\dfrac{1}{52}$

$P\left(A\right)P\left(B\right)=P\left(A\cap B\right)$

$\therefore\, A$ and

$B$ are independent events.

Arun and Tarun appeared for an interview for two vacancies. The probability of Arun's selection is $\dfrac{1}{4}$ and that of Tarun's rejection is $\dfrac23$ . Find the probability that at least one of them will be selected.

Let

$A$ be the event of Arun Selection.

$T$ be event of Tarun`s Selection.

$P(A)=\dfrac 14$

$P(T)=1-\dfrac 23=\dfrac13$

Required probability

$=1-P(\bar A) P(\bar T)=1-\left (1-\dfrac {1}{4} \right)\left (\dfrac {2}{3} \right)=\dfrac {1}{2}$

A random variable $X$ has the following probability distribution :

Values of X: 0 1 2 3 4 5 6 7 8
P(X): a 3a 5a 7a 9a 11a 13a 15a 17a

Determine the value of $k$ , if $a=\dfrac1k$ .

In a probability Distribution, The Sum of all Probabilities is equal to

$1$

$a+3a+5a+7a+9a+11a+13a+15a+17a=1$

$\\81a=1$

$\\a=\dfrac 1{81}$

Now comparing the given answer with

$\dfrac{1}k$

$\therefore k=81$

The probability distribution function of a random variable $X$ is given by
$x_{i}$ : 0 1 2
$p_{i}$ : $c/2$ $c$ $2c$
where $c > 0$ . Find $c$ .

In a probability Distribution The Sum of all Probabilities is equal to

$1$

$\dfrac c2+c+2c=1$

$\\c+2c+4c=1$

$\\7c=1$

$\\c=\dfrac 17$

Two cards are drawn simultaneously from a well-shuffled deck of 52 cards. Find the probability distribution of the number of successes, when getting a spade is considered a success.

In a deck of

$52$ cards, there are

$13$ spades.

Let

$X$ be the random variable denoting the number of success and success here is getting a spade for an event when two cards are drawn simultaneously.

$\therefore\,$ We can say that the number of successes is equal to some spades obtained in each draw.

$\therefore\,X$ can take values

$0, 1$ or

$2.$

$P\left(X=0\right)=\dfrac{^{39}{{C}_{2}}}{^{52}{{C}_{2}}}=\dfrac{39\times 38}{52\times 51}=\dfrac{19}{34}$

[For selecting

$0$ spades, we removed all

$13$ spades from the deck and selected out of

$39$ ]

$P\left(X=1\right)=\dfrac{^{13}{{C}_{1}}\times^{39}{{C}_{1}}}{^{52}{{C}_{2}}}=\dfrac{13\times 39\times 2}{52\times 51}=\dfrac{13}{34}$

[For selecting

$1$ spade, we need to select and

$1$ out of

$13$ spade and not any other spade]

$P\left(X=2\right)=\dfrac{^{13}{{C}_{2}}}{^{52}{{C}_{2}}}=\dfrac{13\times 12}{52\times 51}=\dfrac{2}{34}$

[For selecting

$2$ spades, we need to select and

$2$ out of

$13$ spades]

Now we have

${p}_{i}$ and

${x}_{i}$ .

$\therefore\,$ The probability distribution for

$X$ is

$P\left(X=0\right)=\dfrac{19}{34},\,\,P\left(X=1\right)=\dfrac{13}{34},\,\,P\left(X=2\right)=\dfrac{2}{34}$

An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.

Let

$X$ denote the number of blue balls in a sample of

$3$ balls drawn from a bag containing

$4$ red and

$3$ blue balls. Then,

$X$ can take values

$0,1,2$ and

$3$

So,

$P(X=0)=P$ (no blue ball)

$=\dfrac {4}{7}\times \dfrac {4}{7}\times \dfrac {4}{7}=\dfrac {64}{343}$

$P(X=1)=P(1\ blue\ ball)$

$=\dfrac {3}{7}\times \dfrac {4}{7}\times \dfrac {4}{7}+ \dfrac {4}{7}\times \dfrac {3}{7}\times \dfrac {4}{7}+ \dfrac {4}{7}\times \dfrac {4}{7}\times \dfrac {3}{7}=\dfrac {144}{343}$

$P(X=2)=P(2\ blue\ balls)$

$=\dfrac {3}{7}\times \dfrac {3}{7}\times \dfrac {4}{7}+ \dfrac {4}{7}\times \dfrac {3}{7}\times \dfrac {3}{7}+ \dfrac {3}{7}\times \dfrac {4}{7}\times \dfrac {3}{7}=\dfrac {108}{343}$

$P(X=3)=P(3\ blue\ ball)$

$=\dfrac {3}{7}\times \dfrac {3}{7}\times \dfrac {3}{7}=\dfrac {27}{343}$

Five defective mangoes are accidently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.

Let

$X$ denote the number of defective mangoes in a sample of

$4$ mangoes drawn from a bag containing and

$15$ good mangoes.Then

$X$ can take the values

$0,1,2,3$ and

$4$

Now

$P\left(X=0\right)=P\left(no\,defective\,mangoes\right)$

$\dfrac{^{15}{{C}_{4}}}{^{20}{{C}_{4}}}$

$=\dfrac{\dfrac{15!}{11!4!}}{\dfrac{20!}{16!4!}}$

$=\dfrac{15\times 14\times 13\times 12}{20\times 19\times 18\times 17}$

$=\dfrac{1365}{4845}=\dfrac{91}{323}$

Now

$P\left(X=1\right)=P\left(1\,defective\,mango\right)$

$=\dfrac{^{5}{{C}_{1}}\times^{15}{{C}_{3}}}{^{20}{{C}_{4}}}$

$=\dfrac{\dfrac{5!}{4!\,1!}\times\dfrac{15!}{12!\,3!}}{\dfrac{20!}{16!\,4!}}$

$=\dfrac{5\times 4\times 15\times 14\times 13}{20\times 19\times 18\times 17}$

$=\dfrac{2275}{4945}=\dfrac{455}{969}$

$P\left(X=2\right)=P\left(2\,defective\,mangoes\right)$

$=\dfrac{^{5}{{C}_{2}}\times\,^{15}{{C}_{2}}}{^{20}{{C}_{4}}}$

$=\dfrac{\dfrac{5!}{3!\,2!}\times\dfrac{15!}{13!\,2!}}{\dfrac{20!}{16!\,4!}}$

$=\dfrac{5\times 15\times 14\times 4\times 3\times 2}{20\times 19\times 18\times 17}$

$=\dfrac{1050}{4845}=\dfrac{70}{323}$

$P\left(X=3\right)=P\left(3\,defective\,mangoes\right)$

$=\dfrac{^{5}{{C}_{3}}\times\,^{15}{{C}_{1}}}{^{20}{{C}_{4}}}$

$=\dfrac{\dfrac{5!}{3!\,2!}\times\dfrac{15!}{14!}}{\dfrac{20!}{16!\,4!}}$

$=\dfrac{\dfrac{5\times 4\times 15}{2}}{\dfrac{20\times 19\times 18\times 17}{4\times 3\times 2}}$

$=\dfrac{150}{4845}=\dfrac{10}{323}$

$P\left(X=4\right)=P\left(4\,defective\,mangoes\right)$

$=\dfrac{^{50}{{C}_{4}}}{^{20}{{C}_{4}}}$

$=\dfrac{\dfrac{50!}{46!4!}}{\dfrac{20!}{16!4!}}$

$=\dfrac{\dfrac{50\times 49\times 48\times 47}{4\times 3\times 2}}{\dfrac{20\times 19\times 18\times 17}{4\times 3\times 2}}$

$=\dfrac{5}{4845}$

$=\dfrac{1}{969}$

Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.

Let

$X$ denote the number of white balls in a sample of

$3$ balls drawn from a bag containing

$4$ white and

$6$ red balls.

Then,

$X$ can take the values

$0$ white ball

$1$ or

$2$ or

$3$ white ball.

$\Rightarrow$

$P(X=0)=\dfrac{^6C_3}{^{10}C_3}$

$=\dfrac{\dfrac{6!}{3!3!}}{\dfrac{10!}{3!7!}}$

$=\dfrac{6\times 5\times \times 4}{10\times 9\times 8}$

$=\dfrac{120}{720}$

$=\dfrac{1}{6}$

$\Rightarrow$

$P(X=1)=\dfrac{^4C_1\times\, ^6C_2}{^{10}C_3}$

$=\dfrac{\dfrac{4!}{1!3!}\times \dfrac{6!}{2!4!}}{\dfrac{10!}{3!7!}}$

$=\dfrac{6\times 5\times 4\times 3}{10\times 9\times 8}$

$=\dfrac{360}{720}$

$=\dfrac{1}{2}$

$\Rightarrow$

$P(X=2)=\dfrac{^4C_2\times \,^6C_1}{^{10}C_3}$

$=\dfrac{\dfrac{4!}{2!2!}\times \dfrac{6!}{1!5!}}{\dfrac{10!}{3!7!}}$

$=\dfrac{36}{120}$

$=\dfrac{3}{10}$

$\Rightarrow$

$P(X=3)=\dfrac{^4C_3}{^{10}C_3}$

$=\dfrac{\dfrac{4!}{3!1!}}{\dfrac{10!}{3!7!}}$

$=\dfrac{4}{120}$

$=\dfrac{1}{30}$

$X$	$0$	$1$	$2$	$3$
$P(X)$	$\dfrac{1}{6}$	$\dfrac{1}{2}$	$\dfrac{3}{10}$	$\dfrac{1}{30}$

A bag contains 2 white, 3 red and 4 blue balls. Two balls are drawn at random from the bag. If X denotes the number of white balls among the two balls drawn, describe the probability distribution of X.

1503675_1663651_ans_aef28cd9e4ae4c8fbeba63fa14ca8596.jpg

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of kings.

$P(X=x)=\,^nC_xq^{n-x}p^x$

Here,

$n=$ number of cards drawn

$=2$

$p=$ Probability of getting king

$=\dfrac{4}{52}=\dfrac{1}{13}$

$q=1-p=1-\dfrac{1}{13}=\dfrac{12}{13}$

Hence,

$\Rightarrow$

$P(X=x)=\,^2C_x\left(\dfrac{1}{13}\right)^x\left(\dfrac{12}{13}\right)^{2-x}$ ----- ( 1 )

Since

$2$ cards are drawn,

We can get

$0$ ace,

$1$ king or

$2$ king.

So,

$X$ can be

$0,1,2$

Putting values in ( 1 )

$P(X=0)=\,^2C_0\left(\dfrac{1}{13}\right)^0\left(\dfrac{12}{13}\right)^{2-0}=1\times 1\times \left(\dfrac{12}{13}\right)^2=\dfrac{144}{169}$

$P(X=1)=\,^2C_1\left(\dfrac{1}{13}\right)^1\left(\dfrac{12}{13}\right)^{2-1}=2\times \dfrac{1}{13}\times \dfrac{12}{13}=\dfrac{24}{169}$

$P(X=2)=\,^2C_2\left(\dfrac{1}{13}\right)^2\left(\dfrac{12}{13}\right)^{2-2}=1\times \left(\dfrac{1}{13}\right)^2=\dfrac{1}{169}$

$X$	$0$	$1$	$2$
$P(X)$	$\dfrac{144}{169}$	$\dfrac{24}{169}$	$\dfrac{1}{169}$

The probability distribution of a random variable X is given below:
X: 0 1 2 3
P(X): k $\dfrac{k}{2}$ $\dfrac{k}{4}$ $\dfrac{k}{8}$
Determine the value of X, if $k=\dfrac8X$

In binomial distribution sum of all probabilities is equal to

$1$

$\implies k+\dfrac k2+\dfrac k4+\dfrac k8=1$

$\\8k+4k+2k+k=8$

$\\15k=8$

$\\k=\dfrac 8{15}$

Now comparing the given answer with

$\dfrac{8}X$

$\therefore X=15$

Two balls are drawn at random with replacement from a box containing $10$ black and $8$ red balls. Find the probability that both the balls are red.

Total number of balls

$= 18$

Number of red balls

$= 8$

Number of black balls

$= 10$

Probability of getting a red ball in the first draw

$=\dfrac{8}{18}=\dfrac{4}{9}$

The ball is replaced after the first draw.

$\therefore\,$ Probability of getting a red ball in the second draw

$=\dfrac{8}{18}=\dfrac{4}{9}$

$\therefore\,$ Probability of getting a both the balls red

$=\dfrac{4}{9}\times\dfrac{4}{9}=\dfrac{16}{81}$

From a lot of 10 bulbs, which includes 3 detectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.

There are

$3$ defective bulbs and

$7$ non-defective bulbs.

Let

$X$ denote the random variable of the number of defective bulb.

Then

$X$ can take values

$0, 1, 2$ since bulbs are replaced.

$p=P\left(D\right)=\dfrac{3}{10}$ and

$q=P\left(\bar{D}\right)=1-\dfrac{3}{10}=\dfrac{7}{10}$

We have

$P\left(X=0\right)=\dfrac{^{7}{{C}_{2}}\times^{3}{{C}_{0}}}{^{10}{{C}_{2}}}=\dfrac{7\times 6}{10\times 9}=\dfrac{7}{15}$

$P\left(X=1\right)=\dfrac{^{7}{{C}_{1}}\times^{3}{{C}_{1}}}{^{10}{{C}_{2}}}=\dfrac{7\times 3\times 2}{10\times 9}=\dfrac{7}{15}$

$P\left(X=2\right)=\dfrac{^{7}{{C}_{0}}\times^{3}{{C}_{2}}}{^{10}{{C}_{2}}}=\dfrac{1\times 3\times 2}{10\times 9}=\dfrac{1}{15}$

$\therefore\,$ Required probability distribution is

$P\left(X=0\right)=\dfrac{7}{15}$

$P\left(X=1\right)=\dfrac{7}{15}$

$P\left(X=2\right)=\dfrac{1}{15}$

Five defective bolts are accidentally mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.

$X$ represents the number of defective bolts drawn.

$\therefore\, X$ can take values

$0,1,2$ or

$3$

$\because\,$ there are total

$25$ bolts

$\left(20\,\, good + 5\,\, defectives\right)$ bolts.

$n\left(S\right)=$ total possible ways of selecting

$5$ bolts

$=^{25}{{C}_{4}}$

$P\left(X = 0\right) = P\left(selecting\, no\, defective\, bolt\right) =\dfrac{^{20}{{C}_{4}}}{^{25}{{C}_{4}}}=\dfrac{20\times 19\times 18\times 17}{25\times 24\times 23\times 22}=\dfrac{969}{2530}$

$P\left(X = 1\right) = P\left(selecting \,1\, defective\, bolt \,and \,3 \,good \,bolts\right)$

$=\dfrac{^{5}{{C}_{1}}\times\,^{20}{{C}_{3}}}{^{25}{{C}_{4}}}$

$=\dfrac{\dfrac{5!}{4!}\times\dfrac{20!}{17!\,3!}}{\dfrac{25!}{21!\,4!}}$

$=\dfrac{5\times 4\times\,20\times 19\times 18}{25\times 24\times 23\times 22}$

$=\dfrac{114}{253}$

$P\left(X = 2\right) = P\left(selecting \,2\, defective\, bolt \,and \,2 \,good \,bolts\right)$

$=\dfrac{^{5}{{C}_{2}}\times\,^{20}{{C}_{2}}}{^{25}{{C}_{4}}}$

$=\dfrac{\dfrac{5!}{3!\,2!}\times \dfrac{20!}{18!\,2!}}{\dfrac{25!}{21!\,4!}}$

$=\dfrac{\dfrac{5\times 4}{2}\times \dfrac{20\times 19}{2}}{\dfrac{25\times 24\times 23\times 22}{4\times 3\times 2}}=\dfrac{38}{253}$

$P\left(X = 3\right) = P\left(selecting \,3\, defective\, bolt \,and \,1 \,good \,bolt\right)$

$=\dfrac{^{5}{{C}_{3}}\times\,^{20}{{C}_{1}}}{^{25}{{C}_{4}}}$

$=\dfrac{\dfrac{5!}{3!\,2!}\times \dfrac{20!}{19!\,1!}}{\dfrac{25!}{21!\,4!}}$

$=\dfrac{\dfrac{5\times 4}{2}\times 20}{\dfrac{25\times 14\times 23\times 22}{4\times 3\times 2}}$

$=\dfrac{4}{253}$

$P\left(X = 4\right) = P\left(selecting \,4\, defective\, bolt \,and \,no \,good \,bolt\right)$

$=\dfrac{^{5}{{C}_{4}}}{^{25}{{C}_{4}}}$

$=\dfrac{5\times 4\times 3\times 2}{25\times 24\times 23\times 22}=\dfrac{1}{2530}$

The probability distribution of a random variable X is given below:
X: 0 1 2 3
P(X): k $\dfrac{k}{2}$ $\dfrac{k}{4}$ $\dfrac{k}{8}$
Determine $P(X\leq 2)+P(X > 2)$

First we are going to fing value of

$k$ ,

We know,

$P(X)=1$

$\Rightarrow$

$k+\dfrac{k}{2}+\dfrac{k}{4}+\dfrac{k}{8}=1$

$\Rightarrow$

$\dfrac{8k+4k+2k+k}{8}=1$

$\Rightarrow$

$\dfrac{15k}{8}=1$

$\therefore$

$k=\dfrac{8}{15}$

$\Rightarrow$ As,

$P(X\le 2)=1-P(X=3)$

$=1-\dfrac{k}{8}$

$=1-\dfrac{8}{15\times 8}$ [ Substituting value of

$k$ ]

$=1-\dfrac{1}{15}$

$=\dfrac{14}{15}$

$\Rightarrow$ Also,

$P(X>2)=P(X=3)$

$=\dfrac{k}{8}$

$=\dfrac{8}{15\times 8}$ [ Substituting value of

$k$ ]

$=\dfrac{1}{15}$

$\Rightarrow$

$P(X\le 2)+P(X\ge 2)=\dfrac{14}{15}+\dfrac{1}{15}$

$=\dfrac{15}{15}$

$=1$

A fair coin is tossed four times. Let X denote the longest string of heads occurring. Find the probability distribution, mean and variance of X.

1621889_1664201_ans_0fd65bbf9ec94341aa70558b5fc10cde.jpeg

The probability distribution of a random variable X is given below:
X: 0 1 2 3
P(X): k $\dfrac{k}{2}$ $\dfrac{k}{4}$ $\dfrac{k}{8}$
Determine $P(X\leq 2)$ and $P(X > 2)$

In binomial distribution sum of all probabilities is equal to

$1$

$\implies k+\dfrac k2+\dfrac k4+\dfrac k8=1$

$\\8k+4k+2k+k=8$

$\\15k=8$

$\\k=\dfrac 8{15}$

$P(X\le 2)=P(X=0)+P(X=1)+P(X=2)$

$\\k+\dfrac k2+\dfrac k4=\dfrac {7k}4=\dfrac {14}{15}$ .........

$\left (\because k=\dfrac8{15} \right )$

$P(X>2)=\dfrac k8=\dfrac 1{15}$ ...........

$\left (\because k=\dfrac8{15} \right )$

A box contains 13 bulbs, out of which 5 are defective. 3 bulbs are randomly drawn, one by one without replacement, from the box. Find the probability distribution of the number of defective bulbs.

Let

$X$ denote the number of defective bulbs in a sample of

$3$ bulbs drawn from a bag containing

$13$ bulbs.

$5$ bulbs in the bag turn out to be defective.

Then,

$X$ can take the values

$0,1,2$ and

$3$ .

$P(X=0)=P($ No defective bulb

$)=\dfrac{^{5}C_0\times\,^8C_3}{^{13}C_3}$

$=\dfrac{8\times 7\times 6}{13\times 12\times 11}$

$=\dfrac{28}{143}$

$P(X=1)=P($ One defective bulb

$)=\dfrac{^5C_1\times\,^8C_2}{^{13}C_3}$

$=\dfrac{5\times\dfrac{8\times 7}{2\times 1}}{\dfrac{13\times 12\times 11}{3\times 2\times 1}}$

$=\dfrac{70}{143}$

$P(X=2)=P($ Two defective bulb

$)=\dfrac{^5C_2\times\,^8C_1}{^{13}C_3}$

$=\dfrac{\dfrac{5\times 4}{2\times 1}\times 8}{\dfrac{13\times 12\times 11}{3\times 2\times 1}}$

$=\dfrac{40}{143}$

$P(X=3)=P($ Three defective bulb

$)=\dfrac{^5C_3\times\,^8C_0}{^{13}C_3}$

$=\dfrac{5\times 4\times 3}{13\times 12\times 11}$

$=\dfrac{5}{143}$

$\Rightarrow$ The probability distribution of

$X$ is given by

$X$	$P(X)$
$0$	$\dfrac{28}{143}$
$1$	$\dfrac{70}{143}$
$2$	$\dfrac{40}{143}$
$3$	$\dfrac{5}{143}$

A card is thrown from a pack of $52$ cards so that each cards equally likely to be selected. Determine whether in the following case are the events $A$ and $B$ independent.
$A=$ the card is drawn is a spade, $B=$ the card is drawn in an ace.

$A=$ the card drawn is spade,

$B=$ the card drawn in an ace.

We have,

$P(A)=\dfrac {13}{52}=\dfrac {1}{4}$

$P(B)=\dfrac {4}{52}=\dfrac {1}{13}$

$P(A\cap B)=\dfrac {1}{52}$

$P(A\cap B)=P(A).P(B)$

Hence Independant.

In a factory machine $A$ produce $30\%$ of the total output, machine $B$ produced $25\%$ and the machine $C$ produced the remaining output. If defective items produced by machine $A, B$ and $C$ are $1\% ,\ 1.2\% ,\ 2\%$ respectively. Three machine working together produced $10000$ item in a day. An item is drawn at random from a day's output and found to be defective.
Find the probability that it was produced by machine $B$ ?

Let

$E_1, E_2, E_3$ be events as define below:

$E_1=$ Item is produced by machine

$A$ ,

$E_2=$ Item is produced by machine

$B$

$E_3=$ Item is produced by machine

$C$ and

$A=$ That event that the item is defective.

It is given that,

$P(E_1)=\dfrac {30}{100}, P(E_2)=\dfrac {25}{100}, P(E_3)=\dfrac {45}{100}$ ,

$P(A/E_1)=\dfrac {1}{100}$

$P(A/E_2)=\dfrac {1.2}{100}$

$P(A/E_3)=\dfrac {2}{100}$

Required probaility

$=P(E_2/A)=\dfrac {P(E_2) P(A/E_2)}{P(E_1) P(A/E_1)+P(E_2) P(A/E_2)+P(E_3) P(A/E_3)}$

$\dfrac {\dfrac {25}{100}\times \dfrac {1.2}{100}}{\dfrac {30}{100}\times \dfrac 1{100}+\dfrac {25}{100}\times \dfrac {1.2}{100}+\dfrac {2}{100}\times \dfrac {45}{100}}$

$\dfrac {30}{30+30+90}=\dfrac 15$

A discrete random variable X has the probability distribution given below:
X: 0.5 1 1.5 2
P(X): k $2k^{}$ $3k^{}$ k
Find the value of k.

In probability distribution sum of all Probabilities is equal to

$1$

$\implies k+2k+3k+k=1$

$\implies 7k=1$

$\therefore$

$\\k=\dfrac 17$

A discrete random variable X has the probability distribution given below:
X: 0.5 1 1.5 2
P(X): k $k^{}$ $2k^{}$ k
Determine the value of $k$

In probability distribution sum of all Probabilities is equal to

$1$

$\implies k+k+2k+k=1$

$\therefore 5k=1$

$\therefore k=\dfrac 15$

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence find the mean of the distribution.

Three Cards are drawn from a pack of

$52$ cards.

Let

$X$ be the number of red cards drawn.

In pack of

$52$ cards there are

$26$ red cards and

$26$ black cards.

We can draw

$3$ cards out of

$52$ in

$^{52}C_3$ ways.

$P(X=0)=P($ No red cards drawn

$)=\dfrac{^{26}C_0\times\, ^{26}C_3}{^{52}C_3}$

$=\dfrac{26\times 25\times 24}{52\times 51\times 50}$

$=\dfrac{2}{17}$

$P(X=1)=P($ One red cared drawn

$)=\dfrac{^{26}C_1\times ^{26}C_2}{^{52}C_3}$

$=\dfrac{26\times \dfrac{26\times 25}{2\times 1}}{\dfrac{52\times 51\times 50}{3\times 2\times 1}}$

$=\dfrac{13}{34}$

$P(X=2)=P($ Two red card drawn

$)=\dfrac{^{26}C_2\times\,^{26}C_1}{^{52}C_3}$

$=\dfrac{\dfrac{26\times 25}{2\times 1}\times\dfrac{26}{1}}{\dfrac{52\times 51\times 50}{3\times 2\times 1}}$

$=\dfrac{13}{34}$

$P(X=3)=P($ Three red card drawn

$)=\dfrac{^{26}C_3\times\,^{26}C_0}{^{52}C_3}$

$=\dfrac{26\times 25\times 24}{52\times 51\times 50}$

$=\dfrac{2}{17}$

$x_i$	$P(x_i)$	$x.P(x_i)$
$0$	$\dfrac{2}{17}$	$0$
$1$	$\dfrac{13}{34}$	$\dfrac{13}{34}$
$2$	$\dfrac{13}{34}$	$\dfrac{26}{34}$
$3$	$\dfrac{2}{17}$	$\dfrac{6}{17}$
		$\sum x_i.P(x_i)=\dfrac{3}{2}$

$\Rightarrow$ Mean of the distribution

$=\sum x_iP(x_i)=\dfrac{3}{2}=1.5$

In a factory machine $A$ produce $30\%$ of the total output, machine $B$ produced $25\%$ and the machine $C$ produced the remaining output. If defective items produced by machine $A, B$ and $C$ are $1\% ,\ 12\% ,\ 2\%$ respectively. Three machine working together produced $10000$ item in a day. An item is drawn at random from a day's output and found to be defective. Find the probability that it was produced by machine $B$ .

Let

$D,E_1,E_2$ and

$E_3$ denote the events that the item is defective, machine

$A$ is chosen, machine

$B$ is chosen and machine

$C$ is chosen, respectively.

$\Rightarrow$

$P(E_1)=30\%=\dfrac{30}{100}=0.3$

$\Rightarrow$

$P(E_2)=25\%=\dfrac{25}{100}=0.25$

$\Rightarrow$

$P(E_3)=100\%-(30+25)\%=45\%=\dfrac{45}{100}=0.45$

Now,

$\Rightarrow$

$P\left(\dfrac{D}{E_1}\right)=1\%=\dfrac{1}{100}=0.01$

$\Rightarrow$

$P\left(\dfrac{D}{E_2}\right)=1.2\%=\dfrac{1.2}{100}=0.012$

$\Rightarrow$

$P\left(\dfrac{D}{E_3}\right)=2\%=\dfrac{2}{100}=0.02$

Probability of being produced by

$B,$

$\Rightarrow$

$P\left(\dfrac{E_2}{D}\right)=\dfrac{P(E_2)\times P\left(\dfrac{D}{E_2}\right)}{P(E_2)\times P\left(\dfrac{D}{E_2}\right)+P(E_1)\times P\left(\dfrac{D}{E_1}\right)+P(E_3)\times p\left(\dfrac{D}{E_2}\right)}$

$=\dfrac{0.25\times 0.012}{(0.25\times 0.012)+(0.3\times 0.01)+(0.45\times 0.02)}$

$=\dfrac{0.003}{0.003+0.003+0.009}$

$=\dfrac{3\times 10^{-3}}{(3+3+9)\times 10^{-3}}$

$=\dfrac{3}{15}$

$=0.2$

Two coins are tossed simultaneously 500 times with the following frequencies of different outcomes:
Two heads: 95 times
One tail: 290 times
No head: 115 times
Find the probability of occurrence of each of these events.

$Given$

$Total\quad number\quad of\quad times\quad 2\quad coins\quad are\quad tossed\quad =\quad 500$

$No\quad of\quad times\quad 2\quad Heads\quad comes\quad up\quad =\quad 95$

$No\quad of\quad times\quad 1\quad Tail\quad comes\quad up\quad =\quad 290$

$No\quad of\quad times\quad No\quad Head\quad comes\quad up\quad =\quad 115$

$Probability\quad of\quad 2\quad Heads\quad coming\quad up\quad =\quad \dfrac { Frequency\quad of\quad 2\quad Heads\quad coming\quad up }{ Total\quad number\quad of\quad times\quad 2\quad coins\quad are\quad tossed }$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\dfrac { 95 }{ 500 } \quad =\quad 0.19$

$Probability\quad of\quad 1\quad Tail\quad coming\quad up\quad =\quad \dfrac { Frequency\quad of\quad 1\quad Tail\quad coming\quad up }{ Total\quad number\quad of\quad times\quad 2\quad coins\quad are\quad tossed }$

$Probability\quad of\quad No\quad Heads\quad coming\quad up\quad =\quad \dfrac { Frequency\quad of\quad No\quad Heads\quad coming\quad up }{ Total\quad number\quad of\quad times\quad 2\quad coins\quad are\quad tossed }$

A die is thrown. Find the probability of getting:
a prime number

1872886_1673946_ans_2de25fb44c234d21a5f076343e73d56c.jpg

If $A$ and $B$ are two event such that

$P(A)=\dfrac {6}{11},\ P(B)=\dfrac {5}{11}$ and $P(A\cup B)=\dfrac {7}{11}$ , find $P(A\cap B), \ P(A/B),\ P(B/A)$ .

$P(A)=\dfrac {6}{11}$

$P(B)=\dfrac {5}{11}$

$P(A\cup B)=\dfrac {7}{11}$

$P(A\cap B)=P(A)+P(B)-P(A\cup B)$

$\dfrac 6{11}+\dfrac 5{11}-\dfrac 7{11}$

$1-\dfrac 7{11}=\dfrac 4{11}$

$P(A/B)=\dfrac {P(A\cap B)}{P(B)}$

$\dfrac {\dfrac 4{11}}{\dfrac 5{11}}$

$\dfrac 45$

$P(B/A)=\dfrac {P(A\cap B)}{P(A)}$

$\dfrac {\dfrac 4{11}}{\dfrac 6{11}}$

$\dfrac 23$ .

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:
Outcome: No head One head Two heads Three heads
Frequency 14 38 36 12
If the three coins are simultaneously tossed again, compute the probability of: 2 heads coming up

$Given$

$Number\quad of\quad times\quad 3\quad coins\quad are\quad tossed\quad simultaneously\quad =\quad 100$

$Number\quad of\quad outcomes\quad when\quad no\quad head\quad appears\quad =\quad 14$

$Number\quad of\quad outcomes\quad when\quad 1\quad head\quad appears\quad =\quad 38$

$Number\quad of\quad outcomes\quad when\quad 2\quad head\quad appears\quad =\quad 36$

$Number\quad of\quad outcomes\quad when\quad 3\quad head\quad appears\quad =\quad 12$

$Total\quad number\quad of\quad outcomes\quad =\quad 14+38+36+12\quad =\quad 100$

$Probability\quad of\quad 2\quad heads\quad coming\quad up,\quad if\quad 3\quad coins\quad are$

$simultaneously\quad tossed\quad again$

$=\dfrac { Number\quad of\quad outcomes\quad when\quad 2\quad head\quad appears }{ Total\quad number\quad of\quad outcomes }$

$=\dfrac { 36 }{ 100 } \quad =\quad 0.36$

Three coins are tossed simultaneously 100 times with the following frequencies of different outcomes:
Outcome: No head One head Two heads Three heads
Frequency 14 38 36 12
If the three coins are simultaneously tossed again, compute the probability of: 3 heads coming up

$Given$

$Number\quad of\quad times\quad 3\quad coins\quad are\quad tossed\quad simultaneously\quad =\quad 100$

$Number\quad of\quad outcomes\quad when\quad no\quad head\quad appears\quad =\quad 14$

$Number\quad of\quad outcomes\quad when\quad 1\quad head\quad appears\quad =\quad 38$

$Number\quad of\quad outcomes\quad when\quad 2\quad head\quad appears\quad =\quad 36$

$Number\quad of\quad outcomes\quad when\quad 3\quad head\quad appears\quad =\quad 12$

$Total\quad number\quad of\quad outcomes\quad =\quad 14+38+36+12\quad =\quad 100$

$Probability\quad of\quad 3\quad heads\quad coming\quad up,\quad if\quad 3\quad coins\quad are$

$simultaneously\quad tossed\quad again$

$=\dfrac { Number\quad of\quad outcomes\quad when\quad 3\quad head\quad appears }{ Total\quad number\quad of\quad outcomes }$

$=\dfrac { 12 }{ 100 } \quad =\quad 0.12$

Three coins are tossed together. Find the probability of getting :
at least one head and one tail

Total number of outcomes

$=(HHH),(HHT),(HTH),(THH),(TTH),(TTT),(THT),(HTT)=8$
Favourable outcomes

$=8$
P(at least one head and one tail)

$=8/8=1$

A coin is tossed 1000 times with the following frequencies:
Head: 455, Tail: 545
Compute the probability for each event.

$Given$

$Total\quad outcomes\quad =\quad 1000$

$Frequency\quad of\quad Head\quad =\quad 455$

$Frequency\quad of\quad Tail\quad =\quad 545$

$A)\quad Probability\quad of\quad Head\quad =\quad \dfrac { Number\quad of\quad times\quad Head\quad appeared }{ Total\quad number\quad of\quad times\quad coin\quad tossed }$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$

$=\quad \dfrac { 455 }{ 1000 } \quad =\quad 0.455$

$B)\quad Probability\quad of\quad Tail\quad =\quad \dfrac { Number\quad of\quad times\quad Tail\quad appeared }{ Total\quad number\quad of\quad times\quad coin\quad tossed }$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$

$=\quad \dfrac { 545 }{ 1000 } \quad =\quad 0.545$

Enter 1 if true else enter 0.
Three coins are tossed together. The probability of getting at least two heads is $\dfrac{1}{2}$

Three coins are tossed simultaneously.

When three coins are tossed then the possible outcome will be:

$TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.$

Total number of outcomes

$= 8.$

Possible outcomes for getting at least two heads

$= HHT, HTH, HHH ,THH$

No. of possible outcomes for getting at least two heads

$=4$

$P(E)=\dfrac{No.\,of\,outcomes}{Total\,no.\,of\,outcomes}$

$P$ (Getting at least two heads)

$=\dfrac{4}{8}=\dfrac{1}{2}$

Enter 1 if it is true else enter 0
Three coins are tossed together. The probability of getting no tails is $\dfrac{1}{8}$ .

Three coins are tossed simultaneously.

When three coins are tossed then the possible outcome will be:

$TTT, THT, TTH, THH. HTT, HHT, HTH, HHH.$

Total number of outcomes

$= 8.$

Possible outcomes for getting no tails

$= HHH$

No. of possible outcomes for getting no tails

$=1$

$P(E)=\dfrac{No.\,of\,outcomes}{Total\,no.\,of\,outcomes}$

$P$ (Getting no tails )

$=\dfrac{1}{8}$

A bag contains $3$ white and $2$ black balls and another bag contains $2$ white and $4$ black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.

Consider the following events:

$E_1=$ Selecting first bag.

$E_2=$ Selecting second bag ,

$A=$ ball draw is white.

Then

$P(E_1)=P(E_2)=\dfrac{1}{2}, P(A/E_1)=\dfrac{3}{5}, P(A/E_2)=\dfrac{2}{6}$

Required probability

$P(A)=P(E_1)P(A/E_1)+P(E_2)P(A/E_2)$

$=\dfrac {1}{2}\times \dfrac {3}{5}+\dfrac {1}{2}\times \dfrac {2}{6}=\dfrac {7}{15}$

If a number $x$ is chosen at random from the numbers $-3,-2,-1,0,1,2,3$ . What is probability that $x^2\le 4$ ?

Numbers

$= -3 , -2, -1 , 0 , 1, 2, 3$

Square of the numbers

$= 9, 4, 1, 0, 1, 4, 9$

Total number for which

$x^2 \le 4$ are

$= 5$

Probability

$= \dfrac{Total \, numbers \, satisfying \ the \ condition}{Total \ numbers}$

$= \dfrac{5}{7}$

A bag contains $5$ red balls and some blue balls. If the probability of drawing a blue ball at random from the bag is three times that of a red ball, find the number of blue balls in the bag.

Red ball

$=5$

Let blue ball be

$'x'$

total blue

$5+x$

Probability of drawing blue ball

$P(b)=\dfrac{x}{5+x}$

Probability of drawing red ball

$P(r)=\dfrac{5}{5+x}$

given

$P(b)=3P(r)$

$\dfrac{x}{(5+x)}=\dfrac{3\times 5}{(5+x)}$

$\boxed{x=15}$

Find the probability that 5 Sundays occurs in the month of November of a randomly selected year.

We know that

November has 30 days, which means 4 weeks and 2 days.

Now, 4 weeks will contain 4 Sunday.

The remaining 2 days may be :

1. Sunday and Monday

2. Monday and Tuesday

3. Tuesday and Wednesday

4. Wednesday and Thursday

5. Thursday and Friday

6. Friday and Saturday

7. Saturday and Sunday

Total number of possible outcomes = 7

Now, favourable outcomes are : Sunday and Monday, Saturday and Sunday

Number of favourable outcomes = 2

Required probability (P)= Number of favourable outcomes /Total number of possible outcomes

P = 2/7

Enter 1 if it is true else enter 0.
A bag contains $3$ red balls, $5$ black balls and $4$ white balls. A ball is drawn at random from the bag. The probability that the ball drawn is red is $\dfrac{1}{4}$ .

Number of red balls

$=3$

Number of black balls

$=5$

Number of white balls

$=4$

Total number of balls

$=3+5+4=12$

$\therefore$ the total number of cases

$=12$

SInce there are

$3$ red balls, the number of favourable outcomes is

$3$

Hence, P(a red ball)

$=\dfrac{\text{Number of favourable cases}}{\text{Total number of cases}}=\dfrac{3}{12}=\dfrac{1}{4}$

20% of the bulbs produced by a machine are defective.find the probability distribution of the numbers of defective bulbs in sample of 4 bulbs chosen at random.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{C_4^{80}\times C_0^{20}}{C_4^{100}}$	$\frac{C_3^{80}\times C_1^{20}}{C_4^{100}}$	$\frac{C_2^{80}\times C_2^{20}}{C_4^{100}}$	$\frac{C_1^{80}\times C_3^{20}}{C_4^{100}}$	$\frac{C_0^{80}\times C_4^{20}}{C_4^{100}}$

$let\,total\,bulbs=100$

$defective=20%$

$non-defective=80%$

$Let\text{ the number of defective bulbs be x}$

$\text{i) when no defective bulb }\left( x=0 \right)$

$P\left( x=0 \right)=\frac{C_{4}^{80}\times C_{0}^{20}}{C_{4}^{100}}$

$ii)\text{ when 1 defective bulb }\left( x=1 \right)$

$P\left( x=1 \right)=\frac{C_{3}^{80}\times C_{1}^{20}}{C_{4}^{100}}$

$\text{iii) when 2 defective bulb }\left( x=2 \right)$

$P\left( x=2 \right)=\frac{C_{2}^{80}\times C_{2}^{20}}{C_{4}^{100}}$

$\text{iv) when 3 defective bulb }\left( x=3 \right)$

$P\left( x=3 \right)=\frac{C_{1}^{80}\times C_{3}^{20}}{C_{4}^{100}}$

$\text{iv) when 4 defective bulb }\left( x=4 \right)$

$P\left( x=4 \right)=\frac{C_{0}^{80}\times C_{4}^{20}}{C_{4}^{100}}$

In a shop $X$ , 30 tins of ghee of Type A and 40 tins of ghee of Type B which look alike, are kept for sale while in shop $Y$ , similar 50 tins of ghee of Type A and 60 tins of ghee of Type B are there. One tin of ghee is purchased from one of the randomly selected shops and is found to be of Type B. Find the probability that it is purchased from shop $Y$ .

$E_1 =$ event of selecting shop

$X$

$E_2 =$ event of selecting shop

$Y$

$E =$ event of selecting type

$B$ ghee

Now,

$P(E_1) = P(E_2) = \dfrac{1}{2}$

$P(E/E_1) = \dfrac{40}{70} = \dfrac{4}{7}$ and

$P(E/E_2) = \dfrac{60}{110} = \dfrac{6}{11}$

Now,

$P(E_2/E) = \dfrac{P(E_2) . P(E/E_2)}{P(E_1) . P(E/E_1) + P(E_2) . P(E/E_2)}$

$= \dfrac{\left(\dfrac{1}{2}\right) \times \left(\dfrac{6}{11}\right)}{\left(\dfrac{1}{2} \right) \left(\dfrac{4}{7} \right) + \left(\dfrac{1}{2} \right) \left(\dfrac{6}{11} \right)}$

$= \dfrac{42}{86} = \dfrac{21}{43}$

1503809_1701638_ans_4ca9e5766a824b0cba5118b01afe1a68.png

A coin is tossed twice. If the outcome is at most one tail, what is the probability that both head and tail have appeared?

Given a coin is tossed twice

Then the sample space

$S$ will be

$\left\{H H,H T, T H, T T\right\}$

Given that the outcome is at most one tail

So favourable outcomes will be

$\left\{ H H,T H,H T\right\}$

Let

$A$ be the event of getting at most one tail

$\implies n(A)=3,n(S)=4$

Probability of

$A$ to happen is

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{3}{4}$

Let

$B$ be the event that both head and tail had appeared

then the outcomes of

$B$ is

$\left\{H T, T H\right\}$

$\implies n(B)=2$

Probability of

$B$ to happen is

$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{2}{4}=\dfrac{1}{2}$

Favourable outcomes of

$A\cap B$ is

$\left\{H T, T H\right\}$

$\implies A\cap B=B$

$\implies P(A\cap B)=P(B)=\dfrac{1}{2}$

Required probability is the probability of

$B$ given that

$A$ has happen

$..i.e.,\ P\left (\dfrac{B}{A}\right)$

As we know that

$P\left (\dfrac{B}{A}\right)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{\dfrac{1}{2}}{\dfrac{3}{4}}=\dfrac{2}{3}$

So Probability that both head and tail have appeared if the the outcome is at most one tail is

$\dfrac{2}{3}$

A dice is thrown twice and the sum of the numbers appearing is observed to be $8$ . What is the conditional probability that the number $5$ has appeared at least once ?

Given a dice is thrown twice

$\implies$ Total Number of outcomes is

$n(S)=6\times 6=36$

Let

$A$ be the event that sum of the numbers appearing on the dice is

$8$

The favourable outcomes of

$A$ are

$\left\{(2,6),(3,5),(4,4),(5,3),(6,2)\right\}$

$\implies n(A)=5$

Let

$B$ be the event that the number

$5$ has appeared atleast once

The favourable outcomes of

$B$ are

$\left\{(1,5),(2,5),(2,5),(4,5),(5,5),(6,5),(5,6),(5,4),(5,3),(5,2),(5,1)\right\}$

$\implies n(B)=11$

Hence Probability of

$A$ to happen is

$P(A)=\dfrac{n(A)}{n(S)}=\dfrac{5}{36}$

Probability of

$B$ to happen is

$P(B)=\dfrac{n(B)}{n(S)}=\dfrac{11}{36}$

The outcomes of

$A\cap B$ will be

$\left\{(3,5),(5,3)\right\}$

$\implies P(A\cap B)=\dfrac{n(A\cap B)}{n(S)}=\dfrac{2}{36}$

$P(B/A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{2/36}{5/36}=\dfrac{2}{5}$

$P(B/A)$ is the probability such that atleast once

$5$ has to be appeared given that sum of numbers on the dice is

$8$

So the required conditional probability is

$\dfrac{2}{5}$

There is a box containing $30$ bulbs of which $5$ are defective. If two bulbs are chosen at random from the box in succession without replacing the first, what is the probability that both the bulbs chosen are defective?

Given Number of defective bulbs is

$5$

Total Number of bulbs is

$30$

Let

$A$ be the event of picking defective bulb in the first trial

Probability of event

$A$ to happen

$P(A)=\dfrac{5}{30}=\dfrac{1}{6}$

After picking one defective bulb in the first trial, as the bulb is not replaced

Total Number of bulbs is

$29$

Number of defective bulbs is

$4$

Let

$B$ be the event of picking defective bulb in the second trial

Probability of event

$B$ to happen

$P(B)=\dfrac{4}{29}$

As the two trials does not depend on each other in our case

$P(A\cap B)=P(A)\times P(B)=\dfrac{1}{6}\times \dfrac{4}{29}=\dfrac{2}{87}$

Probability that the both bulbs chosen are defective is

$\dfrac{2}{87}$

A coin is tossed 4 times. Let X denote the number of heads . Find the Probability distribution of X. also Find the mean and variance of X .

1569665_1708240_ans_994f30eed07848f0818d786d25265811.png

A machine operates only when all of its three components functions. The probability of the failures of the first, second and third components are $0.14, 0.10$ and $0.05$ respectively. What is the probability that the machine will fail?

Let

$E_1, E_2, E_3$ be the respective events that the

$1st, 2nd$ and

$3rd$ components function.

Then,

$P(\bar E_1)=0.14, P(\bar E_2)=0.10$ and

$P(\bar E_3)=0.05$

$\Rightarrow P(E_1)=(1-0.14)=0.86, P(E_2)=(1-0.10)=0.90$ and

$P(E_3)=(1-0.05)=0.95$

$\Rightarrow P$ ( machine fails )

$=1-P$ ( machine functions)

$=1-P[(E_1$ and

$E_2$ and

$E_3)]$

$=1-P[(E_1\cap$

$E_2 \cap$

$E_3)]$

$=1-[P(E_1)\times P(E_2)\times P(E_3)]$ .

$=1-(0.86 \times 0.90 \times 0.95)$

$=0.2647$

A sample of $3$ bulbs is tested. A bulb is labelled as $G$ if it is good and $D$ if it is defective. Find the number of all possible outcomes.

All possible outcomes are

$GGG, GGD, GDG, DGG, GDD, DGD, DDG, DDD$ .
Clearly, there are

$8$ possible outcomes.

A coin is tossed (m+n) times (m > n).Show that the probability of at least m consecutive heads is $n+2/2^{m+1}$

Let H and T denote turning up of the head and tail and X denote the turning of head or tail.Then,

$P(H)=P(T)=\dfrac{1}{2}$ and

$P(X)=1$

$P(HH....... m \, times)(XXX.....n\, times )$

$=\dfrac{1}{2}\times \dfrac{1}{2}.....m$ times

$=\dfrac{1}{2^m}$

$P[T{HHH....... m\, times}{XX.....n-1\,times}]$

$=P(T)P(H)+.....+m\,times$

$P(XX.....n-1\,times )=\dfrac{1}{2^{m+1}}$

If the sequence of the heads starts with

$(r+1)^{th}$ ,then the first r-1 throws may be head or tail but

$r^{th}$ throw must be tail and we have

$(XX....(r-1) times)T(HH.....m times )(XX....(n-m-r)times )=\dfrac{1}{2^{m+1}}$

Since all the above cases are mutually exclusive, the required probability is

$\dfrac{1}{2^m}+[\dfrac{1}{2^{m+1}}+\dfrac{1}{2^{m+1}}+....+n\, times]$

$=\dfrac{1}{2^m}+\dfrac{n}{2^{m+1}}+\dfrac{n+2}{2^{m+1}}$

Assume that each born child is equally likely to be a boy or a girl . If a family has two children , what is the conditional probability that both are girls given that the youngest is a girl ?

A family has

$2$ children,
then sample space = S = {BB , BG , GB , GG} where B stands for boy and G for girl.

Let A and B be two event such that

A = Both are girls = {GG}

B = The youngest is a girl = {BG , GG}

$P \left ( \dfrac{A}{B} \right ) = \dfrac{P(A \cap B)}{P(B)}$ [

$\therefore \, A \cap B =$ {GG}]

$P \left ( \dfrac{A}{B} \right ) = \dfrac{\dfrac{1}{4}}{\dfrac{2}{4}} = \dfrac{1}{2}$

For a biased die,the probability for the different face to turn up are given below:
This die is tossed and you are told that either face 1 or face 2 has turned up.Then the probability that it is face 1 is...................

Let

$E_1$ be the event that face 1 has turned up and

$E_2$ be the event that face 1 or 2 has turned up.By the given data,

$P(E_2)=0.1+0.32=0.42,P(E_1\cap E_2)=0.1$

Given that

$E_2$ has happened and we have to find then the probability of happening of

$E_1$ .Therefore, by conditional probability theorem,we have

$P(E_1/E_2)=\dfrac{P(E_1 \cap E_2)}{P(E_2)}$

$=\dfrac{0.1}{0.42}$

$=\dfrac{10}{42}$

$=\dfrac{5}{21}$

In a village there are three mohallas A , B and C . In A , $60$ % persons believe in honesty , while in B $70$ % and in C , $80$ % . A person is selected at random from village and found , he is honest . Find the probability that he belongs to mohalla B .

Let the event be defined as

$E_1 =$ Selection of mohalla

$A$

$E_2 =$ Selection of mohalla

$B$

$E_3 =$ Selection of mohalla

$C$

$A =$ Selection of honest person

$P(E_1) = \dfrac{1}{3} , P(E_2) = \dfrac{1}{3} , P(E_3) = \dfrac{1}{3}$

$P \left ( \dfrac{A}{E_1} \right ) = \dfrac{60}{100} = \dfrac{3}{5}$ ;

$P \left ( \dfrac{A}{E_2} \right ) = \dfrac{70}{100} = \dfrac{7}{10}$

$P \left ( \dfrac{A}{E_3} \right ) = \dfrac{80}{100} = \dfrac{4}{5}$ ;

$P \left ( \dfrac{E_2}{A} \right ) =$ required

$P \left ( \dfrac{E_2}{A} \right ) = \dfrac{P(E_2) . P \left ( \dfrac{A}{E_2} \right )}{P(E_1) . P \left ( \dfrac{A}{E_1} \right ) + P(E_2) . P \left ( \dfrac{A}{E_2} \right ) + P(E_3). P \left ( \dfrac{A}{E_3} \right )}$

$= \dfrac{\dfrac{1}{3} . \dfrac{7}{10}}{\dfrac{1}{3} . \dfrac{3}{5} + \dfrac{1}{3} . \dfrac{7}{10} + \dfrac{1}{3} . \dfrac{4}{5}} = \dfrac{\dfrac{7}{30}}{\dfrac{3}{15} + \dfrac{7}{30} + \dfrac{4}{15}}$

$= \dfrac{7}{6 + 7 + 8} = \dfrac{7}{21} = \dfrac{1}{3}$

Assume that each born child is equally likely to be a boy or a girl . If a family has two children , what is the conditional probability that both are girls given that at least one is a girl ?

A family has

$2$ children,
then sample space = S = {BB , BG , GB , GG} where B stands for boy and G for girl.

Let C be event such that

C = at least one is a girl = {BG , GB , GG}

Now ,

$P(A / C) = \dfrac{P(A \cap C)}{P(C)}$ [

$\therefore \, A \cap C =$ {GG}]

$= \dfrac{\dfrac{1}{4}}{\dfrac{3}{4}} = \dfrac{1}{3}$

A person wants to construct a hospital in a village for welfare. The probabilities are $0.40$ that some bad elements opposite this work , $0.80$ that the hospital will be completed if there is not any oppose of any bad elements and $0.30$ that the hospital will be completed if bad elements oppose. Determine the probability that the construction of hospital will be completed .

Let the event be defined as

$A =$ Construction of hospital will be completed

$E_1 =$ There will be oppose of bad elements

$E_2 =$ There will be no oppose of any bad element

$P(E_1) = 0.40 = \dfrac{4}{10} , P(E_2) = 1 - 0.40 = 0.60 = \dfrac{6}{10}$

$P \left ( \dfrac{A}{E_1} \right ) = \dfrac{30}{100} = \dfrac{3}{10}$ ;

$P \left ( \dfrac{A}{E_2} \right ) = \dfrac{80}{100} = \dfrac{8}{10}$

$P(A) =$ required

$P(A) = P(E_1) . P \left ( \dfrac{A}{E_1} \right ) + P(E_2) . P \left ( \dfrac{A}{E_2} \right )$

$= \dfrac{4}{10} . \dfrac{3}{10} + \dfrac{6}{10} . \dfrac{8}{10} = \dfrac{12}{100} + \dfrac{48}{100} = \dfrac{60}{100} = \dfrac{3}{5}$

In a survey it was found that $30\%$ of the population is using non-biodegradable products while the remaining is using biodegradable products. What is the probability that a person at random uses non biodegradable products?
Which type of products should be used in a society for its proper development biodegradable o non-biodegradable? Justify your answer?

$P$ (Person using non- biodegradable products)

$=(100-30)\%$

$=\dfrac {70}{100}=\dfrac {7}{10}$
Biodegradable products are reusable and less pollution, so such products should be used.

A number $x$ is selected at random from the numbers $1,2,3$ and $4$ . Another number $y$ is selected at random from the numbers $1,4,9$ and $16$ . Find the probability that product of $x$ and $y$ is less than $16$ .

$x$ can be any of

$1,2,3$ or

$4$ .

$y$ can be any one of

$1,4,9$ or

$16$
Total number of cases of product of

$x$ and

$y=16$
Product less than

$16=(1\times 1, 1\times 4, 1\times 9, 2\times 1, 2\times 4, 3\times 1, 3\times 4, 4\times 1)$
Number of cases, where product is less than

$16=8$

$\therefore$ Required probability

$=\dfrac{8}{16}$ or

$\dfrac{1}{2}$

Two dice are thrown together and the total score is noted. The events $E, F$ and $G$ are 'a total of 4 ', 'a total of 9 or more', and 'a total divisible by 5 ', respectively. Calculate $P(E), P(F)$ and $P(G)$ and decide which pairs of events, if any, are independent.

Two dice are thrown together i.e., sample space

$(S)=36 \Rightarrow n(S)=36$

$E=A$ total of

$4=\{(2,2),(3,1),(1,3)\}$

$\Rightarrow n(E)=3$

$\mathrm{F}=\mathrm{A}$ total of 9 or more

$=\{(3,6),(6,3),(4,5),(4,6),(5,4),(6,4),(5,5),(5,6),(6,5),(6,6)\}$

$\Rightarrow n(F)=10$

$G=$ a total divisible by

$5=\{(1,4),(4,1),(2,3),(3,2),(4,6),(6,4),(5,5)\}$

$\Rightarrow n(G)=7$

Here,

$(E \cap F)=\phi$

and

$(E \cap G)=\phi$

Also,

$(F \cap G)=\{(4,6),(6,4),(5,5)\}$

$\Rightarrow n(F \cap G)=3$

and

$(E \cap F \cap G)=\phi$

$\therefore P(E)=\dfrac{n(F)}{n(S)}=\dfrac{3}{36}=\dfrac{1}{12}$

$P(F)=\dfrac{n(F)}{n(S)}=\dfrac{10}{36}=\dfrac{5}{18}$

$P(G)=\dfrac{n(G)}{n(S)}=\dfrac{7}{36}$

$P(F \cap G)=\dfrac{3}{36}=\dfrac{1}{12}$

And

$P(F) \cdot P(G)=\dfrac{5}{18} \cdot \dfrac{7}{36}=\dfrac{35}{648}$

Here, we see that

$P(F \cap G) \neq P(F) . P(G)$

[since, only

$F$ and

$G$ have common events, so only

$F$ and

$G$ are used here].

Hence, there is no pair which is independent.

A factory produces bulbs. The probability that any one bulb is defective is $\dfrac{1}{50}$ and they are packed in boxes of $10 .$ From a single box, find the probability that
(i) none of the bulbs is defective

Let

$X$ is the random variable which denotes that a bulb is defective.
Also,

$n=10, p=\dfrac{1}{50}$ and

$q=\dfrac{49}{50}$ and

$P(X=r)=^{n} C_{r} p^{r} q^{n-r}$
(i) None of the bulbs are defective i.e.,

$r=0$

$P(X=r)=P_{(0)}=^{10} C_{0}\left(\dfrac{1}{50}\right)^{0}\left(\dfrac{49}{50}\right)^{10-0}=\left(\dfrac{49}{50}\right)^{10}$

For a loaded die, the probabilities of outcomes are given as under:
$P(1)=P(2)=0.2, P(3)=P(5)=P(6)=0.1$ and $P(4)=0.3$
If the die were fair, determine whether or not the events A and B are independent.

Referring to the above solution, we have

$A=\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$

$\Rightarrow n(A)=6$ and

$n(S)=6^{2}=36[\text { Where }, \text { s is sample space }]$

$\therefore P(A)=\dfrac{n(A)}{n(B)}=\dfrac{6}{36}=\dfrac{1}{6}$
And

$B=\{(4,6),(6,4),(5,5),(6,5),(5,6),(6,6)\}$

$\Rightarrow n(B)=6$ and

$n(S)=6^{2}=36$

$\therefore P(B)=\dfrac{n(B)}{n(S)}=\dfrac{6}{36}=\dfrac{1}{6}$

Also,

$A \cap B=\{(5,5),(6,6)\}$

$\Rightarrow n(A \cap B)=2$ and

$n(S)=36$

$\therefore(A \cap B)=\dfrac{2}{36}=\dfrac{1}{18}$

Also,

$P(A) . P(B)=\dfrac{1}{36}$

Thus,

$P(A \cap B) \neq P(A) \cdot P(B)\left[\because \dfrac{1}{18} \neq \dfrac{1}{36}\right]$

So, we can say that both A and B are not independent events.

The probability distribution of a random variable X is given below:
$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & k & \dfrac{k}{2} & \dfrac{k}{4} & \dfrac{k}{8} \\\hline\end{array}$
(ii) Determine $P(X \leq 2)$ and $P(X>2)$

We have,

$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & k & \dfrac{k}{2} & \dfrac{k}{4} & \dfrac{k}{8} \\\hline\end{array}$

$\displaystyle \sum_{i=1}^{n} P_{i}=1, i=1,2, \ldots, n$ and

$p_{i} \geq 0$

$\therefore k+\dfrac{k}{2}+\dfrac{k}{4}+\dfrac{k}{8}=1$

$\Rightarrow 8 k+4 k+2 k+k=8$

$k=\dfrac{8}{15}$

$P(X \leq 2)=P(0)+P(1)+P(2)=k+\dfrac{k}{2}+\dfrac{k}{4}$

$=\dfrac{(4 k+2 k+k)}{4}=\dfrac{7 k}{4}=\dfrac{7}{4} \cdot \dfrac{8}{15}=\dfrac{14}{15}$

And

$P(X>2)=P(3)=\dfrac{k}{8}=\dfrac{1}{8} \cdot \dfrac{8}{15}=\dfrac{1}{15}$

A factory produces bulbs. The probability that any one bulb is defective is $\dfrac{1}{50}$ and they are packed in boxes of $10 .$ From a single box, find the probability that
(ii) exactly two bulbs are defective

Let

$X$ is the random variable which denotes that a bulb is defective.
Also,

$n=10, p=\dfrac{1}{50}$ and

$q=\dfrac{49}{50}$ and

$P(X=r)=^{n} C_{r} p^{r} q^{n-r}$
(ii) Exactly two bulbs are defective i.e.,

$r=2$

$P(X=r)=P_{(2)}=^{10} C_{2}\left(\dfrac{1}{50}\right)^{2}\left(\dfrac{49}{50}\right)^{8}$

$=\dfrac{10 !}{8 ! 2 !}\left(\dfrac{1}{50}\right)^{2} \cdot\left(\dfrac{49}{50}\right)^{8}=45 \times\left(\dfrac{1}{50}\right)^{10} \times(49)^{8}$

A discrete random variable X has the probability distribution given as below:
$\begin{array}{|l|ll|l|l|}\hline \mathrm{X} &0.5 & 1 & 1.5 & 2 \\\hline \mathrm{P(X)}& k & k^2 & 2k^2 & k \\\hline\end{array}$
(i) Find the value of k

We have,

$\begin{array}{|l|ll|l|l|}\hline \mathrm{X} &0.5 & 1 & 1.5 & 2 \\\hline \mathrm{P(X)}& k & k^2 & 2k^2 & k \\\hline\end{array}$
(i). We know that,

$\displaystyle \sum_{i=1}^{n} P_{i}=1$ , where

$P_{i} \geq 0$

$\Rightarrow P_{1}+P_{2}+P_{3}+P_{4}=1$

$\Rightarrow k+k^{2}+2 k^{2}+k=1$

$\Rightarrow 3 k^{2}+2 k-1=0$

$\Rightarrow 3 k^{2}+3 k-k-1=0$

$\Rightarrow 3 k(k+1)-1(k+1)=0$

$\Rightarrow(3 k-1)(k+1)=0$

$\Rightarrow k=1 / 3 \Rightarrow k=-1$

$\mathrm{k}=-1$ not possible because

$k$ is

$\geq 0 \Rightarrow \mathrm{k}=1 / 3$

A factory produces bulbs. The probability that any one bulb is defective is $\dfrac{1}{50}$ and they are packed in boxes of $10 .$ From a single box, find the probability that
(iii) more than 8 bulbs work properly

Let

$X$ is the random variable which denotes that a bulb is defective.
Also,

$n=10, p=\dfrac{1}{50}$ and

$q=\dfrac{49}{50}$ and

$P(X=r)=^{n} C_{r} p^{r} q^{n-r}$
(iii) More than 8 bulbs work properly i.e., there is less than 2 bulbs which are defective.
So,

$r<2 \Rightarrow r=0,1$

$\therefore P(X=r)=P(r<2)=P(0)+P(1)$

$=^{10} C_{0}\left(\dfrac{1}{50}\right)^{0}\left(\dfrac{49}{50}\right)^{10-0}+^{10} C_{1}\left(\dfrac{1}{50}\right)^{1}\left(\dfrac{49}{50}\right)^{10-1}$

$=\left(\dfrac{49}{50}\right)^{10}+\dfrac{10 !}{1 ! 9 !} \cdot \dfrac{1}{50} \cdot\left(\dfrac{49}{50}\right)^{9}$

$=\left(\dfrac{49}{50}\right)^{10}+\dfrac{1}{5} \cdot\left(\dfrac{49}{50}\right)^{9}=\left(\dfrac{49}{50}\right)^{9}\left(\dfrac{49}{50}+\dfrac{1}{5}\right)$

$=\left(\dfrac{49}{50}\right)^{9}\left(\dfrac{59}{50}\right)=\dfrac{59(49)^{9}}{(50)^{10}}$

Out of $400$ bulbs in a box, $15$ bulbs are defective. One bulb is taken out at random from the box. Find the probability that the drawn bulb is not defective.

Total number of bulbs in the box

$=400$
Total number of defective bulbs in the box

$=15$
Total number of non-defective bulbs in the box

$=400-15=385$

$P$ (bulb is not defective)

$=\dfrac {Number\ of\ non-defective\ bulbs}{Total\ number\ of\ bulbs}=\dfrac {385}{400}=\dfrac {77}{80}$

For the following probability distribution determine standard deviation of the random variable X.
$\begin{array}{|l|l|l|l|}\hline \mathrm{X} & 2 & 3 & 4 \\\hline \mathrm{P}(\mathrm{X}) & 0.2 & 0.5 & 0.3 \\\hline\end{array}$

We have,

$\begin{array}{|l|l|l|l|}\hline \mathrm{X} & 2 & 3 & 4 \\\hline \mathrm{P}(\mathrm{X}) & 0.2 & 0.5 & 0.3 \\\hline \mathrm{XP}(\mathrm{X}) & 0.4 & 1.5 & 1.2 \\\hline \mathrm{X}^{2} \mathrm{P}(\mathrm{X}) & 0.8 & 4.5 & 4.8 \\\hline\end{array}$

We know that, standard deviation of

$X=\sqrt{V a r X}$

Where,

$\operatorname{Var} X=E\left(X^{2}\right)-[E(X)]^{2}$

$=\sum_{i=1}^{n} x_{i}^{2} P\left(x_{1}\right)-\left[\sum_{i=1}^{n} x_{i} P_{i}\right]^{2}$

$\begin{array}{l}\therefore Var X=[0.8+4.5+4.8]-[0.4+1.5+1.2]^{2} \\=10.1-(3.1)^{2}=10.1-9.61=0.49\end{array}$

$\therefore$ Standard deviation of

$\mathrm{X}=\sqrt{V a r X}=\sqrt{0.49}=0.7$

Two biased dice are thrown together. For the first die $P(6)=\dfrac{1}{2},$ the other scores being equally likely while for the second die, $P(1)=\dfrac{2}{5}$ and the other scores are equally likely. Find the probability distribution of 'the number of one seen'.

For first die,

$P(6)=\dfrac{1}{2}$ and

$P\left(6^{\prime}\right)=\dfrac{1}{2}$

$\Rightarrow P(1)+P(2)+P(3)+P(4)+P(5)=\dfrac{1}{2}$

$\Rightarrow P(1)=\dfrac{1}{10}$

$P\left(1^{\prime}\right)=\dfrac{9}{10}[\because P(1)=P(2)=P(3)=P(4)=P(5)]$

For second die,

$P(1)=\dfrac{2}{5}$ and

$P(1)=1-\dfrac{2}{5}=\dfrac{3}{5}$

Let

$X=$ Number of one's seen

$\text { For } X=0, \quad P(X=0)=P(1^{\prime}) \cdot P(1^{\prime})=\dfrac{9}{10} \cdot \dfrac{3}{5}=\dfrac{27}{50}=0.54$

$P(X=1)=P(1^{\prime}) \cdot P(1^{\prime})+P(1^{\prime}) \cdot P(1^{\prime})=\dfrac{9}{10} \cdot \dfrac{2}{5}+\dfrac{1}{10} \cdot \dfrac{3}{5}$

$=\dfrac{18}{50}+\dfrac{3}{50}=\dfrac{21}{50}=0.42$

$P(X=2)=P(1) \cdot P(1)=\dfrac{1}{10} \cdot \dfrac{2}{5}=\dfrac{2}{50}=0.04$

Hence, the required probability distribution is as below.

$\begin{array}{|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 \\\hline \mathrm{P}(\mathrm{X}) & 0.54 & 0.42 & 0.04 \\\hline\end{array}$

Two dice are tossed. Find whether the following two events A and B areindependent: $A=\{(x, y): x+y=11\}$ and $B=\{(x, y): x \neq 5\},$ Where $(x, y)$ denotes a typical sample point.

We have,

$A=\{(x, y): x+y=11\}$ and

$B=\{(x, y): x \neq 5\}$

$\therefore A=\{(5,6),(6,5)\},$

$B=\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2, 6), (3,1),(3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(6,1),(6,2),(6,3)(6,4),(6,5),(6,6)\}$

$\Rightarrow n(a)=2, n(B)=30$ and

$n(A \cap B)=1$

$P(A)=\dfrac{2}{36}=\dfrac{1}{18}$ and

$P(B)=\dfrac{30}{36}=\dfrac{5}{6}$

$\Rightarrow P(A) \cdot P(B)=\dfrac{5}{108}$ and

$P(A \cap B)=\dfrac{1}{36} \neq P(A) \cdot P(B)$
So, A and B are not independent.

An item is manufactured by three machines A, B and C. Out of the total number of items manufactured during a specified period, 50% are manufactured on A, 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective, and 3% of these produced on C are defective. All the items are stored at one godown. One item is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?

Let

$E_1$ = Event that item is manufactured on A.

$E_2$ = Event that item is manufactured on B.

$E_3$ = Event that item is manufactured on C.
Let E be the event that an item is defective.

$\therefore P(E_1) = \dfrac{50} {100} = \dfrac{1} {2}, P(E_2) = \dfrac{30} {100} = \dfrac{3} {10}, P(E_3) = \dfrac{20} {100} = \dfrac{1} {5}$

$P\left ( \dfrac{E} {E_1} \right ) = \dfrac{2} {100} = \dfrac{1} {50}, P\left ( \dfrac{E} {E_2} \right ) = \dfrac{2} {100} = \dfrac{1} {50}$ and

$P\left ( \dfrac{E} {E_3} \right ) = \dfrac{3} {100}$

$\therefore P\left ( \dfrac{E_1} {E} \right ) = \dfrac{P(E_1) \cdot P\left ( \dfrac{E} {E_1} \right )} {P(E_1) \cdot P\left ( \dfrac{E} {E_1} \right ) + P(E_2) \cdot P\left ( \dfrac{E} {E_2} \right ) + P(E_3) \cdot P\left ( \frac{E} {E_3} \right )}$

$= \dfrac{\dfrac{1} {2} \cdot \dfrac{1} {50}} {\dfrac{1} {2} \cdot \dfrac{1} {50} + \dfrac{3} {10} \cdot \dfrac{1} {50} + \dfrac{1} {5} \cdot \dfrac{3} {100}}$

$= \dfrac{\dfrac{1} {100}} {\dfrac{1} {100} + \dfrac{3} {100} + \dfrac{3} {100}} = \dfrac{\dfrac{1} {100}} {\dfrac{5+3+3} {500}} = \dfrac{5} {11}$

Two natural numbers $\mathbf{r},$ s are drawn one at a time, without replacement from the set $S=\{1,2,3, \ldots, n\}$ . Find $\mathrm{P}[\mathrm{r} \leq \mathrm{p} \mid \mathrm{s} \leq \mathrm{p}]$ , where $p \in S$

$\because$ Set

$S=\{1,2,3, \ldots, n\}$

$\therefore P(r \leq p / s \leq p)=\dfrac{P(p \cap S)}{P(S)}$

$=\dfrac{p-1}{n} \times \dfrac{n}{n-1}=\dfrac{p-1}{n-1}$

The probability distribution of a random variable x is given as under:
$P(X = x) = {\begin{matrix}kx^{2} & for \space x = 1, 2, 3 \\ 2kx & for \space x = 4, 5, 6\\ 0 & otherwise \end{matrix}}$
Where k is a constant. Calculate
(i) E(X)
(ii) $E(3X^{2})$
(iii) $P(X \geq 4)$

We know that,

$\sum P_i = 1$

$\Rightarrow 44k = 1 \Rightarrow k = \frac{1} {44}$

$\therefore \sum XP(X) = k + 8k + 27k + 32k + 50k + 72k + 0$

$190k = 190 \times \frac{1} {44} = \frac{95} {22}$

(i) So,

$E(X) = \sum XP(X) = \frac{95} {22} = 4.32$

(ii) Also,

$E(X^{2}) = \sum X^{2}P(X) = k + 16k + 81k + 128k + 250k + 432k$

$= 908k = 908 \times \frac{1} {44}$

$\left[ \therefore k = \frac{1} {44} \right]$

$= 20.636 = 20.64(approx)$

$E(3X^{2}) = 3E (X^{2}) = 3 \times 20.64 = 61.92 = 61.9$

(iii)

$P(X \geq 4) = P(X = 4) + P(X = 5) + P(X = 6)$

$= 8k + 10k + 12k = 30k = 30k = 30 \cdot \frac {1} {44} = \frac {15} {22}$

1905606_1802789_ans_e9351a44a545486a926ae687f33a267f.PNG

The probability distribution of a random variable X is given below:
$\begin{array}{|l|l|l|l|l|}\hline \mathrm{X} & 0 & 1 & 2 & 3 \\\hline \mathrm{P}(\mathrm{X}) & k & \dfrac{k}{2} & \dfrac{k}{4} & \dfrac{k}{8} \\\hline\end{array}$
(iii) Find $P(X \leq 2)+P(X>2)$

We have,

$\begin{array}{|l|l|l|l|l|}\hline X & 0 & 1 & 2 & 3 \\\hline P(X) & k & \dfrac{k}{2} & \dfrac{k}{4} & \dfrac{k}{8} \\\hline\end{array}$
(iii)

$P(X \leq 2)+P(X>2)=\dfrac{14}{15}+\dfrac{1}{15}=1$

Find the probability distribution of
(iii) number of heads in four tosses of a coin

1960540_1860128_ans_44786f430209412a93088785666c4d2f.png

There are $5\%$ defective items in a large of items. What is the probability that a sample of $10$ items will include not more than one defective item?

2172827_1860109_ans_bf42e80d1bc14a6c89195ae28ccc02ef.jpg

Find the probability distribution of
(ii) number of tails in the simultaneous tosses of three coins

1960491_1860126_ans_5134645b9bdd45b6b565853614cc7e35.png

Let A and B be two events. If P(A/B) = P(A), then A is _________ of B.

$\therefore P(A/B) = \dfrac{P(A \cap B)} {P(B)}$

$\Rightarrow P(A) = \dfrac{P(A \cap B)} {P(B)}$

$\Rightarrow P(A) \cdot P(B) = P(A \cap B)$
So, A is independent of B.

In a bundle of $50$ shirts, $44$ are good, $4$ have minor defects and $2$ have major defects. What is the probability that it is acceptable to a trader who accepts only a good shirt?

We have,

Total number of shirts $= 50$

Total number of elementary events $= 50 = n(S)$

As, trader accepts only good shirts and number of good shirts $= 44$

Event of accepting good shirts $= 44 = n(E)$

Probability of accepting a good shirt $= n(E)/ n(S) = 44/50 = 22/25$

In a hockey team, there are 6 defenders, 4 offenders, and 1 goalie. Out of these, one player is to be selected randomly as a captain. Find the probability of the selection that -A defender will be selected.

Given:

Number of defenders

$=6$

Number of offenders

$=4$
Number of goalee

$=1$
Let

$S$ be the sample space.

$\therefore \mathrm{n}(\mathrm{S})=$ Number of defenders

$+$ Number of offenders

$+$ Number of goalee

$=6+4+1$

$=11$
Event B: A defender will be selected
Given:

$n(B)=6$

$\therefore P(B)=\dfrac{n(B)}{n(S)}$

$=\dfrac{6}{11}$
Hence, the probability that the goalie will be selected is

$=\dfrac{6}{11}$

In a bundle of $50$ shirts, $44$ are good, $4$ have minor defects and $2$ have major defects. What is the probability that it is acceptable to a trader who rejects only a shirt with major defects?

We have,

Total number of shirts $= 50$

Total number of elementary events $= 50 = n(S)$

As, trader rejects shirts with major defects only and number of shirts with major defects $= 2$

Event of accepting shirts $= 50 – 2 = 48 = n(E)$

Probability of accepting shirts $= n(E)/ n(S) = 48/50 = 24/25$

It is known that $10\%$ of certain articles manufactured are defective. What is the probability that in a random sample of $12$ such articles $9$ are defective?

${\textbf{Step -1: Writing probablity distribution}}$

${\text{Let X denote the number of times of defective bulb being selected out of 12,}}$

$\Rightarrow {\text{ n = 12}}$

$\Rightarrow {\text{ p = }}\dfrac{{{\text{10}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{\text{1}}}{{{\text{10}}}}$

$\Rightarrow {\text{ q = 1 - p = 1 - }}\dfrac{{\text{1}}}{{{\text{10}}}}{\text{ = }}\dfrac{{\text{9}}}{{{\text{10}}}}$

$\therefore {\text{ P(X = x) = }}{{\text{ }}^{\text{n}}}{{\text{C}}_{\text{x}}}{{\text{p}}^{\text{x}}}{{\text{q}}^{{\text{n - x}}}}$

${\textbf{Step -2: Finding probablity of selecting 9 defective bulbs}}$

${\text{For 9 defective bulbs, x = 9}}$

$\Rightarrow {\text{ P(X = 9) = }}{{\text{ }}^{{\text{12}}}}{{\text{C}}_{\text{9}}}{\left( {\dfrac{{\text{1}}}{{{\text{10}}}}} \right)^{\text{9}}}{\left( {\dfrac{{\text{9}}}{{{\text{10}}}}} \right)^{\text{3}}}$

$\Rightarrow {\text{ P(X = 9) = 220 }} \times {\text{ }}\dfrac{{{{\text{9}}^{\text{3}}}}}{{{\text{1}}{{\text{0}}^{\text{3}}}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{{\text{1}}{{\text{0}}^{\text{9}}}}}$

$\Rightarrow {\text{ P(X = 9) = }}\dfrac{{{\text{22 }} \times {\text{ }}{{\text{9}}^{\text{3}}}}}{{{\text{1}}{{\text{0}}^{{\text{11}}}}}}$

$\mathbf{{\text{Final Answer: The probablity that 9 bulbs are defective is }}\dfrac{{{\text{22 }} \times {\text{ }}{{\text{9}}^{\text{3}}}}}{{{\text{1}}{{\text{0}}^{{\text{11}}}}}}}$

(i) A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that the bulb is defective?
(ii) Suppose the bulb drawn in (i) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Out of 20 bulbs, contain 4 defective ones.
(i)

$\therefore$ Probability of defective bulb

$=\dfrac{4}{20}$
(ii) Out of

$20,$ one bulb is drawn at random from the lot then remaining bulbs

$=19$ . Out of 19 bulbs, the bulbs which arc not defective

$=19-4=15$

$\therefore$ The probability that the bulb is not defective,

$\mathrm{P}(\mathrm{E})=\dfrac{15}{19}$

$4$ rotten oranges are mixed accidently with $16$ good oranges. Find the probability distribution of the number of rotten oranges in a draw of two oranges.

$4$ bad oranges are mixed with

$16$ good oranges. The number of total oranges

$=4+16=20$

$2$ bad oranges are to be chosen
Probability of choosing a bad orange

$=\cfrac{4}{20}=\cfrac{1}{5}$
Probability of choosing one good orange

$=1-\cfrac{1}{5}=\cfrac{4}{5}$

$X$ is the number of bad oranges

$P(X=0=P(GG)$

$P(X=1)=(\cfrac{4}{5})\times \cfrac{15}{19}=\cfrac{12}{19}$

$P(X=1)=\cfrac{4}{20}\times \cfrac{16}{19}+\cfrac{16}{20}\times \cfrac{4}{19}$

$=\cfrac{32}{95}$

$P(X=2)=P(BB)$

$=\cfrac{4}{20}\times \cfrac{3}{19}=\cfrac{3}{95}$
Probability distribution of bad oranges is

$X$	0	1	2
$P(X)$	$\cfrac{12}{19}$	$\cfrac{32}{95}$	$\cfrac{3}{95}$

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
i) number greater than $4$

1964747_1860145_ans_413eda17c44344438699ce7e88c4ac78.png

A die is thrown twice and the sum of the numbers appearing is observed to be $6$ . What is the conditional probability that the number $4$ has appeared at least once?

${\textbf{Step -1: Find probability to get sum.}}$

${\text{Let, A be the event that the sum of the numbers is 6}}{\text{.}}$

${\text{So the number of ways that the sum of the dice is 6 = }}\left( {{\text{1, 5}}} \right){\text{, }}\left( {{\text{2, 4}}} \right){\text{, }}\left( {{\text{3, 3}}} \right){\text{, }}\left( {{\text{4, 2}}} \right){\text{, }}\left( {{\text{5, 1}}} \right){\text{.}}$

${\text{Therefore, A = }}\left( {{\text{1, 5}}} \right){\text{, }}\left( {{\text{2, 4}}} \right){\text{, }}\left( {{\text{3, 3}}} \right){\text{, }}\left( {{\text{4, 2}}} \right){\text{, }}\left( {{\text{5, 1}}} \right){\text{.}}$

${\text{So as we see that there are 5 possible ways so that the sum of the numbers is 6}}{\text{.}}$

${\text{As we know that when two dice are thrown the possible number of cases are}}\;{\text{6}} \times 6 = 36$

${\text{So the probability to get the sum,}}\;{\text{P(A)}} = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$

$\Rightarrow {\text{P(A)}} = \dfrac{5}{{36}}$

${\textbf{Step -2: Find probability.}}$

${\text{Now let }}\left( {\text{E}} \right){\text{ be the event such that 4 appears once}}{\text{.}}$

${\text{So the possible cases when two dice are throw one by one,}}$

${\text{E = }}\left( {{\text{1, 4}}} \right){\text{, }}\left( {{\text{2, 4}}} \right){\text{, }}\left( {{\text{3, 4}}} \right){\text{, }}\left( {{\text{4, 4}}} \right){\text{, }}\left( {{\text{5, 4}}} \right){\text{, }}\left( {{\text{6, 4}}} \right){\text{, }}\left( {{\text{4, 1}}} \right){\text{, }}\left( {{\text{4, 2}}} \right){\text{, }}\left( {{\text{4, 3}}} \right){\text{, }}\left( {{\text{4, 5}}} \right){\text{, }}\left( {{\text{4, 6}}} \right){\text{.}}$

${\text{So there are 11 possible cases such that 4 appear once}}{\text{.}}$

${\text{Now the intersection of event A and E, i}}{\text{.e}}{\text{. A}} \cap {\text{E = }}\left( {2,4} \right){\text{ and }}\left( {4,2} \right)$

${\text{So, the probability of P}}\left( {{\text{A}} \cap {\text{E}}} \right) = \dfrac{2}{{36}}$

${\text{Now the conditional probability P }}\left( {{\text{E/A}}} \right){\text{ such that the sum of the numbers is 6 }}$

${\text{and digit 4 appeared once is,}}$

$\Rightarrow P(\dfrac{E}{A}) = \dfrac{{{\text{P}}\left( {{\text{A}} \cap {\text{E}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}$

$= \dfrac{{\dfrac{2}{{36}}}}{{\dfrac{5}{{36}}}}$

$= \dfrac{2}{5}$

${\textbf{Hence, the probability is }}\boldsymbol{\dfrac{2}{5}.{\text{ }} }$

Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater that 4' given that 'there is at least one tail?

The results of the experiment are expressed by following diagram:
Sample space of experiment is:

$S=\left\{ (H,H),(H,7),(T,1),(1,2),(1,3),(T,4),(7,5),(T,6) \right\}$ where

$(HH)$ shows that head appears on both throws and

$(T,1)$ shows that Tail on coin and

$I$ on die. So the probabilities of 8 fundamental events

$(H,H),(H,T),(T,1),(T,2),(1,3),(T,4),(7,5),(T,6)$ are

$\cfrac { 1 }{ 4 } ,\cfrac { 1 }{ 4 } ,\cfrac { 1 }{ 12 } ,\cfrac { 1 }{ 12 } ,\cfrac { 1 }{ 12 } ,\cfrac { 1 }{ 12 } ,\cfrac { 1 }{ 12 } ,\cfrac { 1 }{ 12 } ,$
Let

$A:$ Number appears more than 4 on die

$B:$ Minimum one tail appears
Then

$B\left\{ (H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6) \right\}$

$P(B)=P\left[ \left\{ \left( H,T \right) \right\} \right] +P\left[ \left\{ \left( T,1 \right) \right\} \right] +P\left[ P\left( 1,2 \right) +P\left( T,3 \right) \right] +P\left[ \left( T,4 \right) \right] +P\left[ \left\{ T,5 \right\} \right] +P\left[ \left\{ T,6 \right\} \right]$

$=\cfrac { 1 }{ 4 }+\cfrac { 1 }{ 12 } +\cfrac { 1 }{ 12 } +\cfrac { 1 }{ 12 } +\cfrac { 1 }{ 12 } +\cfrac { 1 }{ 12 } +\cfrac { 1 }{ 12 } =\cfrac{3}{4}$
and

$P\left( A\cap B \right) =P\left[ \left\{ \left( T,5 \right) \right\} \right] +P\left[ \left\{ \left( T,6 \right) \right\} \right]$

$=\cfrac{1}{12}+\cfrac{1}{12}=\cfrac{1}{6}$
So,

$P\left( \cfrac { A }{ B } \right) =\cfrac { P\left( A\cap B \right) }{ P(B) } =\cfrac{1/6}{3/4}=\cfrac{2}{9}$

Find which of the following probability distribution is possible for a random variable:
(i)
$X:$ 0 1 2
$P(X):$ $0.4$ $0.4$ $0.2$
(ii)
$X:$ 0 1 2
$P(X):$ $0.6$ $0.1$ $0.2$
(iii)
$X:$ 0 1 2 3 4
$P(X):$ $0.1$ $0.5$ $0.2$ $-0.1$ $0.3$

(i) sum of probabilities

$=0.4+0.4+0.2=1$
Hence given distribution is a probability distribution
(ii) sum of probabilities

$=0.6+0.1+0.2=0.9\ne 1$
Hence given distribution is not a probability distribution
(iii) Here one of the probability is

$-0.1$ which is negative. Hence this distribution is not probability distribution

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
ii) Six appears on at least one die

1964784_1860148_ans_21765c89b3434df49ac19f2416cd759b.jpg

The probability of picking a non - defective items from a sample is $\dfrac{7}{12}.$ Find the probability of picking a defective item.

$P(A)=\dfrac{7}{12}$ and

$P(\bar{A})=?$

$P(A)+P(\bar{A})=1$

$\begin{aligned} \dfrac{7}{12}+\mathrm{P}(\overline{\mathrm{A}}) &=1 \\ \mathrm{P}(\overline{\mathrm{A}}) &=1-\dfrac{7}{12} \end{aligned}$

$\mathrm{P}(\overline{\mathrm{A}})=\dfrac{5}{12}$

A random variable $X$ can take al non-negative values and the probability that $X$ take the value $r$ is proportional to $ar$ where $(0< a< 1)$ . Find $P(X=0)$

Given:

$P(X=r)\propto \cfrac { 1 }{ { \alpha }^{ } }$

$\Rightarrow P(X=r)=k\times \cfrac { 1 }{ { \alpha }^{ r } }$ (where

$k$ is any constant)

$\Rightarrow P(r=0)=k.\cfrac { 1 }{ { \alpha }^{ 0 } }$

$P(r=1)=\cfrac { k }{ { \alpha }^{ } } ;P(r=2)=\cfrac { k }{ { \alpha }^{ 2 } } ;P(r=3)=\cfrac { k }{ { \alpha }^{ 3 } } ...$
But

$P(r=0)+P(r=1)+P(r=2)+...=1$

$\cfrac { k }{ { \alpha }^{ 0 } } +\cfrac { k }{ { \alpha }^{ } } +\cfrac { k }{ { \alpha }^{ 2 } } =1\left[ \because { \alpha }^{ 0 }=1 \right]$

$\quad k\left[ 1+\cfrac { 1 }{ { \alpha }^{ } } +\cfrac { 1 }{ { \alpha }^{ 2 } } +... \right] =1\quad$

$k\left( \cfrac { 1 }{ 1-\alpha } \right) =1\Rightarrow k=\left( 1-\alpha \right)$
Putting the value of

$k$ in

$P(X=r)=k\cfrac { 1 }{ { \alpha }^{ r } }$

$P(X=r)=\cfrac { 1-\alpha }{ { \alpha }^{ r } }$

$(X=0)=\cfrac { 1-\alpha }{ { \alpha }^{ 0 } } \quad$

$\cfrac { 1-\alpha }{ 1 } =1-\alpha \quad$

${ \alpha }^{ 0 }=1,\therefore \quad P(X=0)=1-\alpha$

Let $X$ be random variable which assumes value ${x}_{1},{x}_{2},{x}_{3},{x}_{4}$ such that
$2P(X={x}_{1})=3P(X={x}_{2})=4P(X={x}_{3})=5P(X={x}_{4})$
Find the probability distribution of $X$ .

Given

$2P(X={x}_{1})=3P(X={x}_{2})=P(X={x}_{3})=5P(X={x}_{4})=K$

$\Rightarrow P(X={ x }_{ 1 })=\cfrac { k }{ 2 } ;P(X={ x }_{ 2 })=\cfrac { k }{ 3 } ;P(X={ x }_{ 3 })=k;P(X={ x }_{ 4 })=\cfrac { k }{ 5 }$

$\Rightarrow P(X={ x }_{ 1 })+P(X={ x }_{ 2 })+P(X={ x }_{ 3 })+P(X={ x }_{ 4 })=1$

$\cfrac { k }{ 2 } +\cfrac { k }{ 3 } +k+\cfrac { k }{ 5 } =1\Rightarrow k=\cfrac { 30 }{ 61 }$

$P(X={ x }_{ 1 })=\cfrac { k }{ 2 } =\cfrac { 30 }{ 2\times 61 } =\cfrac { 15 }{ 61 } ;P(X={ x }_{ 2 })=\cfrac { k }{ 3 } =\cfrac { 30 }{ 3\times 61 } =\cfrac { 10 }{ 61 }$

$P(X={ x }_{ 3 })=k=\cfrac { 30 }{ 61 } ;P(X={ x }_{ 4 })=\cfrac { k }{ 5 } =\cfrac { 30 }{ 5\times 61 } =\cfrac { 6 }{ 61 } \quad \quad$
Probability distribution is

$x$	${x}_{1}$	${x}_{2}$	${x}_{3}$	${x}_{4}$
$P(X)$	$\cfrac { 15 }{ 61 }$	$\cfrac { 10 }{ 61 }$	$\cfrac { 30 }{ 61 }$	$\cfrac { 6 }{ 61 }$

From a lot of $10$ items containing $3$ detectives, a sample of $4$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the sample is drawn randomly, find
(i) the probability distribution of $X$
(ii) $P(x\le 1)$
(iii) $P(x< 1)$
(iv) $P(0< x< 2)$

Given, from a lot of

$10$ itmes,

$3$ are defective

Good items

$=10-3=7$

Let

$x$ represents the number of defective items

Clearly values of

$X$ are

$0,1,2,3$

$P(x=0)=P(GGGG)$

$=p$ (good items)

$=\cfrac{7}{10}\times \cfrac{6}{9}\times \cfrac{5}{8}\times \cfrac{4}{7}=\cfrac{1}{6}$

$P(x=1)=P$ (three good and one defective)

$=P(BGGG)+P(GBGG)+P(GGBG)+P(GGGB)$

$=\cfrac{3}{10}\times \cfrac{7}{9}\times \cfrac{6}{8}\times \cfrac{5}{7}+\cfrac{7}{10}\times \cfrac{3}{9}\times \cfrac {6}{8}\times \cfrac{5}{7}+\cfrac{7}{10}\times \cfrac{6}{9}\times \cfrac{3}{8}\times \cfrac{5}{7}+\cfrac{7}{10}\times \cfrac{6}{9}\times \cfrac{5}{8}\times \cfrac{3}{7}$

$\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 8 } +\cfrac { 1 }{ 8 } =\cfrac{4}{8}=\cfrac{1}{2}$

$P(X=2)=P$ (two good and two defective)

$=P(BBGG)+P(BGGB)+P(GBBG)$

$\cfrac { 3 }{ 10 } \times \cfrac { 2 }{ 9 } \times \cfrac { 7 }{ 8 } \times \cfrac { 6 }{ 7 } +\cfrac { 3 }{ 10 } \times \cfrac { 7 }{ 9 } \times \cfrac { 6 }{ 8 } \times \cfrac { 2 }{ 7 } +\cfrac { 7 }{ 10 } \times \cfrac { 3 }{ 9 } \times \cfrac { 2 }{ 8 } \times \cfrac { 6 }{ 7 }$

$\cfrac { 1 }{ 20 }+\cfrac { 1 }{ 20 } +\cfrac { 1 }{ 20 } +\cfrac { 1 }{ 20 } =\cfrac{4}{20}=\cfrac{1}{5}$

$P(X=3)=P(BBBG)+P(BGBB)+P(BBGB)+P(GBBB)$

$=\cfrac { 3 }{ 10 } \times \cfrac { 2 }{ 9 } \times \cfrac { 1 }{ 8 } \times \cfrac { 7 }{ 7 } +\cfrac { 3 }{ 10 } \times \cfrac { 7 }{ 9 } \times \cfrac { 1 }{ 7 } +\cfrac { 3 }{ 10 } \times \cfrac { 2 }{ 9 } \times \cfrac { 7 }{ 8 } \times \cfrac { 1 }{ 7 } +\cfrac { 7 }{ 10 } \times \cfrac { 3 }{ 9 } \times \cfrac { 2 }{ 8 } \times \cfrac { 1 }{ 7 }$

$=\cfrac { 1 }{ 120 } +\cfrac { 1 }{ 120 } +\cfrac { 1 }{ 120 } +\cfrac { 1 }{ 120 } =\cfrac{4}{120}=\cfrac{1}{30}$

Probability distribution is following:

(i)

$X$ 0 1 2 3

$P(X)$

$\cfrac{1}{6}$

$\cfrac{1}{2}$

$\cfrac{3}{10}$

$\cfrac{1}{30}$

(ii)

$P(x\le 1)=\cfrac{2}{3}$

(iii)

$P(x< 1)=P(X=0)=\cfrac{1}{6}$

(iv)

$P(0< X< 2)=P(X=1)=\cfrac{1}{2}$

An urn contains $4$ white and $3$ red balls. Find the probability distribution of the number of red balls if $3$ balls are drawn at random.

Three bals are taken out of an urn
Sample space is

$=S(RRR,RRW,RWR,WRR,RWW,WRW,WWR,WWW)$
Where

$R$ represents red ball and

$W$ , white ball
Let

$X=$ number of red balls.
so possible value of

$X$ are

$3,2,1,2,1,0$ or

$0,1,2,3$ (no red ball)

$P(X=0)=P$ (No red ball)

$=P(WWW)$

$=\cfrac{4}{7}\times \cfrac{3}{6}\times \cfrac{2}{5}$ [

$\because$

$P(R)=\cfrac{3}{7}(P(W)=\cfrac{4}{7})$ ]

$=\cfrac{4}{35}$

$P(X=1)=P(RWW,WRW,WWR)=P(RWW)+P(WRW)+P(WWR)$

$=\cfrac{3}{7}\times \cfrac{4}{6}\times \cfrac{3}{5}+\cfrac{4}{7}\times \cfrac{3}{6}\times \cfrac{3}{5}+\cfrac{4}{7}\times \cfrac{3}{6}\times \cfrac{3}{5}$

$=\cfrac{6}{35}+\cfrac{6}{35}+\cfrac{6}{35}=\cfrac{18}{35}$

$P(X=2)=P(RRW,RWR,WRR)=P(RRW)+P(RWR)+P(WRR)$

$=\cfrac{3}{7}\times \cfrac{2}{6}\times \cfrac{4}{5}+\cfrac{3}{7}\times \cfrac{4}{6}\times \cfrac{2}{5}+\cfrac{4}{7}\times \cfrac{3}{6}\times \cfrac {2}{5}$

$=\cfrac{4}{35}+\cfrac{4}{35}+\cfrac{4}{35}=\cfrac{12}{35}$
Required probability distribution

$X$	0	1	2	3
$P(X)$	$\cfrac{4}{35}$	$\cfrac{18}{35}$	$\cfrac{12}{35}$	$\cfrac{1}{35}$

Ten eggs are drawn successively, with replacement, from a lot containing $10$ % defective eggs. Find the probability that there is at least one defective egg.

Probability of defective eggs

$=10$ %

$\quad p=\cfrac { 10 }{ 100 } =\cfrac { 1 }{ 10 }$
(Probability of good eggs)

$=q$

$q=1-p=1-\cfrac { 10 }{ 100 } =\cfrac { 9 }{ 10 }$
(Probability of at least one egg defective out of

$10$ )

$p(1)+p(2)+p(3)+...$

$=p(0)+p(1)+p(2)+...+p(10)-p(0)$

$=[p(0)+p(1)+p(2)+...+p(10)]-p(0)$

$=1-p(0)=1-{ \left( \cfrac { 9 }{ 10 } \right) }^{ 10 }$

An equipment will work only when its three components A ,B and C are in working condition . Probability that A will not work is 0 . 15 and (M) 5 of B and 0 .10 to C .What is the probability that equipment will not be in working condition before year ending ?

According to question

$P(\bar{A}) = 0 . 015 \Rightarrow P (A) = 0 . 85$

$P(\bar{B}) = 0 .05 \Rightarrow P(B) = 0 . 95$

$P(\bar{C}) = 0.10 \Rightarrow P(C) = 0.90$
Prob of damage equipment before year ending
= 1 -P (A) . P(B). P(C)
= 1 - 0 . 72675 = 0 . 27325

There are 48 mobile phones in a group out of which 42 are good, 3 are some defective and other 3 are largely defected. Varnika will purchase the mobile phone only when it is good, but a businessman will buy the mobile phone only when there is defective largely. A phone is chosen at random from these. What is the probability that the chosen phone is:
(i) Accepted by Varnika
(ii) Accepted by businessman.

Total number of mobile phones in a group is,

$n\left( S \right) =48$

1) Varnika will purchase the mobile phone only when it is good.

Total number of mobiles which are in good condition =

$42$

Let

$A$ = An event such that mobile phone is accepted by Varnika

$\therefore n\left( A \right) =42$

$\therefore P\left( A \right) =\dfrac { n\left( A \right) }{ n\left( S \right) }$

$\therefore P\left( A \right) =\dfrac { 42 }{ 48 } =\dfrac { 7 }{ 8 }$

2) Businessman will buy the mobile phone only when it is defective largely.

Total number of mobiles which are largely defective=

$3$

Let

$B$ = An event such that mobile phone is accepted by businessman

$\therefore n\left( B \right) =3$

$\therefore P\left( B \right) =\dfrac { n\left( B \right) }{ n\left( S \right) }$

$\therefore P\left( B \right) =\dfrac { 3 }{ 48 } =\dfrac { 1 }{ 16 }$

By chance 12 defected pens are mixed into 132 good pens. On seeing them it cannot be decided which pen is defected and which is good. If at random one pen is drawn out them, what is the probability that pen is good?

2014136_1877818_ans_d91a8a0e7baa4eb49816e787aa4047dd.png

A die is manufactured in a such a way that an even number is twice likely to occur as an odd number. If the die is tossed twice, find the probability distribution of the random variable $X$ representing the perfect square in the two tosses.

Let

$X$ represents the number of perfect squares

$S=$ sample space

$=(1,2,3,4,5,6)$

$n(S)=6$

$n(X)=$ Possible number of perfect squares

$=\left\{ 1,4 \right\} =2$
Probability of getting perfect square

$=\cfrac{n(X)}{n(S)}=\cfrac{2}{6}$
And probability of not getting perfect square

$=1-\cfrac{2}{6}=\cfrac{4}{6}$
When die is tossed twice

$n(S)=6\times 6=36$

$P(X=0)=8$ (no perfect square)

$=\times \cfrac { 4 }{ 6 } =\cfrac { 4 }{ 9 }$

$P(X=1)=P$ (one perfect square)

$=\cfrac { 2 }{ 6 } \times \cfrac { 4 }{ 6 } +\cfrac { 4 }{ 6 } \times \cfrac { 2 }{ 6 } =\cfrac { 4 }{ 9 }$

$P(X=2)=P$ (both are perfect square)

$P(X=2)=\cfrac { 2 }{ 6 } \times \cfrac { 2 }{ 6 } =\cfrac{1}{9}$
Probabiity distribution is following:

$X$	0	1	2
$P(X)$	$\cfrac{4}{9}$	$\cfrac{4}{9}$	$\cfrac{1}{9}$

20 ball pens are defective among a heap of 144 ball pens. You would like to buy a good pen not bad. The shopkeeper takes out a ball pen at random from these and gives you. What is the probability that:
(i) You will buy that ball pen.
(ii) You will not buy that ball pen.

Total number of ball pens =

$144$

Defective ball pens =

$20$

Ball pens in good condition =

$144-20=124$

Let

$A$ = An event such that you will buy good ball pen

$\therefore n\left( A \right) =124$

$P\left( A \right) =\dfrac { n\left( A \right) }{ n\left( S \right) }$

$\therefore P\left( A \right) =\dfrac { 124 }{ 144 } =\dfrac { 31 }{ 36 }$

2) Let

$\overline { A }$ = Complementary event of event

$A$

$\therefore \overline { A }$ = An event such that you will buy defective ball pen

$\therefore n\left( \overline { A } \right) =20$

$P\left( \overline { A } \right) =\dfrac { n\left( \overline { A } \right) }{ n\left( S \right) }$

$\therefore P\left( \overline { A } \right) =\dfrac { 20 }{ 144 } =\dfrac { 5 }{ 36 }$

(i) Among 20 bulbs, 4 bulbs are defective. A bulb is drawn at random from these. What is the probability that the bulb drawn will be defective?
(ii) Let the bulb drawn is neither defective not it is mixed with the remaining bulbs again. Now a bulb is taken out from these remaining bulbs. What is the probability of the bulb drawn is not defective?

1) Total number of bulbs are given by,

$n\left( S \right) =20$

Let

$A$ = an event such that bulb is defective

$\therefore n\left( A \right) =4$

$P\left( A \right) =\dfrac { n\left( A \right) }{ n\left( S \right) }$

$\therefore P\left( A \right) =\dfrac { 4 }{ 20 } =\dfrac { 1 }{ 5 }$

2) From 20 bulbs, one non defective bulb is taken out.

$\therefore n\left( S \right) =19$

Number of defective bulbs =

$4$

Number of non defective bulbs =

$19-4=15$

Let

$B$ = An event such that bulb drawn is non defective

$\therefore n\left( B \right) =15$

$P\left( B \right) =\dfrac { n\left( B \right) }{ n\left( S \right) }$

$\therefore P\left( B \right) =\dfrac { 15 }{ 19 }$

To explain the computation of conditional probability of a given event $B$ when event $A$ has already occurred.

$\textbf{Step 1: Define conditional probability.}$

$\text{Conditional probability is the probability of any event given that, a particular event }$

$\text{is already occured.}$

$\text{Let, A and B are two events.}$

$\text{If A is already occured, then probability of occurrence of B will be considered as conditional}$

$\text{probability as it may depend upon the occurrence of A.}$

$\text{And it is represented as :}\mathbf{P(B|A)}$

$\text{The formula to calculate conditional probability is given by :}$

$\mathbf{P(B|A)=\dfrac{P(A\cap B)}{P(A)}}$

Fifteen coupons are numbered from 1 toSeven coupons are selected at random one at a time with replacement. The probability that the largest number appearing on a selected coupon is 9, is

Total number of ways

$={15}^{7}$

According to the question we must select

$7$ coupons numbered from

$1$ to

$9$ such that

$9$ is selected at least once.

Total number of favourable ways

$={9}^{7}-{8}^{7}$

Required probability

$=\dfrac{{9}^{7}-{8}^{7}}{{15}^{7}}$

Die $A$ has $4$ red and $2$ white faces whereas die $B$ has $2$ red and $4$ white faces. A coin is flipped once. If it shows a head, the game continues by throwing die $A$ , if it shows tail, then die $B$ is to be used. If the probability that die $A$ is used is $\displaystyle\frac{64}{66}$ where it is given that red turns up every time in first n throws, then $n$ is

Let

$R$ be the event that a red face appears in each of the first

$n$ throws

$E_{1}:$ Die

$A$ is used when head has already fallen.

$E_{2}:$ Die

$B$ is used when tail has already fallen.

$P\left (\displaystyle{R}|{E_{1}} \right )=\left (\displaystyle\frac{2}{3} \right )^{n}$ and

$P\left (\displaystyle{R}|{E_ {2}} \right )=\left (\displaystyle\frac{1}{3} \right )^{n}$

Given,

$P(E_{1}|R)=\displaystyle \frac{64}{66}$

$\Rightarrow \displaystyle\frac{P(E_{1})P\left (\displaystyle{R}|{E_{1}} \right )}{P(E_{1}).P({R}|{E_{1}})+P(E_{2}).P\displaystyle ({R}|{E_{2}})}=\displaystyle\frac{64}{66}$

$\displaystyle\frac{\displaystyle\frac{1}{2}\left (\displaystyle\frac{2}{3} \right )^{n}}{\displaystyle\frac{1}{2}.\left (\displaystyle\frac{2}{3} \right )^ {n}+\displaystyle\frac{1}{2}.\left (\displaystyle\frac{1}{3} \right )^{n}}=\displaystyle\frac{32}{33}$ (Bayes' theorem)

$\Rightarrow \displaystyle\frac{2^{n}}{2^ {n}+1}=\displaystyle\frac{32}{33}$

This can be written as

$\dfrac{2^n}{2^n + 1} = \dfrac{2^5}{2^5 + 1}$

$\Rightarrow n=5$

A lot contains $20$ articles. The probability that the lot contains exactly $2$ defective articles is $0\cdot4$ and that lot contains exactly $3$ defective articles is $0\cdot6$ . Articles are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. The probability that testing procedure ends at the twelfth testing is $\cfrac {11k}{1900}$ . Find the value of $k$ ?

Let

${ A }_{ 1 }=$ the event that 'the lot contain 2 defective articles'

${ A }_{ 2 }=$ the event that 'the lot contains 3 defective articles'

$A=$ the event that ' the testing procedure ends with the twelfth testing'.
Now

$P\left( { A }_{ 1 } \right) =0.4$ and

$P\left( { A }_{ 2 } \right) =0.6$
The testing procedure ending at the twelfth testing means that, one defective article must be found in the first eleven testing and the remaining one must be found at the twelfth testing, in case the lot contain 2 defective articles.
So

$\displaystyle P\left( \frac { A }{ { A }_{ 1 } } \right) =\frac { _{ }^{ 2 }{ { C }_{ 1 } }\times ^{ 18 }{ { C }_{ 10 } } }{ ^{ 20 }{ { C }_{ 11 } } } \times \frac { 1 }{ 9 }$
Similarly

$\displaystyle P\left( \frac { A }{ { A }_{ 2 } } \right) =\frac { _{ }^{ 3 }{ { C }_{ 2 } }\times ^{ 17 }{ { C }_{ 7 } } }{ ^{ 20 }{ { C }_{ 11 } } } \times \frac { 1 }{ 9 }$
Thus the required probability

$\displaystyle =P\left( A \right) =P\left( A\cap { A }_{ 1 } \right) +P\left( A\cap { A }_{ 2 } \right) =P\left( { A }_{ 1 } \right) .P\left( \frac { A }{ { A }_{ 1 } } \right) +P\left( { A }_{ 2 } \right) .P\left( \frac { A }{ { A }_{ 2 } } \right)$

$\displaystyle =0.4\times \frac { _{ }^{ 2 }{ { C }_{ 1 } }\times ^{ 18 }{ { C }_{ 10 } } }{ ^{ 20 }{ { C }_{ 11 } } } \times \frac { 1 }{ 9 } +0.6\times \frac { _{ }^{ 3 }{ { C }_{ 2 } }\times ^{ 17 }{ { C }_{ 7 } } }{ ^{ 20 }{ { C }_{ 11 } } } \times \frac { 1 }{ 9 }$

$\displaystyle =0.4\times \frac { 11 }{ 190 } +0.6\times \frac { 11 }{ 228 } =\frac { 44 }{ 1900 } +\frac { 66 }{ 2280 } =\frac { 99 }{ 1900 } \\ \therefore k=1$

The probability that the total is $6$ is $\cfrac{k}{108}$ .Find $k$ ?

The probability that the total is

66 when three dice are rolled

Favourable outcomes are

$(1,2,3) no. of ways of appearance of 1,2,3=3!=6$

(1,1,4) no. of ways of appearance of

$1,1,4=3!/2!=3$

(2,2,2) no. of ways of appearance of

$2,2,2=3!/3!=1$

$Total=10$

$Total outcome=6*6*6=216$

$Probability=10/216=5/108$

$k=5$

A man takes a step forward with probability $0.4$ and backward with probability $0.6$ . Let $k$ be the probability that at the end of eleven steps he is one step away from the starting point.Find the first digit after the decimal of $k$ ?

$Either\quad he\quad took\quad 6\quad forward\quad steps\quad and\quad 5\quad backward\quad steps\quad or\quad 6\quad backward\quad steps\quad and\quad 5\quad forward\quad steps\\ =11C6*(0.4)^{ 6 }*(0.5)^{ 5 }+11C6*(0.4)^{ 5 }*(0.6)^{ 5 }\\ =11C5*(0.4)^{ 5 }*(0.6)^{ 5 }*(0.4+0.6)...[11C6=11C5]\\ =(11*10*9*8*7)/(1*2*3*4*5)*(0.24)^{ 5 }\\ =0.368.$

A lot contains $20$ articles. The probability that the lot contains exactly $2$ defective articles is $0.4$ and the probability that the lot contains exactly $3$ defective articles is $0.6$ . Article are drawn from the lot at random one by one without replacement and are tested till all defective articles are found. What is the probability that the testing procedure ends at the twelfth testing?

${\textbf{Step -1: Assume events, probability and write their values.}}$

${\text{Let}}$

$A_1 =$

${\text{The event that the lot contain 2 defective articles.}}$

${\text{Let}}$

$A_2 =$

${\text{The event that the lot contain 3 defective articles.}}$

${\text{Let}}$

$A =$

${\text{The event that the testing procedure ends with the twelfth testing.}}$

${\text{Now,}}$

$P(A_1) = 0.4$

${\text{and}}$

$P(A_2) = 0.6$

${\text{The testing procedure ending at the twelfth testing means that one defective article must be}}$

${\text{found in the first eleven testing and the remaining one must be found at the twelfth testing,}}$

${\text{in case the lot contains 2 defective articles.}}$

${\textbf{Step -2: Find probability using combination method.}}$

${\text{So,}}$

$P( \dfrac{A}{A_1}) = \dfrac{^2C_1 \times ^{18}C_{10}}{^{20}C_{11}} \times \dfrac{1}{9}$

${\text{Similarly,}}$

$P( \dfrac{A}{A_2}) = \dfrac{^3C_2 \times ^{17}C_{7}}{^{20}C_{11}} \times \dfrac{1}{9}$

${\textbf{Step -3: Find probability.}}$

${\text{Thus, the required probability is:}}$

$P(A) = P(A \cap A_!) + P(A \cap A_2)$

$= P(A_1).P( \dfrac{A}{A_1})$

$+ P(A_2).P( \dfrac{A}{A_2})$

$=0.4 \times \dfrac{^2C_1 \times ^{18}C_{10}}{^{20}C_{11}} \times \dfrac{1}{9} + 0.6 \times \dfrac{^3C_2 \times ^{17}C_{7}}{^{20}C_{11}} \times \dfrac{1}{9}$

$=0.4 \times \dfrac{11}{190} + 0.6 \times \dfrac{11}{228}$

$= \dfrac{44}{1900} + \dfrac{66}{2280}$

$= \dfrac{99}{1900}$

${\textbf{Thus, the probability that the testing procedure ends at the twelfth testing is}}$ $\boldsymbol{\dfrac{99}{1900}}.$

A box contains $2$ fifty paise coins, $5$ twenty five paise coins and a certain fixed number $n(\ge 2)$ of ten and five paise coins. Five coins are taken out of the box at random.The probability that the total value of these five coins is less than Re. $1$ & $50$ paise is $1-\left\{ \cfrac { 10(k+n) }{ { _{ }^{ n+7 }{ C } }_{ 5 } } \right\}$ .Find $k$ ?

There are

$(n+7)$ coins in the box out of which five coins can be taken out in

$^{ n+7 }{ C }_{ 5 }$ ways.
The total value of

$5$ coins can be equal to or more than one rupee and fifty paise in the following ways.
1) when one

$50$ paise coin and four

$25$ paise coins are chosen.
2) when two

$50$ paise coins and three

$25$ paise coins are chosen.
3) when two

$50$ paise, two

$25$ paise coins and one from

$n$ coins of ten and five paise.
Therefore the total number of ways of selecting five coins so that the total value of the coins is not less than one rupee and fifty paise is,

$\left( ^{ 2 }{ C }_{ 1 }.^{ 5 }{ C }_{ 5 }.^{ n }{ C }_{ 0 } \right) +\left( ^{ 2 }{ C }_{ 2 }.^{ 5 }{ C }_{ 3 }.^{ n }{ C }_{ 0 } \right) +\left( ^{ 2 }{ C }_{ 2 }.^{ 5 }{ c }_{ 2 }.^{ n }{ C }_{ 1 } \right) \\ =10+10+10n=10\left( n+2 \right)$
So, the number of ways of selecting five coins, so that the total value of the coins is less than one rupee and fifty paise is

$^{ n+7 }{ C }_{ 5 }-10\left( n+2 \right)$
Therefore required probability

$\displaystyle =\frac { ^{ n+7 }{ C }_{ 5 }-10\left( n+2 \right) }{ ^{ n+7 }{ C }_{ 5 } } =1-\frac { 10\left( n+2 \right) }{ ^{ n+7 }{ C }_{ 5 } }$

$\Rightarrow k=2$

Four small squares on a chess board are selected at random. the probability that they form a square of the size 2 X 2 is $\dfrac{a}{b}$ , where HCF of a and b is 1, then $a+b=?$

Total number of ways of selecting

$4$ squares is

$64$ .
Number of ways of selecting square of size

$2\times 2$

$=7\times 7 =49$

$\therefore$ Required probability

$=\displaystyle\frac { 49 }{ _{ }^{ 64 }{ { C }_{ 4 } } } =\frac { 7 }{ 90768 }$
Therefore

$a+b=7+90768=90775$

There is $30$ % chance that it rains on any particular day. What is the probability that there is at least one rainy day within a period of $7$ days? Given that there is at least one rainy day, The probability that these are at least two rainy days is $(1-{ \left( 0.7 \right) }^{ 7 });\cfrac { 1-k{ \left( 0.7 \right) }^{ 7 } }{ 1-{ \left( 0.7 \right) }^{ 7 } }$ . Find the value of $k$ .

Let A= event that there is at least one rainy day in a week.
And B= event that there are at least two rainy days in a week.

$\displaystyle \therefore P\left( \frac { B }{ A } \right) =\frac { p\left( A\cap B \right) }{ P\left( A \right) } =\frac { P\left( B \right) }{ P\left( A \right) }$ or

$B\subseteq A$
=(1-P(one rainy day or no rainy day))/(1-P(no rainy day))

$\displaystyle =\frac { 1-\left[ ^{ 7 }{ { C }_{ 1 } }{ \left( 0.3 \right) }^{ 2 }{ \left( 0.7 \right) }^{ 2 }+^{ 7 }{ { C }_{ 0 } }{ \left( 0.7 \right) }^{ 2 } \right] }{ 1-{ \left( 0.7 \right) }^{ 7 } } =\frac { 1-\left( 2.8 \right) { \left( 0.7 \right) }^{ 6 } }{ 1-{ \left( 0.7 \right) }^{ 7 } }$

In a test an examinee either guesses or copies or knows the answer to a multiple choice question with 4 choices. The probability that he makes a guess is 1/3 & the probability that he copies the answer is 1/The probability that his answer is correct given that he copied it, is 1/Find the probability that he knew the answer to the question given that he correctly answered it.
If expressed in the form of $a/b$ (simplest form), $b-a = ?$

Let define the following events:

${ E }_{ 1 }:$ Examine guesses the answer to the question.

${ E }_{ 2 }:$ Examine copies the answer to the question.

${ E }_{ 3 }:$ Examine know the answer to the question.

$E:$ answer is correct
Here

$\displaystyle P\left( { E }_{ 1 } \right) =\frac { 1 }{ 3 } ,P\left( { E }_{ 2 } \right) =\frac { 1 }{ 6 }$

$\displaystyle \therefore P\left( { E }_{ 3 } \right) =1-\left( \frac { 1 }{ 3 } +\frac { 1 }{ 6 } \right) =\frac { 1 }{ 2 }$
Since the question is a multiple choice question with four choice so the probability that the answer is correct
when is guessed is

$\displaystyle P\left( \frac { E }{ { E }_{ 1 } } \right) =\frac { 1 }{ 4 }$
Also the probability that his answer is correct, given that he copied it is

$\displaystyle \frac { 1 }{ 8 }$
i.e

$\displaystyle P\left( \frac { E }{ { E }_{ 2 } } \right) =\frac { 1 }{ 8 }$
Moreover his answer is correct, given he knew the answer is a sure event, so

$\displaystyle P\left( \frac { E }{ { E }_{ 3 } } \right) =1$
Hence by Baye's theorem the required probability is

$\displaystyle P\left( \dfrac { { E }_{ 3 } }{ E } \right) =\dfrac { P\left( \dfrac { E }{ { E }_{ 3 } } \right) .P\left( { E }_{ 3 } \right) }{ \sum _{ i=1 }^{ 3 }{ P\left( \dfrac { E }{ { E }_{ i } } \right) .P\left( { E }_{ i } \right) } } =\dfrac { 1.\dfrac { 1 }{ 2 } }{ \dfrac { 1 }{ 4 } .\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 8 } .\dfrac { 1 }{ 6 } +\dfrac { 1 }{ 2 } .1 } =\dfrac { 24 }{ 29 }$

If two events $A$ and $B$ such that $P({ A }^{ c })=0.3,P(B)=0.4$ and $P(A\cap { B }^{ c })=0.5$ , $P[B/A\cup { B }^{ c })]=\frac{1}{k}$ ,then value of k is

$P\left( A\cup { B }^{ c } \right) =P\left( A \right) +P\left( { B }^{ c } \right) -P\left( A\cap { B }^{ c } \right) \\ =1-P\left( { A }^{ c } \right) +P\left( { B }^{ c } \right) -P\left( A\cap { B }^{ c } \right) \\ =0.7+0.6-0.5=0.8$
Therefore,

$P\left( B|A\cup { B }^{ c } \right) =\cfrac { P\left( B\cap \left( A\cup { B }^{ c } \right) \right) }{ P\left( A\cup { B }^{ c } \right) } =\cfrac { P\left[ \left( B\cap A \right) \cup \left( B\cap { B }^{ c } \right) \right] }{ P\left( A\cup { B }^{ c } \right) } \\ =\cfrac { P\left( A\cap B \right) }{ P\left( A\cup { B }^{ c } \right) } =\cfrac { P\left( A \right) -P\left( A\cap { B }^{ c } \right) }{ P\left( A\cup { B }^{ c } \right) } =\cfrac { 0.7-0.5}{ 0.8 } =0.25$

The probabilities of three events $A,B$ and $C$ are $P(A)=0.6, P(B)=0.4$ and $P(C)=0.5$ . If $P(A\cup B)=0.8, P(A\cap C)=0.3, P(A\cap B \cap C)=0.2$ and $P(A\cup B \cup C) \ge 0.85$ , $0.04k \le P(B\cap C) \le 0.07k$ . Find the value of $k$ .

$P\left( A\cap B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cup B \right) =0.6+0.4-0.8=0.2$

$P\left( A\cup B\cup C \right) =P\left( A \right) +P\left( B \right) +P\left( C \right) -P\left( A\cap B \right) +P\left( A\cap B\cap C \right) -P\left( A\cap B \right) -P\left( B\cap C \right) \\ \Rightarrow P\left( B\cap C \right) =1.2-P\left( A\cup B\cup C \right) \\ \because 0.85\le P\left( A\cup B\cup C \right) \le 1\\ \therefore \left( 1 \right) \Rightarrow P\left( B\cap C \right) \le 1.2-1-0.85$
and

$P\left( B\cap C \right) \ge 1.2-1\Rightarrow 0.2\le P\left( B\cap C \right) \le 0.35.$

An urn contairis 2 white and 2 black balls. A ball is drawn at random.If it is white, it is not replaced into the urn, otherwise it is replaced along with another ball of the same colour.The process is repeated.The probability that the third ball drawn is black is $1-\displaystyle \frac{k}{30}$ . Find the value of $k$ .

The first ball can be drawn in the following four possible ways;
(i) Both the first and the second balls drawn are white.
(ii)The first ball drawn is white and the second ball drawn is black.
(iii) The first ball drawn is black and the second ball drawn is white.
(iv) Both the first and the second balls drawn are black.
Let the events in (i),(ii),(iii),and (iv)be respectively denoted by

$\displaystyle E_{1},E_{2},E_{3}and E_{4}.$
Also let e denote the event tbat the third ball drawn in black.
Then

$\displaystyle P\left ( E _{1}\right )=\frac{2}{4}\times \frac{1}{3}=\frac{1}{6},P\left ( E_{2} \right ) =\frac{2}{4}\times \frac{2}{3}=\frac{1}{3},$

$\displaystyle P\left ( E _{3}\right )=\frac{2}{4}\times \frac{2}{5} =\frac{1}{5},P\left ( E_{4} \right )=\frac{2}{4}\times \frac{3}{5}=\frac{3}{10},$
Also

$\displaystyle P\left ( E/E_{1} \right )=\frac{2}{5}=1,$
since when the event

$\displaystyle E_{1}$ has already happened,
i.e,the first two balls drawn are both white,
they are not replaced and so there are left 2 black balls in the urn so that the probability that the third ball drawn in this case is black =2/2=1.
Again

$\displaystyle P\left ( E/E_{1} \right )=3/4$
Note that when the event

$\displaystyle E_{2}$ has already happened there are 3 black and one white balls in the uren.
So in this case tbe probability that the third ball drawn is black =3/4.
Similarly

$\displaystyle P\left ( E/E_{3} \right )=3/4$ and

$P\left ( E/E_{4} \right )=2/3.$
Now by the theorem of total probability for compound events ,we have

$\displaystyle P\left ( E \right )=P\left ( E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{4} \right )$

$\displaystyle P\left ( E \right ) =P\left ( E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{4} \right )$

$\displaystyle =\frac{1}{6}\times 1+\frac{1}{3}\times \frac{3}{4}+\frac{1}{5}\times \frac{3}{4}+\frac{3}{10}\times\frac{2}{3}=\frac{1}{6}+\frac{1}{4}+\frac{3}{20}+\frac{1}{5}=\frac{23}{30}.$

An urn contains five balls. Two balls are drawn and are found to be white. The probability that all the balls are white is $\displaystyle \frac{1}{k}$ . Find the value of $k$ .

Let

$\displaystyle A_{i}\left ( i= 1,2,3,4 \right )$ be the event that the urn contains

$2, 3, 4$ or

$5$ white ball an db the event that two white balls are drawn.

We have

$t$ find

$\displaystyle P\left ( A_{4}/B \right ).$

Since the four events

$\displaystyle A_{1}, A_{2}, A_{3}, A_{4}$ are equally likely,

we have

$\displaystyle P\left ( A_{1} \right )= P\left ( A_{2} \right )= P\left ( A_{3} \right )= P\left ( A_{4} \right )= \dfrac{1}{4}.$

$\displaystyle P\left ( B/A_{1} \right )$ is the probability of event that the urn contains

$2$ white balls and both have been drawn.

Hence

$\displaystyle P\left ( B/A_{1} \right )= \dfrac{^{2}C_{2}}{^{5}C_{2}}= \dfrac{1}{10}.$

Similarly,

$\displaystyle P\left ( B/A_{2} \right )= \dfrac{^{3}C_{2}}{^{5}C_{2}}= \dfrac{3}{10}.$

$\displaystyle P\left ( B/A_{3} \right )= \dfrac{^{4}C_{2}}{^{5}C_{2}}= \dfrac{6}{10}= \dfrac{3}{5}.$ and

$\displaystyle P\left ( B/A_{4} \right )= \dfrac{^{5}C_{2}}{^{5}C_{2}}= 1.$

$\displaystyle \therefore$ By Baye's theorem

$\displaystyle P\left ( A_{4}/B \right )= \dfrac{P\left ( A_{4} \right )P\left ( B/A_{4} \right )}{\sum_{i= 1}^{4}P\left ( A_{1} \right )P\left ( B/A_{i} \right )}$

$\displaystyle = \dfrac{\dfrac{1}{4}.1}{\dfrac{1}{4}\left ( \dfrac{1}{10}+\dfrac{3}{10}+\dfrac{3}{5}+1 \right )}= \dfrac{1}{2}.$

It is known that an urn containing altogether 10 balls was filled in the following manner: A coin was tossed 10 times, and according as it showed heads or tails, one white or on black ball was put into the urn. Balls are drawn from this urn one at a time, 10 times in succession (with replacements) and every one turns out to be white. The chance that the urn contains nothing but white balls is $\displaystyle \frac{1}{2^{k}}$ . Find the value of $k$ .

Let

$\displaystyle B_{k}$ denote the event that the urn contains

$k$ white balls, where

$k$ assumes values from

$0$ to

$10$ .

$10$ throws, if the coin show

$k$ heads and

$10-k$ tails, then

$\displaystyle P\left ( B_{k} \right )= ^{10}C_{k}\left ( \dfrac{1}{2} \right )^{k}\left ( \dfrac{1}{2} \right )^{10-k}= ^{10}C_{k}\dfrac{1}{2^{10}}$
Let

$A$ denote the event of drawing a white ball, then.

$\displaystyle P\left ( A\mid B_{k} \right )= \dfrac{k}{10}$ where

$k=0, 1, 2, ..., 10$
Hence using Baye's formula

$\displaystyle p\left ( B_{10}\mid A \right )= \dfrac{P\left ( B_{10} \right )P\left ( A\mid B_{10} \right )}{\displaystyle\sum_{k= 0}^{10}P\left ( B_{k} \right )P\left ( A\mid B_{k} \right )}$

$=\displaystyle\dfrac{\left(\ ^{10}C_{10}\dfrac1{2^{10}}\right)\dfrac{10}{10}}{\displaystyle\sum_{k=0}^{10}\ ^{10}C_k\dfrac1{2^{10}}\dfrac k{10}} = \dfrac1{2^9}$

applying the formula in the denominator

$\displaystyle n.2^{n-1}= \sum_{k= 1}^{n} k.^{n}C_{k}$ of binomial theorem with

$n=10.$

A box contains $100$ tickets, numbered $1, 2, ...., 100$ . Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more thanThe minimum number of them is $2$ with probability $(15-k)/15$ . Find the value of $k.$

Let

$A$ be the event that the maximum number on the two chosen tickets is not more than

$10$ ,
that is, the no. on them

$\displaystyle \leq$

$10$ and

$B$ the event that the minimum no.
On them is

$5$ , that is, the no them is

$\displaystyle \geq$

$5$ .
We have to find

$P(B/A)$ .
Now

$\displaystyle P\left ( B/A \right )=\frac{P\left ( A\cap B \right )}{P\left ( A \right )}=\frac{n\left ( A\cap B \right )}{n\left ( A \right )}.$
Now the number of ways of getting a number

$r$ on the two tickets is the
coeff. of

$\displaystyle x^{r}$ in the expansion of

$\displaystyle \left ( x^{1}+x^{2}+....+x^{100} \right )^{2}=x^{2}\left ( 1+x+....+x^{99} \right )^{2}$

$\displaystyle x^{2}\left ( \frac{1-x^{100}}{1-x} \right )^{2}=x^{2}\left ( 1-2x^{100}+x^{200} \right )\left ( 1-x \right )^{-2}$

$\displaystyle x^{2}\left ( 1-2x^{100}+x^{200} \right )\left ( 1+2x+3x^{2}+\cdots \left ( r+1 \right )x^{r} +\cdots \right )$
Thus coeff. of

$\displaystyle x^{2}=1, of x^{3}=2, of x^{4}=3,\cdots ,of x^{10}$ is

$9$ .
Hence

$n(A)=1+2+3+4+5+6+7+8+9=45$ and

$\displaystyle n\left ( A\cap B \right )=4+5+6+7+8+9=39.$
[Note that in finding n(A), we have to add the coefficients of

$\displaystyle x^{2},x^{3},\cdots ,x^{10}$ and in

$n\displaystyle \left ( A\cap B \right )$ we add the coefficients of

$\displaystyle x^{2},x^{3},\cdots x^{10}$ ].
Hence required prob.

$\displaystyle =P\left ( B/A \right )=\frac{39}{45}=\frac{13}{15}.$

A bag $A$ contains $2$ white and $3$ red balls and a bag $B$ contains $4$ white and $5$ red balls. One ball is drawn at random from one of the bags and is found to be red. The probability that it was drawn from bag $B$ is $\dfrac {25}X$ , then find the value of X.

Let

$\displaystyle E_{1}$ be the event that the ball is drawn from bag

$A$ ,

$\displaystyle E_{2}$ the event that it is drawn from bag

$B$ and

$E$ that the ball is red.

We have to find

$\displaystyle P\left ( E_{2}\mid E\right ).$
Since both the bags are equally likely to be selected, we have

$\displaystyle P\left ( E_{1} \right )= P\left ( E_{2} \right )= \frac{1}{2}.$

Also

$\displaystyle P\left ( E/E_{1} \right )= \dfrac35$ and

$\displaystyle P\left ( E/E_{2} \right )= \dfrac{5}{9}$ .

Hency by Baye's theorem, we have

$\displaystyle P\left ( E_{2}/E \right )= \dfrac{P\left ( E_{2} \right )P\left ( E/E_{2} \right )}{P\left ( E_{1} \right )P\left ( E/E_{1} \right )+P\left ( E_{2} \right )P\left ( E/E_{2} \right )}$

$\displaystyle = \dfrac{\dfrac{1}{2}.\dfrac{5}{9}}{\dfrac{1}{3}.\dfrac{3}{5}+\dfrac{1}{2}.\dfrac{5}{9}}= \dfrac{25}{52}.$

Now comparing the given answer with

$\dfrac{25}X$

$\therefore X=52$

Note. If in the problem, you read' is found to be' then it is the problem on Baye's Theorem.

A lot contains 20 articles. The probability that the lot contains exactly 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.Articles are drawn from the lot at random one by one without replacement and tested till all the defective articles are found. The probability that the testing procedure ends at the twelfth testing $\displaystyle \frac{100-k}{1900}$ . Find the value of $k$ .

${\textbf{Step -1: Assume events, probability and write their values.}}$

${\text{Let}}$

$A_1 =$

${\text{The event that the lot contain 2 defective articles.}}$

${\text{Let}}$

$A_2 =$

${\text{The event that the lot contain 3 defective articles.}}$

${\text{Let}}$

$A =$

${\text{The event that the testing procedure ends with the twelfth testing.}}$

${\text{Now,}}$

$P(A_1) = 0.4$

${\text{and}}$

$P(A_2) = 0.6$

${\text{The testing procedure ending at the twelfth testing means that one defective article must be}}$

${\text{found in the first eleven testing and the remaining one must be found at the twelfth testing,}}$

${\text{in case the lot contains 2 defective articles.}}$

${\textbf{Step -2: Find probability using combination method.}}$

${\text{So,}}$

$P( \dfrac{A}{A_1}) = \dfrac{^2C_1 \times ^{18}C_{10}}{^{20}C_{11}} \times \dfrac{1}{9}$

${\text{Similarly,}}$

$P( \dfrac{A}{A_2}) = \dfrac{^3C_2 \times ^{17}C_{7}}{^{20}C_{11}} \times \dfrac{1}{9}$

${\textbf{Step -3: Find probability.}}$

${\text{Thus, the required probability is:}}$

$P(A) = P(A \cap A_!) + P(A \cap A_2)$

$= P(A_1).P( \dfrac{A}{A_1})$

$+ P(A_2).P( \dfrac{A}{A_2})$

$=0.4 \times \dfrac{^2C_1 \times ^{18}C_{10}}{^{20}C_{11}} \times \dfrac{1}{9} + 0.6 \times \dfrac{^3C_2 \times ^{17}C_{7}}{^{20}C_{11}} \times \dfrac{1}{9}$

$=0.4 \times \dfrac{11}{190} + 0.6 \times \dfrac{11}{228}$

$= \dfrac{44}{1900} + \dfrac{66}{2280}$

$= \dfrac{99}{1900}$

$= \dfrac{100 - 1}{1900}$

$\text{Given,}$

$\Rightarrow P(k)= \dfrac{100 - k}{1900}$

$\text{By comparing both the results we find the value of k as 1.}$

${\textbf{Hence, the value of k is 1.}}$

Let 0.8 the probability that a man aged 90 years will die in a year., If p is the probability that out of 4 men $A_1, A_2, A_3$ and $A_4$ each aged 90, $A_1$ will die in a year and will be the first to die, find 10000p

Let

$E_i$ denote the event that

$A_i$ dies in a year.
Then

$P(E_i)=0.8$ and

$P(E_i)=0.2$ for

$i=1, 2, 3, 4$ .
Now,P (none of

$A_1, A_2, A_3, A_4$ dies in a year)

$=P(E'_1\cap E'_2\cap E'_3\cap E'_4)=(0.2)^4$
Let E denote the event that at least one of

$A_1, A_2, A_3, A_4$ dies in a year.
Then

$P(E)=1-P(E'_1\cap E'_2\cap E'_3\cap E'_4)=1-(0.2)^4=0.9984$ .
If F denote the event that

$A_1$ is first to die.
Then

$P(F|E)=1/4$ . Also,

$\displaystyle p=\frac {1}{4}(0.9984)=0.2496$

Two integers $r$ and $s$ are drawn one at a time without replacement from the set $1, 2, .... n$ . If $p_k=P(r\leq k\ |\ s\leq k)$ , find $4p_7$ if $n=25$

We have

$P(r\leq k|s\leq k)=\displaystyle \frac {P[(r\leq k)and (s\leq k)]}{P(s\leq k)}$
Also,

$P(s\leq k)=\displaystyle \frac {k}{n}$ and

$P[(r\leq k) and (s\leq k)]=\displaystyle \frac {^kC_2}{^nC_2}=\frac {k(k-1)}{n(n-1)}$
Thus,

$p_k=(k-1)/(n-1)$
For

$n=25$ and

$k=7$

$p_{ 7 }=\cfrac { (7-1) }{ (25-1) } =\cfrac { 6 }{ 24 } =\cfrac { 1 }{ 4 } \\ \Rightarrow 4p_{ 7 }=1$

Two numbers are selected at random from the number $1, 2, ..., n$ . Let $p$ denote the probability that the difference between the first and second is not less than $m$ (where $0 < m < n$ ). If $n=25$ and $m=10$ , find $5p$

Let the first number be

$x$ and the second be

$y$ .
Let

$A$ denote the event that the difference between the first and second numbers is at least

$m$ .
Let

$E_x$ denote the event that the first number chosen is

$x$ .
We must have

$x-y\geq m$ or

$y\leq x-m$ .

Therefore

$x > m$ and

$y < n - m$ .
Thus,

$P(E_x)=0$ for

$0< x\leq m$ and

$P(E_x)=1/n$ for

$m < x\leq n$ .

Also,

$P(A|E_x)=\dfrac{(x-m)}{(n-1)}$ .
Therefore,

$\displaystyle P(A)=\sum_{x=1}^nP(E_x) P(A|E_x)$

$\displaystyle =\sum_{x=m+1}^nP(E_x) P(A|E_x)=\sum_{x=m+1}^n\frac {1}{n}\cdot \frac {x-m}{n-1}$

$\displaystyle =\frac {1}{n(n-1)}[1+2+...+(n-m)]=\frac {(n-m)(n-m+1)}{2n(n-1)}$
Putting

$n=25$ and

$m=10$ ,

$\displaystyle P(A)=\frac { (25-10)(25-10+1) }{ 2.25(25-1) } =\frac { 15.16 }{ 2.25.24 } =\frac { 1 }{ 5 }$

$\Rightarrow 5p=1$

A certain drug, manufactured by a Company is tested chemically for its toxic nature. Let the event "THE DRUG IS TOXIC" be denoted by H & the event "THE CHEMICAL TEST REVEALS THAT THE DRUG IS TOXIC" be denoted by S. Let $P(H) = a, P(S/H) = P (\overline S / H) = 1 - a$ . Then show that the probability that the drug is not toxic given that the chemical test reveals that it is toxic, is $\displaystyle \frac{k_1}{k_2}$ (HCF of k_1 and k_2 is 1) then find the value of $k_1+k_2$ for $a=3$ , for

$P(\overline H/S)=\dfrac {P(\overline HS)}{P(S)}$ .....(1)
also $\dfrac {P(HS)}{P(H)}=1-a\Rightarrow P(HS)=a(1-a)$ .......(2)
also $\dfrac {P(\overline S\overline H)}{P(\overline H)}=1-aP(\overline S\overline H)=(1-a)^2$
$P(\overline S\cap \overline H)=P(\overline {S\cup H})=(1-a)^2$
$\Rightarrow 1 [P(S) + P(H) P(SH)] = (1 a)^2$
$\Rightarrow [P(S) + a a + a^2] = 2a a^2$
$P(S) = 2a(1 a)$ ... (3)
also $P(\overline HS) = P(S) P(SH)$
$= 2a(1 a) a(1 a) = a(1 a)$ ... (4)
from (3) & (4)

$P(\overline H / S)=\dfrac {1}{2}$

Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than 4' given that 'there is at least one tail'.

The sample space of experiment may be described as

$S={(H,H),(H,T),(T,1),(T,2) (T,3) (T,4) (T,5) (T,6)}$
where

$(H,H)$ denotes that both the tosses result into head and

$(T,i)$ denote the first toss result into a tail and the number 'i' appeared on the die for

$i=1,2,3,4,5,6$ .
Thus, the probabilities assigned to 8 elementary events are

$\displaystyle \frac {1}{4}, \frac {1}{4}, \frac {1}{12}, \frac {1}{12}, \frac {1}{12}, \frac {1}{12}, \frac {1}{12}, \frac {1}{12}$ respectively.
Let F be the event that 'there is atleast one tail' and E be the event 'the die shows a number greater than 4'. Then,

$F={(H,T),(T,1),(T,2),(T,3),(T,4),(T,5),(T,6)}$

$E={(T,5),(T,6)}$ and

$E\cap F={(T,5),(T,6)}$
Now,

$P(F)=P({(H,T)})+P({(T,1)})+P({(T,2)})+P({(T,3)})+P({(T,4)})+P({(T,5)})+P({(T,6)})$

$=\displaystyle \frac {1}{4}+ \frac {1}{12}+ \frac {1}{12}+ \frac {1}{12}+ \frac {1}{12}+ \frac {1}{12}+ \frac {1}{12}= \frac {3}{4}$
and

$P\left( E\cap F \right) =\quad P\left( \left\{ \left( T,5 \right) \right\} \right) +P\left( \left\{ \left( T,6 \right) \right\} \right) =\displaystyle \frac { 1 }{ 12 } +\frac { 1 }{ 12 } =\frac { 1 }{ 6 } \\$
Hence,

$P\left( { E }|{ F } \right) =\displaystyle \frac { P\left( E\cap F \right) }{ P\left( F \right) } =\frac { \dfrac { 1 }{ 6 } }{ \dfrac { 3 }{ 4 } } =\frac { 2 }{ 9 }$

A man is known to speak the truth $3$ out of $5$ times. He throws a die and reports that it is $'6'$ . The probability that it is actually ' $6$ ' is $\dfrac{3}{k}$ . Find the value of $k$ .

Let

$E$ be the event that the man reports that one occurs in throwing of the die and let

$S_{1}$ be the event that one occurs and

$S_{2}$ be the event that one does not occur.
Then,

$P(S_{1})=\displaystyle \frac {1}{6}$

$P(S_{2})=\displaystyle \frac {5}{6}$

$P\left( { E }|{ { S }_{ 1 } } \right)$ = Probability that the man reports that

$6$ occurs when six has actually occured on die =

$\displaystyle \frac {3}{5}$ = probability that man speaks truth.

$P\left( { E }|{ { S }_{ 2 } } \right)$ = Probability that the man reports that

$6$ occurs when

$6$ has not actually occured on the die =

$1-\displaystyle \frac {3}{5}= \frac {2}{5}$ = probability that man does not speak truth.
By Baye's theorem,

$P\left( { { S }_{ 1 } }|{ E } \right) =\displaystyle \frac { P\left( { S }_{ 1 } \right) P\left( { E }|{ { S }_{ 1 } } \right) }{ P\left( { S }_{ 1 } \right) P\left( { E }|{ { S }_{ 1 } } \right) +P\left( { S }_{ 2 } \right) P\left( { E }|{ { S }_{ 2 } } \right) } =\displaystyle \frac { \frac { 1 }{ 6 } \times \frac { 3 }{ 5 } }{ \frac { 1 }{ 6 } \times \frac { 3 }{ 5 } +\frac { 5 }{ 6 } \times \frac { 2 }{ 5 } } =\frac { 3 }{ 13 } \\$
Hence, required probability is

$\displaystyle \frac {3}{13}$ .

A prisoner escapes from a jail and is equally likely to choose one of the four roads I, II, III or IV to reach away from the hands of law. If he choose I road, he is successful with probability 1/6 and for II, III and IV this is 1/8, 1/10 and 1/If the prisoner is successful, the probability that he chose road

$P(E_{ 1 })=P(E_{ 2 })=P(E_{ 3 })=P(E_{ 4 })=\cfrac { 1 }{ 4 }$
And

$P(S|E_{ 1 })=\cfrac { 1 }{ 6 } ,P(S|E_{ 2 })=\cfrac { 1 }{ 8 } ,P(S|E_{ 3 })=\cfrac { 1 }{ 10 } ,P(S|E_{ 4 })=\cfrac { 1 }{ 12 }$
Using bayes theorem
A)

$P\left( E_{ 1 }|S \right) =\cfrac { P(S|E_{ 1 }).P(E_{ 1 }) }{ P(S|E_{ 1 }).P(E_{ 1 })+P(S|E_{ 2 }).P(E_{ 2 })+P(S|E_{ 3 }).P(E_{ 3 })+P(S|E_{ 4 }).P(E_{ 4 }) } =\cfrac { 20 }{ 57 }$
B)

$P\left( E_{ 2 }|S \right) =\cfrac { P(S|E_{ 2 }).P(E_{ 2 }) }{ P(S|E_{ 1 }).P(E_{ 1 })+P(S|E_{ 2 }).P(E_{ 2 })+P(S|E_{ 3 }).P(E_{ 3 })+P(S|E_{ 4 }).P(E_{ 4 }) } =\cfrac { 15 }{ 57 }$
C)

$P\left( E_{ 3 }|S \right) =\cfrac { P(S|E_{ 3 }).P(E_{ 3 }) }{ P(S|E_{ 1 }).P(E_{ 1 })+P(S|E_{ 2 }).P(E_{ 2 })+P(S|E_{ 3 }).P(E_{ 3 })+P(S|E_{ 4 }).P(E_{ 4 }) } =\cfrac { 12 }{ 57 }$
D)

$P\left( E_{ 4 }|S \right) =\cfrac { P(S|E_{ 4 }).P(E_{ 4 }) }{ P(S|E_{ 1 }).P(E_{ 1 })+P(S|E_{ 2 }).P(E_{ 2 })+P(S|E_{ 3 }).P(E_{ 3 })+P(S|E_{ 4 }).P(E_{ 4 }) } =\cfrac { 10 }{ 57 }$

Match the follwoing

(A) Total number of ways

$\displaystyle ={ 6 }^{ 5 }$

To find the favourable number of ways, a total of

$12$ in

$5$ throws can be obtained in the following two ways only

(i) One blank and four

$3'$ s. or, (ii) Three

$2'$ s and two

$3'$ s.

The number of ways in case (i)

$\displaystyle =^{ 5 }{ { C }_{ 1 } }=5$

and the number of ways in case (ii)

$\displaystyle =^{ 5 }{ { C }_{ 2 } }=10.$

$\therefore m=$ the favourable number of ways.

$\displaystyle =5+10=15.$

Hence, the required probability

$\displaystyle =\frac { 15 }{ { 6 }^{ 5 } }=\frac { 5 }{ 2592 } .$

(B) Required probability

$\displaystyle =\frac { ^{ 39 }{ { C }_{ 1 } } }{ ^{ 52 }{ { C }_{ 1 } } } \times \frac { ^{ 39 }{ { C }_{ 1 } } }{ ^{ 52 }{ { C }_{ 1 } } } \times \frac { ^{ 13 }{ { C }_{ 1 } } }{ ^{ 52 }{ { C }_{ 1 } } } =\frac { 3 }{ 4 } \times \frac { 3 }{ 4 } \times \frac { 1 }{ 4 } =\frac { 9 }{ 64 } .$

$W$ denote the event that

$A$ draws a white ball and

$T$ the event that

$A$ speaks truth.

In the usual notations, we are given that

$\displaystyle P\left( W \right) =\frac { 1 }{ 9 } ,P\left( \frac { T }{ W } \right) =\frac { 5 }{ 6 }$ so that

$\displaystyle P\left( \overline { W } \right) =1-\frac { 1 }{ 9 } =\frac { 8 }{ 9 }$ and

$\displaystyle P\left( \frac { T }{ \overline { W } } \right) =1-\frac { 5 }{ 6 } =\frac { 1 }{ 6 } .$

Using Baye's theorem required probability is given by

$\displaystyle P\left( \frac { W }{ T } \right) =\frac { P\left( W\cap T \right) }{ P\left( T \right) }$

$\displaystyle =\frac { P\left( W \right) \times P\left( \frac { T }{ W } \right) }{ P\left( W \right) \times P\left( \frac { T }{ W } \right) +P\left( \overline { W } \right) \times P\left( \frac { T }{ \overline { W } } \right) }$

$\displaystyle =\dfrac { \left( \dfrac { 1 }{ 9 } \right) \times \left( \dfrac { 5 }{ 6 } \right) }{ \left( \dfrac { 1 }{ 9 } \right) \times \left( \dfrac { 5 }{ 6 } \right) +\left( \dfrac { 8 }{ 9 } \right) \times \left( \dfrac { 1 }{ 6 } \right) } =\dfrac { 5 }{ 13 } .$

A random variable $X$ has the following probability distribution:
$X$
$0$
$1$
$2$
$3$
$4$
$5$
$6$
$7$
$P(X)$
$0$
$k$
$2k$
$2k$
$3k$
$k^{2}$
$2k^{2}$
$7k^{2}+k$
Determine
(i) $k$
(ii) $P\left ( X< 3 \right )$
(iii) $P\left ( X>6 \right )$
(iv) $P\left ( 0< X< 3 \right )$

(i) Since X is random variable.

$\sum { P(X_i)=1 }$

$0+k+2k+2k+3k+k^2+2k^2+7k^2+k=1\\\implies 10k^2+9k-1=0\\\implies 10k^2+10k-k-1=0\\\implies 10k(k+1)-1(k+1)=0\\\implies (k+1)(10k-1)=0\\\implies k=-1, k=\dfrac 1{10}$

$k=-1$ not possible, hence

$k=\dfrac 1{10}$

(ii)

$P(X<3)=P(X=0)+P(X=1)+P(X=2)\\\implies = 0+k+2k=3k\\\implies P(X<3)=3\times\dfrac 1{10}= \dfrac 3{10}$

(iii)

$P(X>6)=P(X=7)=7k^2+k=\dfrac 7{100}+\dfrac 1{10}\\\implies P(X>6)=\dfrac {7+10}{100}=\dfrac {17}{100}$

(iv)

$P(0<X<3)=P(X=1)+P(X=2)=k+2k=3k\\\implies P(0<X<3)=\dfrac 3{10}$

A family has three children. Event $A$ is that family has at most one boy, Event $'B'$ is that family has at least one boy and one girl, Event $'C'$ is that the family has at most one girl. Find whether events $'A'$ and $'B'$ are independent. Also find whether $A, B, C$ are independent or not.

A family of three children can have
(i) All

$3$ boys
(ii)

$2$ boys+

$1$ girl
(iii)

$1$ boy+

$2$ girls
(iv)

$3$ girls
(i)

$P$ (3 boys)

$\displaystyle =^{3}\textrm{C}_{0}\left ( \frac{1}{2} \right )^{3}=\frac{1}{8}$ (Since each child is equally likely to be aboy or a girl)
(ii)

$P$ (2 boys+1 girl)

$\displaystyle =^{3}\textrm{C}_{1}\times \left ( \frac{1}{2} \right )^{2}\times \frac{1}{2}=\frac{3}{8}$ (Note that there are three cases BBG, BGB, GBB)
(iii)

$P$ (1 boy + 2 girls)

$\displaystyle =^{3}\textrm{C}_{2}\times \left ( \frac{1}{2} \right )^{2}\times \left ( \frac{1}{2} \right )^{2}=\frac{3}{8}$
(iv)

$P$ (3 girls)

$\displaystyle =\frac{1}{8}$
Event

$'A'$ is associated with (iii) & (iv). Hence

$\displaystyle P\left ( A \right )=\frac{1}{2}$
Event

$'B'$ is associated with (ii) & (iii). Hence

$\displaystyle P\left ( B \right )=\frac{3}{4}$
Event

$'C'$ is associated with (i) & (ii). Hence

$\displaystyle P\left ( C \right )=\frac{1}{2}$

$\displaystyle P\left ( A\cap B \right )=P\left ( iii \right )=\frac{3}{8}=P\left ( A \right ).P\left ( B \right ).$ Hence

$A$ and

$B$ are independent of each other

$P\left ( A\cap C \right )=0\neq P\left ( A \right ).P\left ( C \right ).$ Hence

$A, B, C$ are not independent

An urn contains $5$ red and $5$ black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, $2$ additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red ?

${\textbf{Step -1: Define and identify probability distribution of random variable.}}$

${\text{The urn contains 5 red and 5 black balls.let a red ball be drawn in the first attempt.}}$

$\therefore$

${\text{P( drawing a red ball)}}$

$= \dfrac{5}{{10}}$

$=\dfrac{1}{{2}}$

${\text{ If two red balls are added to urn, then the urn contains 7red and 5 black balls.}}$

$\therefore$

${\text{P( drawing a red ball)}}$

$= \dfrac{7}{{12}}$

${\text{ Let a black ball be drawn in the first attempt.}}$

$\therefore$

${\text{P( drawing a black ball in first attempt )}}$

$= \dfrac{5}{{10}}$

$=\dfrac{1}{{2}}$

${\text{ If two black balls are added to urn, then the urn contains 5 red and 7 black balls.}}$

$\therefore$

${\text{P( drawing a red ball )}}$

$= \dfrac{5}{{12}}$

${\textbf{Step -2: Calculate the probability of drawing second ball as red.}}$

${\textbf{}}$

$=$

$\dfrac{1}{2}\times\dfrac{7}{{12}}\times\dfrac{1}{2}\times\dfrac{5}{{12}}$

$=$

$\dfrac{1}{2}\left( {\dfrac{7}{{12}}\times\dfrac{5}{{12}}} \right)$

$=\dfrac{1}{2}\left( {\dfrac{{7 + 5}}{{12}}} \right)$

$= \dfrac{1}{2} \times {1}$

$=\dfrac{1}{2}$

$= {0.5}$

${\textbf{Hence , The probability of drawing second ball as red is 0.5.} }$

One card is drawn at random from a well-shuffled deck of $52$ cards. In how many of the following cases are the events $E$ and $F$ independent?
(i) $E:$ 'the card drawn is a spade'
$F:$ 'the card is drawn is an ace'
(ii) $E:$ 'the card drawn is black'
$F:$ 'the card drawn is a king'
(iii) $E:$ 'the card drawn is a king or queen'
$F:$ 'the card drawn is a queen or jack'.

In a deck of

$52$ cards,

$13$ cards are spades and

$4$ cards are aces.

$\therefore P\left( E \right) = P\left( the card drawn is a spade \right) = \displaystyle\frac { 13 }{ 52 } = \displaystyle\frac { 1 }{ 4 }$

$\therefore P\left( F \right) = P\left( the card drawn is an ace \right) = \displaystyle\frac { 4 }{ 52 } = \displaystyle\frac { 1 }{ 13 }$
In the deck of cards, only

$1$ card is an ace of spades.

$P\left( EF \right) = P\left( the card drawn is spade and an ace \right) = \displaystyle\frac { 1 }{ 52 }$

$P\left( E \right) \times P\left( F \right) = \displaystyle\frac { 1 }{ 4 } \cdot \displaystyle\frac { 1 }{ 13 } = \displaystyle\frac { 1 }{ 52 } = P\left( EF \right)$

$\Rightarrow P\left( E \right) \times P\left( F \right) =P\left( EF \right)$
Therefore, the events

$E$ and

$F$ are independent.
(ii) In a deck of

$52$ cards,

$26$ cards are black and

$4$ cards are kings.

$\therefore P\left( E \right) =P\left( the card drawn is black \right) =\displaystyle\frac { 26 }{ 52 } = \displaystyle\frac { 1 }{ 2 }$

$\therefore P\left( F \right) =P\left( the card drawn is a king \right) = \displaystyle\frac { 4 }{ 52 } = \displaystyle\frac { 1 }{ 13 }$
In the pack of

$52$ cards,

$2$ cards are black as well as kings.

$\therefore P\left( EF \right) =P\left( the card drawn is a black king \right) = \displaystyle\frac { 2 }{ 52 } = \displaystyle\frac { 1 }{ 26 }$

$P\left( E \right) \times P\left( F \right) = \displaystyle\frac { 1 }{ 2 } \cdot \displaystyle\frac { 1 }{ 13 } = \displaystyle\frac { 1 }{ 26 } = P\left( EF \right)$
Therefore, the given events

$E$ and

$F$ are independent.
(iii) In a deck of

$52$ cards,

$4$ cards are kings,

$4$ cards are queens, and

$4$ cards are jacks.

$\therefore P\left( E \right) =P\left( the card drawn is a king or a queen \right) = \displaystyle\frac { 8 }{ 52 } = \displaystyle\frac { 2 }{ 13 }$

$\therefore P\left( F \right) =P\left( the card drawn is a queen or a jack \right) = \displaystyle\frac { 8 }{ 52 } = \displaystyle\frac { 2 }{ 13 }$
There are

$4$ cards which are king or queen and queen or jack.

$\therefore P\left( EF \right) =P\left( the card drawn is a king or a queen, or queen or a jack \right) = \displaystyle\frac { 4 }{ 52 } = \displaystyle\frac { 1 }{ 13 }$

$P\left( E \right) \times P\left( F \right) = \displaystyle\frac { 2 }{ 13 } \cdot \displaystyle\frac { 2 }{ 13 } = \displaystyle\frac { 4 }{ 169 } \neq \displaystyle\frac { 1 }{ 13 }$

$\Rightarrow P\left( E \right) \cdot P\left( F \right) \neq P\left( EF \right)$
Therefore, the given events

$E$ and

$F$ are not independent.

A random variable $X$ has the following probability distribution:
Determine
(i) $k$
(ii) $P\left(X < 3\right)$
(iii) $P\left(X > 6\right)$
(iv) $P\left(0 < X < 3\right)$

(i) It is known that the sum of probabilities of a probability distribution of random variables is one.

$\therefore 0 + k + 2k + 3k + {k}^{2} + 2{k}^{2} + \left(7{k}^{2} + k\right) = 1$

$\Rightarrow 10{k}^{2} + 9k - 1 = 0$

$\Rightarrow \left(10k - 1\right) \left(k + 1\right) = 0$

$\Rightarrow k = -1, \displaystyle\frac{1}{10}$

$k = -1$ is not possible as the probability of an event is never negative.

$\therefore k = \displaystyle\frac{1}{10}$
(ii)

$P\left(X < 3\right) = P\left(X = 0\right) + P\left(X = 1\right) + P\left(X = 2\right)$

$= 0 + k + 2k$

$= 3k$

$= 3 \times \displaystyle\frac{1}{10}$

$= \displaystyle\frac{3}{10}$
(iii)

$P\left(X > 6\right) = P\left(X = 7\right)$

$= 7{k}^{2} + k$

$= 7 \times {\displaystyle\frac{1}{10}}^{2} + \displaystyle\frac{1}{10}$

$= \displaystyle\frac{7}{100} + \displaystyle\frac{1}{10}$

$= \displaystyle\frac{17}{100}$
(iv)

$P \left(0 < X < 3\right) = P\left(X = 1\right) + P\left(X = 2\right)$

$= k + 2k$

$= 3k$

$= 3 \times \displaystyle\frac{1}{10}$

$= \displaystyle\frac{3}{10}$

Suppose we have four boxes $A, B, C$ and $D$ containing colored marbles as given :
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box $A$ , box $B$ , box $C$ ?

Let

$R$ be the event of drawing the red marble.
Let

${ E }_{ A }, { E }_{ B }$ , and

${ E }_{ C }$ respectively denote the events of selecting the box

$A, B$ , and

$C$ .
Total number of marbles

$= 40$
Number of red marbles

$= 15$

$\therefore P\left( R \right) =\displaystyle\frac { 15 }{ 40 } =\displaystyle\frac { 3 }{ 8 }$

Probability of drawing the red marble from box

$A$ is given by

$P\left( { E }_{ A }|R \right)$ .

$\therefore P\left( { E }_{ A }|R \right) =\displaystyle\frac { P\left( { E }_{ A }\cap R \right) }{ P\left( R \right) } =\displaystyle\frac { \displaystyle\frac { 1 }{ 40 } }{ \displaystyle\frac { 3 }{ 8 } } =\displaystyle\frac { 1 }{ 15 }$
Probability that the red marble is from box

$B$ is

$P\left( { E }_{ B }|R \right)$ .

$\Rightarrow P\left( { E }_{ B }|R \right) =\displaystyle\frac { P\left( { E }_{ B }\cap R \right) }{ P\left( R \right) } =\displaystyle\frac { \displaystyle\frac { 6 }{ 40 } }{ \displaystyle\frac { 3 }{ 8 } } =\displaystyle\frac { 2 }{ 5 }$
Probability that the red marble is from box

$C$ is

$P\left( { E }_{ C }|R \right)$ .

$\Rightarrow P\left( { E }_{ C }|R \right) =\displaystyle\frac { P\left( { E }_{ C }\cap R \right) }{ P\left( R \right) } =\displaystyle\frac { \displaystyle\frac { 8 }{ 40 } }{ \displaystyle\frac { 3 }{ 8 } } =\displaystyle\frac { 8 }{ 15 }=0.53$

Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduce the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

$\textbf{Step 1: Calculate Probability of Reduction in Chances of Heart Attack Because of Drugs and Yoga}$

$\text{Let A denote the events that a Person has a heart attack}$

$\text{Let $E_{1}$ denote the event that a Person Followed the Course of Yoga}$

$\text{Let $E_{2}$ denote the event that a Person Followed the Drugs Prescription}$

$\text{Since,chances of a Patient having a heart attack is 40%-}$

$\therefore P\left( A \right) =0.40$

$\Rightarrow P\left( { E }_{ 1 } \right) =P\left( { E }_{ 2 } \right) =\displaystyle\frac { 1 }{ 2 }$

$\textbf{[Given]}$

$\text{Since,Meditation and Yoga course reduces the risk of Heart Attack by 30%}$

$\therefore P\left( A|{ E }_{ 1 } \right) =0.40\times 0.70=0.28$

$\text{Since,Certain Drugs reduces the risk of Heart Attack by 25%-}$

$\therefore P\left( A|{ E }_{ 2 } \right) =0.40\times 0.75=0.30$

$\textbf{Step 2: Find Probability that the Patient followed a course of meditation and yoga}$

$P\left( { E }_{ 1 }|A \right)$

$\text{Denotes Probability that the Patient suffering a heart attack followed a }$

$\text{course of Meditation and Yoga is given by }$

$\Rightarrow P\left( { E }_{ 1 }|A \right) =\displaystyle\frac { P\left( { E }_{ 1 } \right) P\left( A|{ E }_{ 1 } \right) }{ P\left( { E }_{ 1 } \right) P\left( A|{ E }_{ 1 } \right) +P\left( { E }_{ 2 } \right) P\left( A|{ E }_{ 2 } \right) }$

$=\displaystyle\frac { \displaystyle\frac { 1 }{ 2 } \times 0.28 }{ \displaystyle\frac { 1 }{ 2 } \times 0.28+\displaystyle\frac { 1 }{ 2 } \times 0.30 }$

$=\displaystyle\frac { 14 }{ 29 }$

$=0.48$

$\textbf{The Probability that the Patient followed a course of meditation and yoga is 0.48}$

An urn contains $3$ white and $6$ red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution

$P\left( w \right) =\dfrac{3}{9}=\dfrac{1}{3}$

$P\left( R \right) =\dfrac{6}{9}=\dfrac{2}{3}$

Probability distribution of number of red balls.

$\Rightarrow$ Mean

$=\sum { xP\left( x \right) } =nP$ ( in case of binomial distribution )

$\Rightarrow \mu =4\times \dfrac{2}{3}=\dfrac{8}{3}$

Variance

$\left( { \sigma }^{ 2 } \right) =\sum x^2P\left(x\right) -\left[ \sum { xP\left( x \right) } \right] ^{ 2 }=npq$ ( in case of binomial distribution )

$=4\times \dfrac{1}{3}\times \dfrac{2}{3}$

$=\dfrac{8}{9}$

Hence, the answer is

$\dfrac{8}{9}.$

Of the students in a college, it is known that $60$ % reside in hostel and $40$ % are day scholars .Previous year results report that $30$ % of all students who reside in hostel attain 'A' grade and $20$ % of day scholars attain 'A' grade in their annual examination. At the end of the year, one student is chose at random from the college and he has an 'A' grade, what is the probability that the student is a hostlier?

Let $E_1, E_2$ and $A$ be events such that

$E_1$ = students is a hostlier

$E_2$ = students is a day scholar

$A$ = getting A grade

Now from equation,

$P(E_1)=\dfrac{60}{100}=\dfrac{6}{10}$

$P(E_2)=\dfrac{40}{100}=\dfrac{4}{10}$

$P(\dfrac{A}{E_1})=\dfrac{30}{100}=\dfrac{3}{10}$

$P(\dfrac{A}{E_2})=\dfrac{20}{100}=\dfrac{2}{10}$

We have to find $P(\dfrac{E_1}{A})$

Now, $P(\dfrac{E_1}{A})=\dfrac{P(E_1).P\left ( \dfrac{A}{E_1}\right )}{P(E_1).P\left ( \dfrac{A}{E_1}\right )+P(E_2).P\left ( \dfrac{A}{E_2}\right )}$

$=\dfrac{\dfrac{6}{10}.\dfrac{3}{10}}{\dfrac{6}{10}.\dfrac{3}{10}+\dfrac{4}{10}.\dfrac{2}{10}}$

$=\dfrac{\dfrac{18}{100}}{\dfrac{18}{100}+\dfrac{8}{100}}\Rightarrow \dfrac{18}{100}\times \dfrac{100}{26}\Rightarrow \dfrac{18}{26}\Rightarrow \dfrac{9}{13}$

Assume that the chances of a patient having a heart attack is 40%. Assuming that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation

Let

$A, E_1, E_2$ denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation and the person adopted the drug prescription respectively.

$\therefore P(A)=0.40$

$P(E_1)=P(E_2)=\dfrac 12$

$P(A/E_1)=0.40\times 0.70=028\\ P(A/E_2)=0.4\times 0.75=0.30$

Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by

$P(E_1/A)=\dfrac {P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}\\\implies =\dfrac {\dfrac 12\times 0.28}{\dfrac 12\times 0.28+\dfrac 12\times 0.30}\\\implies P(E_1/A)=\dfrac {14}{29}$

The given table categorises the house types and the number of bedrooms.

Development Houses
2 Br   3 Br 4 Br Total
Single-Family   5   19 34   58
Townhouse 24 42 30   96
Total 29 61 64 154

A daycare centre is planning to distribute pamphlets to the families with children living houses with three or more bedrooms, due to budget. In addition to sending out flyers, it also decides to send out invitations for a free day of daycare, to the family residing in two categories with the most houses. If a house that already received a flyer is chosen at random to receive the second stage of the marketing, then the probability that the house belongs to one of these two groups is $\dfrac pq$ , where $p,q$ are co-primes. What is the value of $q-p$ ?

The total number of cases where the houses have

$3$ or more bedrooms is

$61+64=125$ .

The two categories with most houses are also the houses with atleast

$3$ bedrooms.

The total number of houses surveyed is

$154$ .

Hence, the required probability is

$\dfrac{125}{154}$ .

Since

$125$ and

$154$ are non co primes, hence

$p=125$ and

$q=154$ .

Therefore

$q-p=154-125=49$

A bag $X$ contains $4$ white balls and $2$ black balls, while another bag $Y$ contains $3$ white balls and $3$ black balls. Two balls are drawn (without replacement) at random from one of the bags and were found to be one white and one black. Find the probability that the balls were drawn from bag $Y.$

Bag

$X$ contains

$4$ white balls and

$2$ black balls , Bag

$Y$ contains

$3$ white balls and

$3$ black balls.

Number of ways of selecting

$1$ white ball and

$1$ black ball from bag

$X$ is

$4 \times 2 = 8$ ways.

Number of ways of selecting

$1$ white ball and

$1$ black ball from bag

$Y$ is

$3 \times 3 = 9$ ways.

Total number of ways is

$8+9 = 17.$

The probability that the balls are drawn from bag

$Y$ is

$\dfrac{9}{17}.$

Bag I contains $3$ red and $4$ black balls and Bag II contains $4$ red and $5$ black balls. Two balls are transferred at random from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black

${\textbf{Step -1: Consider all possible events for transferring balls.}}$

${\text{Given that, }}$

${\text{Bag I contains 3 red and 4 black balls}}{\text{.}}$

${\text{Bag II contains 4 red and 5 black balls}}{\text{.}}$

${\text{Let }}{{\text{E}}_1},{{\text{E}}_2},{{\text{E}}_3},{\text{E}},$ ${\text{be events such that,}}$

${{\text{E}}_{\text{1}}}{\text{: Both transferred balls from Bag I to Bag II are red}}{\text{.}}$

${{\text{E}}_2}{\text{: Both transferred balls from Bag I to Bag II are black}}{\text{.}}$

${{\text{E}}_3}{\text{: 1 red ball and 1 black ball are transferred from Bag I to Bag II}}{\text{.}}$

${\text{E: Drawing a red ball from Bag II}}{\text{.}}$

${\textbf{Step -2: Find the possibilities of all the above mentioned events. }}$

${\text{Now,}}$

${\text{P}}\left( {{{\text{E}}_{\text{1}}}} \right) = \dfrac{{{}^3{C_2}}}{{{}^7{C_2}}}$ $\left( {{\text{Out of 7 balls in bag I, 3 of them are red.}}} \right)$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_{\text{1}}}} \right) = \dfrac{{\dfrac{{3!}}{{2!\left( {3 - 2} \right)!}}}}{{\dfrac{{7!}}{{2!\left( {7 - 2} \right)!}}}}$ $\left( {{\text{Using formula: }}{{}^n{{\text{C}}_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}}} \right)$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_{\text{1}}}} \right) = \dfrac{{3! \times 5! \times 2!}}{{2! \times 1! \times \left( {7 \times 6 \times 5!} \right)}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_{\text{1}}}} \right) = \dfrac{1}{7}$

${\text{P}}\left( {{{\text{E}}_2}} \right) = \dfrac{{{}^4{C_2}}}{{{}^7{C_2}}}$ $\left( {{\text{Out of 7 balls in bag I, 4 of them are black}}{\text{.}}} \right)$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}} \right) = \dfrac{{\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{\dfrac{{7!}}{{2!\left( {7 - 2} \right)!}}}}$ $\left( {{\text{Using formula: }}{}^n{{\text{C}}_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right)$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}} \right) = \dfrac{{4! \times 5! \times 2!}}{{2! \times 2! \times \left( {7 \times 6 \times 5!} \right)}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}} \right) = \dfrac{2}{7}$

${\text{P}}\left( {{{\text{E}}_3}} \right) = \dfrac{{{}^3{C_1} \times {}^4{C_1}}}{{{}^7{C_2}}}$

$\left( {{\text{Out of 7 balls in bag I, 3 of them are red and 4 of them are black}}{\text{.}}} \right)$

${\text{P}}\left( {{{\text{E}}_3}} \right) = \dfrac{{3 \times 4}}{{{}^7{C_2}}}$ $\left( {{\text{Using property: }}{{}^n{{\text{C}}_1} = 1}} \right)$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_3}} \right) = \dfrac{{12}}{{\dfrac{{7!}}{{2!\left( {7 - 2} \right)!}}}}$ $\left({{\text{Using formula: }}{}^n{{\text{C}}_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}} \right)$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_3}} \right) = \dfrac{{12 \times 5! \times 2!}}{{7 \times 6 \times 5!}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_3}} \right) = \dfrac{4}{7}$

${\text{Now,}}$

${\text{Two balls are transferred from a bag I to bag II thus, the total number of balls}}$

${\text{in bag II will be 11}}{\text{.}}$

${\text{E: Drawing a red ball from Bag II}}{\text{.}}$

$\therefore {\text{P}}\left( {{\text{E/}}{{\text{E}}_{\text{1}}}} \right){\text{: Where two balls are added to bag II which are red}}{\text{.}}$

${\text{The total number of red balls in bag II will be 4 + 2 = 6}}{\text{.}}$

$\therefore {\text{P}}\left( {{\text{E/}}{{\text{E}}_{\text{1}}}} \right) = \dfrac{6}{{11}}$

${\text{Similarly, P}}\left( {{\text{E/}}{{\text{E}}_2}} \right){\text{: Drawing red ball from bag II}}$

$\therefore {\text{P}}\left( {{\text{E/}}{{\text{E}}_2}} \right) = \dfrac{4}{{11}}$

$\therefore {\text{P}}\left( {{\text{E/}}{{\text{E}}_3}} \right){\text{: One of the balls transferred is red, so in bag the total number of red balls}}$

\ ${\text{in a bag II is 4 + 1 = 5}}{\text{.}}$

$\therefore {\text{P}}\left( {{\text{E/}}{{\text{E}}_3}} \right) = \dfrac{5}{{11}}$

${\textbf{Step -3: Using Bayes theorem find the required probability.}}$

${\text{Bayes theorem: The relationship between the probability of hypothesis }}\left( {\text{H}} \right)$

${\text{before getting the evidence }}\left( {\text{E}} \right)$ ${\text{is P}}\left( {\text{H}} \right)$

${\text{and the probability of the hypothesis after getting the evidence P}}\left( {{\text{H/E}}} \right){\text{ is,}}$

${\text{P}}\left( {{\text{H/E}}} \right){\text{ = }}\dfrac{{{\text{P}}\left( {{\text{E/H}}} \right)}}{{{\text{P}}\left( {\text{E}} \right)}}$

${\text{Here we have,}}$ ${\text{P}}\left( {{{\text{E}}_2}{\text{/E}}} \right){\text{ = }}\dfrac{{{\text{P}}\left( {{\text{E/}}{{\text{E}}_2}} \right){\text{P}}\left( {{{\text{E}}_2}} \right)}}{{{\text{P}}\left( {{\text{E/}}{{\text{E}}_1}} \right){\text{P}}\left( {{{\text{E}}_1}} \right) + {\text{P}}\left( {{\text{E/}}{{\text{E}}_2}} \right){\text{P}}\left( {{{\text{E}}_2}} \right) + {\text{P}}\left( {{\text{E/}}{{\text{E}}_3}} \right){\text{P}}\left( {{{\text{E}}_3}} \right)}}$

${\text{Substituting the known values in the above equation we get,}}$

${\text{P}}\left( {{{\text{E}}_2}{\text{/E}}} \right){\text{ = }}\dfrac{{\dfrac{2}{7} \times \dfrac{4}{{11}}}}{{\left( {\dfrac{1}{7} \times \dfrac{6}{{11}}} \right) + \left( {\dfrac{2}{7} \times \dfrac{4}{{11}}} \right) + \left( {\dfrac{4}{7} \times \dfrac{5}{{11}}} \right)}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}{\text{/E}}} \right){\text{ = }}\dfrac{{\dfrac{8}{{77}}}}{{\left( {\dfrac{6}{{77}}} \right) + \left( {\dfrac{8}{{77}}} \right) + \left( {\dfrac{{20}}{{77}}} \right)}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}{\text{/E}}} \right){\text{ = }}\dfrac{{\dfrac{8}{{77}}}}{{\left( {\dfrac{{6 + 8 + 20}}{{77}}} \right)}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}{\text{/E}}} \right){\text{ = }}\dfrac{8}{{77}} \times \dfrac{{77}}{{34}}$

$\Rightarrow {\text{P}}\left( {{{\text{E}}_2}{\text{/E}}} \right){\text{ = }}\dfrac{4}{{17}}$

${\textbf{Final Answer: The probability that the transferred balls were both black is}}$ $\mathbf{\dfrac{4}{{17}}.}$

In a factory which manufactures bolts, machines A, B and C manufacture $30 \%, 50 \%$ and $20 \%$ of the bolts respectively. Of their output $3 \% ,4 \%$ and $1 \%$ respectively are defective bolts. A bolt is drawn at random from the product and is found to be defective. Find the probability that this is not manufactured by machine B.

$=$ bolt manufactured by company A.

$=$ bolt manufactured by company B.

$=$ bolt manufactured by company C.

$=$ defective bolt drawn.

$P(x) =\dfrac{3}{10}, P(y)=\dfrac{1}{2}, P(z)=\dfrac{1}{5}$

$P(w|x)=\dfrac{3}{100}, P(w|y)=\dfrac{4}{100}, P(w|z)=\dfrac{1}{100}$

Probability that defective bolt is drawn from machine B.

$P(y|w)=\dfrac { P(y).P(w|y) }{ P(x).P(w|x)+P(y).P(w|y)+P(z).P(w|z) }$

$P(y|w)=\dfrac{20}{31}$

Probability that it is not manufactured by B

$=1-\dfrac{20}{31}=\dfrac{11}{31}$ .

Assume that the chances of a patient having a heart attack is 40%. Assuming that a mediation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chance by 25%. At a time a patient can choose any one of the two options with equal possibilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of mediation and yoga. Interpret the result and state which of the above stated methods is more beneficial for the patient.

Let E be the event that a patient had a heart attack.
Let

$E_1$ be the event that a patient used Drug A, therefore,

$P(E_1)=\dfrac {1}{2}$

Let

$E_2$ be the event that a patient used Drug B, therefore,

$P(E_2)=\dfrac {1}{2}$

Required probability:

$P\left(\dfrac {E_1}{E}\right)$

$P\left(\dfrac {E}{E_1}\right)=\dfrac {40}{100}\left ( 1-\dfrac {30}{100} \right )=\dfrac {28}{100}$

$P\left(\dfrac {E}{E_2}\right)=\dfrac {40}{100}\left ( 1-\dfrac {25}{100} \right )=\dfrac {30}{100}$

Also

$P(E_1)=P(E_2)=\dfrac {1}{2}$

$P($ Probability that the person who had heart attach followed meditation and yoga

$) =$

$P\left(\dfrac {E_1}{E}\right)$

$P\left(\dfrac {E_2}{E}\right)=\dfrac {P\left(\dfrac {E}{E_2}\right).P(E_2)}{\displaystyle \sum_{i=1}^{4}\left(P\left(\dfrac {E}{E_i}\right)\right).P(E_i)}$

$=\dfrac {\dfrac {28}{100}\times \dfrac {1}{2}}{\dfrac {28}{100}\times \dfrac {1}{2}+\dfrac {30}{100}\times \dfrac {1}{2}}$

$=\dfrac {\dfrac {28}{200}}{\dfrac {28}{200}+\dfrac {30}{200}}=\dfrac {28}{28+30}=\dfrac {14}{29}$

There are three coins. One is a two-headed coin (having head on both faces), another is a biased coin that comes up heads $75$ % of the times and third is also a biased coin that comes up tails $40$ % of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?

If there are

$3$ coins.
Let these are

$A, B, C$ respectively.
For coin A

$\rightarrow$ Probability of getting head

$P(H/A)=1$

For coin B

$\rightarrow$ Probability of getting head

$P(H/B)=\dfrac{3}{4}$

For coin C

$\rightarrow$ Probability of getting head

$P(H/C)=0.6$
we have to find

$P\left (\dfrac{A}{H}\right )$ = Probability of getting H by coin A.
So, we can use formula

$P\left ( \dfrac{A}{H} \right )=\cfrac{P\left ( \cfrac{H}{A} \right ).P(A)}{P\left ( \cfrac{H}{A} \right ).P(A)+P\left ( \cfrac{H}{B} \right ).P(B)+P\left ( \cfrac{H}{C} \right ).P(C)}$

Here,

$P\left ( \dfrac{H}{A} \right )=1,P\left ( \dfrac{H}{B} \right )=\dfrac{3}{4}, P\left ( \dfrac{H}{C} \right )=0.6$

Put value in formula so,

$P\left ( \dfrac{A}{H} \right )=\cfrac{1.\cfrac{1}{3}}{1.\cfrac{1}{3}+\cfrac{3}{4}.\cfrac{1}{3}+0.6.\cfrac{1}{3}}$

$=\cfrac{1}{1+0.75+0.6}$

$=\cfrac{100}{235}$

$=\cfrac{20}{47}$

Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls? Given that :
(i) the youngest is a girl.
(ii) at least one is a girl.

Sample space

$S = \{GG, GB, BG, BB \}$
(i) A = Both are girls
B = Youngest is a girl

$P \left (\dfrac{A}{B} \right )=\cfrac{P(A\cap B)}{P(B)}=\dfrac{\cfrac{1}{4}}{\cfrac{2}{4}}=\dfrac{1}{2}$
(ii) A: Both are girls
B: atleast one is a girl

$P\left (\dfrac{A}{B} \right )=\dfrac{P(A\cap B)}{P(B)})=\dfrac{\cfrac{1}{4}}{\cfrac{3}{4}}=\dfrac{1}{3}$

A card from a pack of $52$ playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and are found to be all spades. Find the probability of the lost card being a spade.

Let

$E_1 =$ Event that lost card is a spade.

$E_2 =$ Event that lost card is a non-spade.

$A =$ Event that three spades are drawn without replacement from 51 cards.

$P(E_1)=\dfrac{13}{52}=\dfrac{1}{4}$

$P(E_2)==1-\dfrac{1}{4}=\dfrac{3}{4}$

$P\left ( \dfrac{A}{E_1} \right )=\dfrac{^{12}C_{3}}{^{51}C_3}, P\left ( \dfrac{A}{E_2} \right )=\dfrac{^{13}C_{3}}{^{51}C_3}$

$P\left ( \dfrac{E_1}{A} \right )=\dfrac{\dfrac{1}{4}.\dfrac{^{12}C_{3}}{^{51}C_3}}{\dfrac{1}{4}.\dfrac{^{12}C_{3}}{^{51}C_3}+\dfrac{3}{4}.\dfrac{^{13}C_{3}}{^{51}C_3}}$

$= \dfrac{10}{49}$

What is the probability that if you roll a standard die three times, you will get three different numbers?

Total number of possible outcomes is

$6 \times 6 \times 6 = 216$
First die we can get any number, so number of ways is

$6$
Second die can get any number except the one which first die got, so the number of ways is

$5$
Third die can get any number except the one which first and second dies got, so the number of ways is

$4$
Therefore the total number of ways such that all three dies got different numbers is

$4 \times 5 \times 6 = 120$
The probability is

$\dfrac {120}{216} =\dfrac {5}{9}$ .

In an automobile factory, certain parts are to be fixed into the chassis in a section before it moves into another section. On a given day, one of the three persons $A,B$ and $C$ carries out this task. $A$ has $45$ % chance, $B$ has $35$ % chance and $C$ has $20$ % chance of doing the task. The probability that $A,B$ and $C$ will take more than the allotted time is $\cfrac { 1 }{ 6 } ,\cfrac { 1 }{ 10 }$ and $\cfrac{1}{20}$ respectively. If it is found that the time taken is more than the allotted time, what is the probability that $A$ has done the task?

Let

${E}_{1},{E}_{2},{E}_{3}$ denote the events of carrying out the task by

$A,B$ and

$C$ respectively.
Let

$H$ denote the event of taking more time.
Then,

$P({E}_{1})=0.45,P({E}_{2})=0.35,P({E}_{3})=0.20$

$P\left( \cfrac { H }{ { E }_{ 1 } } \right) =\cfrac { 1 }{ 6 } ,P\left( \cfrac { H }{ { E }_{ 2 } } \right) =\cfrac { 1 }{ 10 } ,P\left( \cfrac { H }{ { E }_{ 3 } } \right) =\cfrac { 1 }{ 20 }$
Then using Bayes' theorem

$P\left( \cfrac { { E }_{ 1 } }{ H } \right) =\cfrac { P\left( { E }_{ 1 } \right) .P\left( \cfrac { H }{ { E }_{ 1 } } \right) }{ P\left( { E }_{ 1 } \right) .P\left( \cfrac { H }{ { E }_{ 1 } } \right) +P\left( { E }_{ 2 } \right) .P\left( \cfrac { H }{ { E }_{ 2 } } \right) +P\left( { E }_{ 3 } \right) .P\left( \cfrac { H }{ { E }_{ 3 } } \right) }$

$=\cfrac { \cfrac { 45 }{ 600 } }{ \cfrac { 45 }{ 600 } +\cfrac { 35 }{ 1000 } +\cfrac { 20 }{ 2000 } } =\cfrac { \cfrac { 3 }{ 40 } }{ \cfrac { 3 }{ 40 } +\cfrac { 7 }{ 200 } +\cfrac { 1 }{ 100 } }$

$=\cfrac{3}{40}\times \cfrac{200}{24}=\cfrac{5}{8}$

For A, B, and C, the chances of being selected as the manager of a firm are 4 : 1 : 2 respectively. The probabilities for them to introduce a radical change in the marketing strategy are 0.3, 0.8 and 0.5 respectively. If a change takes place; find the probability that it is due to the appointment of B.

Let

$E_1, E_2, E_3$ and E be the events as defined below:

$E_1$ = A is selected as Manager

$E_2$ = B is selected as Manager

$E_3$ = C is selected as Manager
and E = radical change occurs in marketing strategy

$P(E_1) = \dfrac{4}{4 + 1 + 2} = \dfrac{4}{7}$

$P(E_2) = \dfrac{1}{4 + 1 + 2} = \dfrac{1}{7}$

$P(E_3) = \dfrac{2}{4 + 1 + 2} = \dfrac{2}{7}$

Given,

$P\left (\dfrac{E}{E_1} \right ) = 0.3$

$P\left( \dfrac{E}{E_2} \right ) = 0.8$

$P\left( \dfrac{E}{E_3} \right ) = 0.5$

We want to find the probability that the radical change in marketing strategy occured due to the appointment of B i.e., we want to find

$P(E_2 / E)$

By Baye's theorem,

$P \left( \dfrac{E_2}{E} \right ) = \dfrac{P(E_2) . P \left( \dfrac{E}{E_2} \right )}{P(E_1) P\left( \dfrac{E}{E_1} \right ) + P(E_2) . P \left( \dfrac{E}{E_2} \right )+ P(E_3) P \left( \dfrac{E}{E_3} \right )}$

$= \dfrac{\dfrac{1}{7} \times 0.8}{\dfrac{4}{7} \times 0.3 + \dfrac{1}{7} \times 0.8 + \dfrac{2}{7} \times 0.5}$

$= \dfrac{0.8}{1.2 + 0.8 + 1}$

$= \dfrac{0.8}{3} = \dfrac{8}{30} = \dfrac{4}{15}$

An urn contains $2$ white and $2$ black balls. A ball is drawn at random. If it is white, it is not replaced into the urn. Otherwise, it is replaced with another ball of the same colour. The process is repeated. Find the probability that the third ball drawn is black.

${\textbf{Step -1:Consider all the possible cases.}}$

${\text{We have given, an urn contains 2 white and 2 black ball}}{\text{. }}$

${\text{There are four possibilities for drawing first two balls :}}$

${\text{First ball}}$	${\text{Second ball}}$	${\text{Event}}$
${\text{White}}$	${\text{White}}$	${E_1}$
${\text{White}}$	${\text{Black}}$	${E_2}$
${\text{Black}}$	${\text{White}}$	${E_3}$
${\text{Black}}$	${\text{Black}}$	${E_4}$

${\textbf{Step -2:Find the required probability.}}$

${{P(}}{{\text{E}}_1}) = \dfrac{2}{4} \times \dfrac{1}{3} = \dfrac{1}{6}.$

${{P(}}{{\text{E}}_2}) = \dfrac{2}{4} \times \dfrac{2}{3} = \dfrac{1}{3}.$

${{P(}}{{\text{E}}_3}) = \dfrac{2}{4} \times \dfrac{2}{5} = \dfrac{1}{5}.$

${{P(}}{{\text{E}}_4}) = \dfrac{2}{4} \times \dfrac{3}{5} = \dfrac{3}{{10}}.$

$P\left( {E|{E_1}} \right) = 1;$ ${\text{because there is no white ball to be selected}}{\text{.}}$

$P\left( {E|{E_2}} \right) = \dfrac{3}{4};$ ${\text{because there are 3 black balls and 1 white ball}}{\text{.}}$

$P\left( {E|{E_3}} \right) = \dfrac{3}{4};$ ${\text{because again there are 3 black balls and 1 white ball}}{\text{.}}$

$P\left( {E|{E_4}} \right) = \dfrac{4}{6} = \dfrac{2}{3};$ ${\text{because there are 4 black balls and 2 white ball}}{\text{.}}$

${\text{The four events }}$ ${E_1}$ ${E_2}$ ${E_3}$ ${\text{and}}$ ${E_4}$ ${\text{are mutually exclusive and exhaustive}}{\text{.}}$

${\text{So, by the theorem of total probability}}$

$\Rightarrow P\left( E \right) = P\left( {{E_1}} \right)P\left( {E|{E_1}} \right) + P\left( {{E_2}} \right)P\left( {E|{E_2}} \right) + P\left( {{E_3}} \right)P\left( {E|{E_3}} \right) + P\left( {{E_4}} \right)P\left( {E|{E_4}} \right)$

$= \dfrac{1}{6} \times 1 + \dfrac{1}{3} \times \dfrac{3}{4} + \dfrac{1}{5} \times \dfrac{3}{4} + \dfrac{3}{{10}} \times \dfrac{2}{3}$

$= \dfrac{1}{6} + \dfrac{1}{4} + \dfrac{3}{{20}} + \dfrac{1}{5}$

$= \dfrac{{10 + 15 + 9 + 12}}{{60}}$

$= \dfrac{{46}}{{60}}$

$= \dfrac{{23}}{{30}}$

${\textbf{Hence, probability that the third ball drawn is black is}}$ $\mathbf{\dfrac{{23}}{{30}}.}$

In a bolt factory, three machines $A, B$ and $C$ manufacture $25$ %, $35$ % and $40$ % of the total production respectively. Of their respective outputs, $5$ %, $4$ % and $2$ % are defective. A bolt is drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine $C$ .

Let

$E_{1}$ : the event that bolt is produced by machine

$A$ .

$E_{2}$ : the event that bolt is produced by machine

$B$ .
and

$E_{3}$ : the event that bolt is produced by machine

$C$ .
Here

$E_{1}, E_{2}$ and

$E_{2}$ are mutually exclusive and exhaustive events.
We have,

$P(E_{1}) = 25$ %

$= \cfrac {25}{100}$

$P(E_{2}) = 35$ %

$= \cfrac {35}{100}$

$P(E_{3}) = 40$ %

$= \dfrac {40}{100}$

Let

$E$ : the event that bolt chosen is found to be defective.

$\therefore P\left (\dfrac {E}{E_{1}}\right ) = 5$ % =

$\dfrac {5}{100}, P\left (\dfrac {E}{E_{2}}\right ) = 4$ %

$= \dfrac {4}{100}$ and

$P\left (\dfrac {E}{E_{3}}\right ) = 2$ %

$= \dfrac {2}{100}$

$P$ (defective bolt is produced by machine

$C$ )

$P\left (\dfrac {E_{3}}{E}\right ) = \dfrac {P(E_{3})\cdot P\left (\dfrac {E_{3}}{E}\right )}{P(E_{1})\cdot P\left (\dfrac {E}{E_{1}}\right ) + P(E_{2}) \cdot P\left (\dfrac {E}{E_{2}}\right ) + P(E_{3})\cdot P\left (\dfrac {E}{E_{3}}\right )}$

$=\dfrac {\dfrac {40}{100}\times \dfrac {2}{100}}{\dfrac {25}{100}\times \dfrac {5}{100} + \dfrac {35}{100}\times \dfrac {4}{100} + \dfrac {40}{100} \times \dfrac {2}{100}}$

$= \cfrac {80}{125 + 140 + 80}$

$= \cfrac {80}{345} = 0.23$

Box I contains two white and three black balls. Box II contains four white and one black balls and box III contains three white and four black balls. A dice having three red, two yellow and one green face, is thrown to select the box. If red face turns up, we pick up box I, if a yellow face turns up we pick up box II, otherwise, we pick up box III. Then, we draw a ball from the selected box. If the ball drawn is white, what is the probability that the dice had turned up with a red face?

Let

$E_1, E_2, E_3$ and

$A$ be the events defined as follows :

$E_1$ : red face turns up

$E_2$ : yellow face turns up

$E_3$ : green face turns up

$A$ : white ball is drawn
We wish to calculate the probability that dice had turn up with red face when A has occured (white ball is selected)
i.e.,

$(E_1|A)$

By Bayes' Theorem,

$P(E_1|A)=\dfrac{P(E_1)\times P(A|E_1)}{P(E_1)\times P(A|E_1)+P(E_2)\times P(A|E_2)+P(E_3)\times P(A|E_3)}$

$=\dfrac{\dfrac{1}{2}\times \dfrac{2}{5}}{\dfrac{1}{2}\times \dfrac{2}{5}+\dfrac{1}{3}\times \dfrac{4}{5}+\dfrac{1}{6}\times \dfrac{3}{7}}$

$=\dfrac{\dfrac{1}{5}}{\dfrac{1}{5}+\dfrac{4}{15}+\dfrac{1}{14}}$

$=\dfrac{\dfrac{1}{5}}{\dfrac{113}{210}}=\dfrac{1}{5}\times \dfrac{210}{113}=\dfrac{42}{113}$ .

From a lot of 30 bulbs which include 6 defectives, a sample of 2 bulbs are drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

The question says that out of $30$ bulbs, $6$ are defective.

So, Number of non-defective bulbs = $30 - 6 =24$ .

$2$ bulbs are drawn from the lot at random with replacement.

Let $X$ be the random variable that denotes the number of defective bulbs in the selected bulbs.

Hence,

$P(X=0) =$ P ( $2$ non- defective and $0$ defective)

$=C(2,0)$ . $\dfrac{24}{30} .$ $\dfrac{24}{30}$ = $\dfrac{16}{25}$

$P(X=1) =$ P ( $1$ non- defective and $1$ defective)

$=C(2,1)$ . $\dfrac{24}{30} .$ $\dfrac{6}{30}$ = $\dfrac{8}{25}$

$P(X=2) =$ P ( $0$ non- defective and $2$ defective)

=C(2,2). $\dfrac{6}{30} .$ $\dfrac{6}{30}$ = $\dfrac{1}{25}$

Therefore, the required probability distribution is as follows.

X	0	1	2
P(X)	$\dfrac{16}{25}$	$\dfrac{8}{25}$	$\dfrac{1}{25}$

A fair die is rolled. If face $1$ turns up, a ball is drawn from Bag $A$ . If face $2$ or $3$ turns up, a ball is drawn from Bag $B$ . If face $4$ or $5$ or $6$ turns up, a ball is drawn from Bag $C$ . Bag $A$ contains $3$ red and $2$ white balls, Bag $B$ contains $3$ red and $4$ white balls and Bag $C$ contains $4$ red and $5$ white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red; what is the probability that it is drawn from Bag $B$ ?

$\textbf{Step 1: Finding the individual probabilities}$

$\text{Probability that the ball is drawn from bag A}=P(E_1)=\dfrac{1}{6}$

$\text{Probability that the ball is drawn from bag B}=P(E_2)=\dfrac{2}{6}=\dfrac{1}{3}$

$\text{Probability that the ball is drawn from bag C}=P(E_3)=\dfrac{3}{6}=\dfrac{1}{2}$

$\text{Let }P(A)\text{ be the probability of drawing a red ball}$

$\text{Probability that the red ball was drawn from bag A}=P(A|E_1)=\dfrac{3}{5}$

$\text{Probability that the red ball was drawn from bag B}=P(A|E_2)=\dfrac{3}{7}$

$\text{Probability that the red ball was drawn from bag C}=P(A|E_3)=\dfrac{4}{9}$

$\textbf{Step 2: Using Bayes' Theorem to find the required probability}$

$\text{We need to find the probability that if a red ball is drawn, the probability that it is }$

$\text{drawn from bag B}=P(E_2|A)$

$\text{Using Bayes' theorem}$

$P(E_2|A)=\dfrac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)}$

$P(E_2|A)=\dfrac{\dfrac{1}{3}\times\dfrac{3}{7}}{\dfrac{1}{6}\times{3}{5}+\dfrac{1}{3}\times\dfrac{3}{7}+\dfrac{1}{2}\times\dfrac{4}{9}}$

$=\dfrac{90}{293}$

$\textbf{Hence , if the drawn ball is red, the probability that it was drawn from bag B is }\mathbf{\dfrac{90}{293}}$

State and prove Baye's theorem.

Bayes theorem is stated as

$P(A/B)=\dfrac{P(B/A)P(A)}{P(B)}$

Proof

$:$

We can do it from set theory applied to conditional probability

$P(A/B)=\dfrac { P(A\cap B) }{ P(B) }$

Likewise

$P(B/A)=\dfrac { P(A\cap B) }{ P(A) }$

Rearrange to get common terms as follows ,

$P(A\cap B)=P(A/B)P(B)=P(B/A)P(A)$

Therefore we get

$P(A/B)=\dfrac{P(B/A)P(A)}{P(B)}$

This is basic statement of Bayes theorem

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will at least $4$ successes.

Let

$p$ denote the probability of success. Then the probability of failure is

$1-p$ .
Now

$p=2(1-p)$ or

$3p=2$ or

$p=\dfrac{2}{3}$ , therefore

$q=\dfrac{1}{3}$ .
Hence if the number of trials is 6, then the probability of getting atleast 4 success is (by application of Bernoulli trials)

$Pr=(\dfrac{1}{3})^{6}[\:^{6}C_{4}2^{4}+\:^{6}C_{5}2^{5}+\:^{6}C_{6}2^{6}]$

$= \dfrac{2^4}{3^6}(15+12+4)$

$=\dfrac{2^4}{3^6} (31)$ Ans.

In a class of 75 students, 15 are above average, 45 are average and the rest below average achievers. The probability that an above average achieving student fails is 0.005, that an average achieving student fails is 0.05 and the probability of a below average achieving student failing is 0.If a student is know to have passed, what is the probability that he is a below average achiever?

Let,

$E_1$ : Event that student is above average

$E_2$ : Event that student is average

$E_3$ : Event that student is below average.
A : Event that student is known to have passed

$P(E_1) = \dfrac{1}{5} , P (E_2) = \dfrac{3}{5}, P (E_3) = \dfrac{1}{5}$

$P(A / E_1) = 1 - 0.005 = 0.995$

$P(A/E_2) = 1 - 0.05 = 0.95$

$P(A/E_3) = 1 - 0.15 = 0.85$
Using Baye's theorem,

$P(E_3/A) = \dfrac{P(E_3) . P(A/E_3)}{P(E_1) . P(A/E_1) + P(E_2) . P(A/E_2) + P(E_3) . P(A/ E_3)}$

$= \dfrac{\dfrac{1}{5} \times \dfrac{85}{100}}{\dfrac{1}{5} \times \dfrac{995}{1000} + \dfrac{3}{5} \times \dfrac{95}{100} + \dfrac{1}{5} \times \dfrac{85}{100}} = \dfrac{\dfrac{17}{100}}{\dfrac{199}{1000} + \dfrac{57}{100} + \dfrac{17}{100}}$

$= \dfrac{\dfrac{17}{100}}{\dfrac{199 + 570 + 170}{1000}}$

$= \dfrac{\dfrac{17}{100}}{\dfrac{939}{1000}} = \dfrac{17 \times 100}{100 \times 939} = \dfrac{170}{939} = 0.1810$

A dice is thrown twice. A success is an even number on each throw. Find the probability distribution of the number of successes.

Total number of cases on throwing a dice

$S=\left\{ 1,2,3,4,5,6 \right\}$

$n(S)=6......(1)$
number of favourable cases

$\quad E=\left\{ 2,4,6, \right\}$

$n(E)=3....(2)$

$P(E)=\cfrac { n(E) }{ n(S) } =\cfrac { 3 }{ 6 } =\cfrac { 1 }{ 2 }$

$\quad P=\cfrac { 1 }{ 2 }$
probability of failure

$=9=1-\cfrac { 1 }{ 2 } =\cfrac { 1 }{ 2 }$

$n=2$
for

$x=0$ (no success)

$P(x=0)={ _{ }^{ 2 }{ C } }_{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 0 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2-0 }={ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }=\cfrac { 1 }{ 4 }$
for

$x=1$ (one success)

$P(x=1)={ _{ }^{ 2 }{ C } }_{ 1 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 1 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2-1 }=\cfrac { 1 }{ 2 }$
for

$x=2$ (two successes)

$P(x=2)={ _{ }^{ 2 }{ C } }_{ 2 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 2 } \right) }^{ 2-2 }=\cfrac { 1 }{ 4 }$
Probability distribution

$x=r$	$0$	$1$	$2$
$P(x=r)$	$\dfrac14$	$\dfrac12$	$\dfrac14$

$\sum { P(x=r) } =\cfrac { 1 }{ 4 } +\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 4 } =1\quad$

A bag contain 6 Red and 4 White balls. 4 balls are drawn one by one without replacement and were found to be atleast 2 white. If the next draw of a ball from this bag will give a white ball then the probability that the drawn of four balls initially contain two white and two red balls.

A bag contains

$6$ red and

$4$ white balls

$P(2$ red balls

$\xi$

$2$ white balls

$)$

$\Rightarrow \dfrac{{^6C_2}{^4C_2}}{^{10}C_4}=\dfrac{6\times15}{210}=\dfrac{15}{14}$

Total outcomes

${^{10}C_4}$

Hence, the answer is

${^{10}C_4}.$

The range of a random variable $X$ is $\left\{ 0,1,2 \right\}$ and given that $P(X=0)=3{ c }^{ 3 },\quad P(X=1)=4c-10{ c }^{ 2 },\quad P(X=2)=5c-1$ , find (i) the value of $c$
(ii) $P(X<1),\quad P(1<X\le 2),\quad P(0<X\le 3)\quad$

$p\left( X=0 \right) +p\left( X=1 \right) +p\left( X=2 \right) =1$

$3{ c }^{ 3 }+4c-10{ c }^{ 2 }+5c-1=1$

$3{ c }^{ 3 }+9c-10{ c }^{ 2 }-2=0$

$\left( i \right)$ Find c

$\left( c-1 \right) \left( 3{ c }^{ 2 }-7c+2 \right) =0$

$3{ c }^{ 2 }-7c+2=0\quad \because c\neq 1$

For a quadratic equation of the form $ax^{ 2 }+bx+c=0$

$c =\dfrac { -b\pm \sqrt { b^{ 2 }-4ac } }{ 2a }$

Here $a=3, b=-7$ and $c=2$

$c=\dfrac { +7\pm \sqrt { 49-24 } }{ 6 }$

$c=\dfrac { 7\pm 5 }{ 6 }$

$c=2$ or $c=\dfrac { 1 }{ 3 }$

but $c\neq 2$

$\therefore c=\dfrac { 1 }{ 3 }$

$0<c<1$

Put value of c in given equations and up to given random range

$p\left( x=0 \right) =3{ c }^{ 3 }$

$=3{ \left( \dfrac { 1 }{ 3 } \right) }^{ 3 }$

X 0 1 2

$p\left( x \right)\quad \quad \dfrac { 1 }{ 9 } \quad \quad \dfrac { 2 }{ 9 } \quad \quad \dfrac { 6 }{ 9 }$

$\left( ii \right)$ Find $p\left( x<1 \right)$

$p\left( x<1 \right) =p\left( x=0 \right) =\dfrac { 1 }{ 9 }$

$\left( ii \right)$ Find $p\left( 1<x<2 \right)$

$p\left( 1<x<2 \right) =p\left( x=2 \right) =\dfrac { 6 }{ 9 }$

$\left( iv \right)$ Find $p\left( 0<x\le 3 \right)$

$p\left( 0<x\le 3 \right) =p\left( 1 \right) +p\left( 2 \right) +p\left( 3 \right)$

$=\dfrac { 2 }{ 9 } +\dfrac { 6 }{ 9 } +$ does not exist

$=\dfrac { 8 }{ 9 }$

A fair coin is tossed $9$ times. Find the probability that it shows head exactly $5$ times.

Let

$X=$ number of head shows

$n=9, p=\dfrac{1}{2}, q=\dfrac{1}{2}$

We know that,

$P(X=x)={}^nC_xp^n \cdot q^{n-x} \:where\: x=0,1,...,n$

$\therefore P(X=5)={}^9C_5 \left(\dfrac{1}{2}\right)^5 \cdot \left(\dfrac{1}{2}\right)^{9-5}$

$={}^9C_5 \left(\dfrac{1}{2}\right)^5 \cdot \left(\dfrac{1}{2}\right)^{4}$

$={}^9C_5 \left(\dfrac{1}{2}\right)^9$

$=\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}\times \dfrac{1}{2^9}$

$=\dfrac{126}{512}$

$\therefore P(X=5)=0.2460$

Suppose a girl throws a die. If she gets a $5$ or $6$ , she tosses a coin $3$ times and notes the number of heads. If she gets $1, 2, 3$ and $4$ she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head. What is the probability that she threw $1, 2, 3$ or $4$ with the die?

${\textbf{Step -1: Assume the events and find the probability of occuring them}}{\textbf{.}}$

${\text{Let's suppose the event}}{\text{.}}$

${\text{A}} = {\text{The event of appearing 1,2,3,4 on dice while throwing a dice}}{\text{.}}$

${\text{B is the event of appearing 5 or 6}}{\text{.}}$

${\text{C is event of exactly one head is being obtained}}{\text{.}}$

${\text{Total outcomes}} = 6$

${\text{we know , Probability}} = \dfrac{{{\text{Favourable outcome}}}}{{{\text{Total Outcomes}}}}.$

${\text{P(A) probability of appearing 1,2,3,4 on dice}} = \dfrac{4}{6} = \dfrac{2}{3}.$

${\text{Probability that 5 or 6 appear on dice , }}$

${\text{P(B)}} = \dfrac{2}{6} = \dfrac{1}{3}.$

${\text{Probability of exactly one head is obtained if 1,2,3,4 appear on dice}}{\text{.}}$

${\text{P}}\left( {{\text{C}}\left| {\text{A}} \right.} \right) = \dfrac{1}{2}.$

${\text{Probability of exact by one head is obtained if 5 or 6 appears on dice}}{\text{.}}$

$\Rightarrow {\text{P}}\left( {{\text{C}}\left| {\text{B}} \right.} \right) = \dfrac{3}{8}.$

${\textbf{Step - 2: Find the required probability}}{\textbf{.}}$

${\text{We have to find , if the girl obtained exactly one head in the toss , she threw 1,2,3 or 4 with dice}}{\text{.}}$

${\text{Assume that to be P}}\left( {{\text{A}}\left| {\text{C}} \right.} \right)$

$\Rightarrow {\text{P}}\left( {{\text{A}}\left| {\text{C}} \right.} \right) = \dfrac{{{\text{P(A)}} + {\text{P}}\left( {{\text{C}}\left| {\text{A}} \right.} \right)}}{{{\text{P(A)}}.{\text{P}}\left( {{\text{C}}\left| {\text{A}} \right.} \right) + {\text{P(B)}}{\text{.P}}\left( {{\text{C}}\left| {\text{B}} \right.} \right)}}$

$\Rightarrow {\text{P}}\left( {{\text{A}}\left| {\text{C}} \right.} \right) = \dfrac{{\dfrac{2}{3} \times \dfrac{1}{2}}}{{\dfrac{2}{3} \times \dfrac{1}{2} + \dfrac{1}{3} \times \dfrac{3}{8}}}$

$\Rightarrow {\text{P}}\left( {{\text{A}}\left| {\text{C}} \right.} \right) = \dfrac{{\dfrac{1}{3}}}{{\dfrac{1}{3} \times \dfrac{3}{8}}}$

$\Rightarrow {\text{P}}\left( {{\text{A}}\left| {\text{C}} \right.} \right) = \dfrac{{\dfrac{1}{3}}}{{\dfrac{{11}}{{24}}}}$

$\Rightarrow {\text{P}}\left( {{\text{A}}\left| {\text{C}} \right.} \right) = \dfrac{8}{{11}}.$

${\textbf{Hence , the probability that show 1,2,3,4 which the dice is }}\mathbf{\dfrac{8}{{11}}.}$

Consider the experiment of tossing a coin. If the coin shows head, toss it again, but if it shows tail, then throw a die. Find the conditional probability of the event that 'the die shows a number greater than $4$ ', given that 'there is atleast one tail'.

The sample space for the given event is,

$\left\{ {{\rm{HH,HT,T1,T2,T3,T4,T5,T6}}} \right\}$

Let E be the event that dice show the number greater than 4 and f be the event of getting at least one tail.

${\rm{E}} = \left\{ {{\rm{T5,T6}}} \right\}$ and ${\rm{F}} = \left\{ {{\rm{HT,T1,T2,T3,T4,T5,T6}}} \right\}$

${\rm{P}}\left( {\rm{E}} \right) = {\rm{P}}\left( {{\rm{T5}}} \right) + {\rm{P}}\left( {{\rm{T6}}} \right)$

$= \dfrac{1}{{12}} + \dfrac{1}{{12}}$

$= \dfrac{1}{6}$

${\rm{P}}\left( {\rm{F}} \right) = {\rm{P}}\left( {{\rm{HT}}} \right) + {\rm{P}}\left( {{\rm{T1}}} \right) + {\rm{P}}\left( {{\rm{T2}}} \right) + {\rm{P}}\left( {{\rm{T3}}} \right) + {\rm{P}}\left( {{\rm{T4}}} \right) + {\rm{P}}\left( {{\rm{T5}}} \right) + {\rm{P}}\left( {{\rm{T6}}} \right)$

$= \dfrac{1}{4} + \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{12}} + \dfrac{1}{{12}}$

$= \dfrac{3}{4}$

${\rm{E}} \cap {\rm{F}} = \left\{ {{\rm{T5,T6}}} \right\}$

${\rm{P}}\left( {{\rm{E}} \cap {\rm{F}}} \right) = \dfrac{1}{6}$

${\rm{P}}\left( {{\rm{E}}|{\rm{F}}} \right) = \dfrac{{{\rm{P}}\left( {{\rm{E}} \cap {\rm{F}}} \right)}}{{{\rm{P}}\left( {\rm{F}} \right)}}$

$= \dfrac{{\dfrac{1}{6}}}{{\dfrac{3}{4}}}$

$= \dfrac{2}{9}$

Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is the probability that they came from Bag III.

Bag I- 1 W, 2 B, 3 R

Bag II- 2 W, 1 B, 1 R

Bag III= 4 W, 3 B, 2 R

Probability that first bag is chosen and one is white and other red=

$\frac { 1 }{ 3 } *\frac { 1*3 }{ 6*6 }$

Probability that second bag is chosen and one is white and other red=

$\frac { 1 }{ 3 } *\frac { 1*2 }{ 4*4 }$

Probability that thirdbag is chosen and one is white and other red=

$\frac { 1 }{ 3 } *\frac { 4*2 }{ 9*9 }$

using Bayes rule

Probability that it came from third bag=

$\frac { \frac { 1 }{ 3 } *\frac { 4*2 }{ 9*9 } }{ \frac { 1 }{ 3 } *\frac { 1*3 }{ 6*6 } +\frac { 1 }{ 3 } *\frac { 1*2 }{ 4*4 } +\frac { 1 }{ 3 } *\frac { 4*2 }{ 9*9 } } =\frac { 64 }{ 199 }$

Thirty-two player ranked $1$ to $32$ are play knockout tournament. Assume that in every match between any two players the better ranked player wins, the probability that ranked $1$ and ranked $2$ players are winner and runner up respectively is $p$ , then the value of $[2/p]$ is, where [.] represents the greatest integer function.

Yes, the player ranked number 1 will win all of his matches, and will eventually win the tournament.

probability is that the players ranked 1 and 2 will meet in the final (because only in this scenario, the ranked 1 player will win, and the ranked 2 player will be the runner-up),for the two players to meet in the final, we need to avoid that they meet before the final

In round 1, 1 can meet all 31 player except 2 $p=30/31$

In round 2, 1 can meet all 14 player except 2 $p=14/15$

In round 3, 1 can meet all 6 player except 2 $p=6/7$

In round 4, 1 can meet all 2 player except 2 $p=2/3$

Therefore required probability $=(30/31)\times (14/15)\times (6/7)\times (2/3)=16/31$

$2/p=62/16=3.7$

$[2/p]=3$

A bas contains $4$ white, $7$ black and $5$ red balls. Three balls are drawn one after the other without replacement. Find the probability that the balls drawn are white, black and red respectively.

A bas contains 44 white, 77 black and 55 red balls Three balls are drawn one after the other without replacement probability that the balls drawn are white black red=

$\dfrac { 3!\times{ 4_{ C } }_{ 1 }\times{ 7_{ C } }_{ 1 }\times{ 5_{ C } }_{ 1 } }{ { { 16 }_{ C } }_{ 3 } } \quad =\dfrac { 6\times4\times7\times5\times6 }{ 16\times15\times14 } \quad =\quad 2/3\quad$

$10$ boys and $5$ girls study in a class. Three students are selected at random, one after the other. Find probability that,
(1) First two are boys and the third is a girl,
(2) First and third are boys and second is a girl

$(i)\$ first

$2$ are boys and third is a girl

$p(bbg)=\dfrac {10}{15}\times \dfrac {9}{14}\times \dfrac {5}{13}$

$p(bbg)=\dfrac {15}{91}$ Answer

$(ii)\$ first and third are boys and second is given

$p(bgb)=\dfrac {10}{15}\times \dfrac {5}{14}\times \dfrac {9}{13}$

$p(bgb)=\dfrac {15}{91}$ Answer

For the following p.d.f.s of X, find $P(X < 1)$ and $P(|x| < 1)$
$f(x)=\dfrac{x^2}{18}$ , $-3 < x < 3$ ;
$=0$ , otherwise.

$f(x) = \begin{cases} 0 & x \le -3 \\ x^{2}/18 & -3 < x < 3 \\ 0 & x \ge 3 \end{cases}$

$\int_{-\infty}^{\alpha} f(x) dx = \int_{-3}^{3} \dfrac{x^{2} }{18} dx = \left[\dfrac{x^{3}}{54} \right]_{-3}^{3} = 1$

$P(x < 1)= \dfrac{ \int_{-\infty}^{1} f(x) dx }{ \int_{-\infty}^{\infty} } = \dfrac{ \int_{-3}^{1} \left( \dfrac{x^{2}}{18} \right) dx }{1} = \left[ \dfrac{x^{3}}{54} \right]_{-3}^{1} = \dfrac{14}{27}$

$P(|x| < 1) = \dfrac{\in_{-1}^{1} f (x) dx}{\int_{-\infty}^{\infty} f(x) dx }= \dfrac{[x^{3}/54]_{-1}^{1}}{1}= \dfrac{1}{24}$

If a fair coin is tossed $10$ times find the probability of (a) exactly six head
(b) At least six heads
(c) at most six heads.

a) Probability of exactly six heads=

${ { 10 }_{ C } }_{ 6 }{ (0.5) }^{ 10 }$

b)Probability of at least 6 heads=

P(6 heads)+P(7 heads)+P(8 heads)+P(9 heads)+P(10 heads)

${ { 10 }_{ C } }_{ 6 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 7 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 8 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 9 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 10 }{ (0.5) }^{ 10 }$ =386/1024

c))Probability of at most 6 heads=

1-P(7 heads)+P(8 heads)+P(9 heads)+P(10 heads)

$1-{ { 10 }_{ C } }_{ 7 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 8 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 9 }{ (0.5) }^{ 10 }+{ { 10 }_{ C } }_{ 10 }{ (0.5) }^{ 10 }$ =848/1024

If $P(not\ A)=0.7,P(B)=0.7$ and $P(B/A)=0.5$ , find $P\left(A\bigcup B \right)$ .

$P\left( \frac { B }{ A } \right) =\frac { P\left( A\bigcap B \right) }{ P\left( A \right) }$

$0.5=\frac { P\left( A\bigcap B \right) }{ 0.3 }$ \\

$P\left( A\bigcap B \right) =0.15$

$P\left( \frac { A }{ B } \right) =\frac { P\left( A\bigcap B \right) }{ P\left( B \right) } =\frac { 0.15 }{ 0.7 } =0.214$

$P\left( A\bigcup B \right) =0.3+0.7-0.15=0.85$

A family has two children. What is the probability that both the children are boys given that at least one of them is a boy?

To find P (both children are males, if it is known that at least one of the children is male).

A: Event that both children are male, and B: event that at least one of them is a male.

A: {MM} and B: {MF, FM, MM} →→ P (A ∩∩ B) = {MM}

Probability that both are males, if we know one is a female =P(A/B)=P(A∩B)/P(B)

(P(A∩B)P(B))

Given S = {MM, MF, FM, FF}, we can see that: P (A) = 1414; P (B) = 3434; P (A ∩∩ B) = 1414

Therefore P(B/A)=P(A∩B)/P(A) =(1/4)/(3/4)=1/3

In a class of $60$ boys and $20$ girls, half of the boys and half of the girls know cricket. Find the probability of the event that a person selected from the class is either a boy or a girl who knows cricket.

${\textbf{Step - 1: Find probability of boys and girls knowing cricket}}$

${\text{Total number of student = 60 + 20 = 80}}$

${\text{Probability of selecting boys = }}\dfrac{{60}}{{80}} = \dfrac{3}{4}$

${\text{Number of girls in class = 20}}$

${\text{Numbers of girls knowing cricket = 10}}$

${\text{Probability that selected one is girl knowing cricket = }}\dfrac{{10}}{{80}}{\text{ = }}\dfrac{1}{8}$

${\textbf{Step - 2: Find probability}}$

${\text{Probability of either boy or girls knowing cricket is}}$

${\text{Total probability = }}\dfrac{3}{4} + \dfrac{1}{8}$

${\text{ = }}\dfrac{{6 + 1}}{8}$

${\text{ = }}\dfrac{7}{8}$

${\textbf{Hence, the answer is }}\mathbf{\dfrac{7}{8}.}$

Box - $1$ contains $5$ red and $2$ blue balls white box - $2$ contains $2$ red and $6$ blue balls. A fair coin is tossed if it turns up head, a ball is drawn from the box - $1$ , else a ball is drawn from the box - $2$ . Find the probability of each of the following:
A red ball is drawn
Ball drawn is from box- $1$ if it is blue

2064966_1201291_ans_eaec1b53d34f4d70afdc494be49e1ee5.jpg

Following is the distribution function F (x) of a discrete r. v. X
x 1 2 3 4 5 6
F(x) 0.2 0.37 0.48 0.62 0.85 1
(i) Find the probability distribution of X
(ii) Find P (x $\leq$ 3), (2 < x < 5)

Probability distribution of x

$\begin{array}{l} P\left( { x\le 3 } \right) \\ \Rightarrow P\left( { x\le 3 } \right) =P\left( { x=3 } \right) +P\left( { x=2 } \right) +P\left({x=1}\right) \\ \Rightarrow P\left( { x\le 3 } \right) =0.48+0.37+0.2=0.87 \end{array}$

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In a bulb factory, three machines $A,B,C$ manufacture $60\%,25\%$ and $15\%$ of the total production respectively. Of their respective outputs $1\%,2\%$ and $1\%$ are defective. A bulb is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by machine $C$ .

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A letter is known to have come either from TATANAGAR or CALCUTTA. On the envelop just two consecutive letter TA are visible. What is the probability that the letter has come from CALCUTTA?

Let

${E}_{1}$ be the event that the letter has come from

$TATANAGAR$ and

${E}_{2}$ be the event that it has come from

$CALCUTTA$

Let

$A$ denote the event that the two consecutive letters on the envelope are

$TA$

$\therefore {E}_{1}$ and

${E}_{2}$ are mutually and exhaustive events.

$P\left({E}_{1}\right)=\dfrac{1}{2}$ and

$P\left({E}_{2}\right)=\dfrac{1}{2}$

$P\left(A/{E}_{1}\right)=$ Probability that the consecutive letters

$TA$ visible on the envelope belong to

$TATANAGAR$

$=\dfrac{2}{8}=\dfrac{1}{4}$ (There are

$8$ consecutive letters namely

$TA,AT,RA,AN,NA,AG,GA,AR$ out of which

$2$ cases are favourable.)

Similarly,

$P\left(A/{E}_{2}\right)=$ Probability that the consecutive letters

$TA$ visible on the envelope belong to

$CALCUTTA$

$=\dfrac{1}{7}$ (There are

$7$ consecutive letters namely

$CA,AL,LC,CU,UT,TT,TA$ out of which

$1$ case is favourable.)

$\therefore P\left(\dfrac{{E}_{2}}{A}\right)=\dfrac{P\left({E}_{2}\right)\dfrac{{E}_{2}}{A}}{P\left({E}_{1}\right)\dfrac{{E}_{1}}{A}+P\left({E}_{2}\right)\dfrac{{E}_{2}}{A}}$

$\dfrac{\dfrac{1}{2}\times\dfrac{1}{7}}{\dfrac{1}{2}\times\dfrac{1}{4}+\dfrac{1}{2}\times\dfrac{1}{7}}=\dfrac{1}{7}\times\dfrac{28}{11}=\dfrac{4}{11}$

There are 3 boxes, the first one containing 1 white, 2 red and 3 black balls; the second one containing 2 white, 3 red and 1 black ball and the third one containing 3 white, 1 red and 2 black balls. A box is chosen at random and from it two balls are drawn at random. One ball is red and the other, white. What is the probability that they come from the second box?

To solve this question, We have to use Bayes Theorem to get reverse probability.

Let first ball drawn is red and second ball drawn is white.

Probability of choosing a box at random

$=\dfrac{1}{3}$ .

probabilty of drawing a red ball from box

$-1 = p\left(R|{B}_{1}\right) =\dfrac{1}{3}$

probabilty of drawing a red ball from box

$-2 = p\left(R|{B}_{2}\right) =\dfrac{1}{2}$

probabilty of drawing a red ball from box

$-3 =\left(R|{B}_{3}\right) =\dfrac{1}{6}$

probability of getting the ball from box

$-2$ if red ball is selected is given by

$P\left({B}_{2}|R\right)=\dfrac{P\left({B}_{2}\right)\times p\left(R|{B}_{2}\right)}{P\left({B}_{1}\right)\times p\left(R|{B}_{1}\right)+P\left({B}_{2}\right)\times p\left(R|{B}_{2}\right)+P\left({B}_{3}\right)\times p\left(R|{B}_{3}\right)}$

$=\dfrac{\dfrac{1}{3}\times\dfrac{1}{2}}{\dfrac{1}{3}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{2}+\dfrac{1}{3}\times\dfrac{1}{6}}$

$=\dfrac{\dfrac{1}{6}}{\dfrac{1}{9}+\dfrac{1}{6}+\dfrac{1}{18}}$

$=\dfrac{\dfrac{1}{6}}{\dfrac{2+3+1}{18}}$

$=\dfrac{1}{6}\times\dfrac{18}{6}=\dfrac{1}{2}$ ......

$(1)$

Now let us consider white ball is drawn second time

probabilty of drawing a white ball from box

$-1 = p\left(W|{B}_{1}\right) =\dfrac{1}{6}$

probabilty of drawing a white ball from box

$-2 = p\left(W|{B}_{2}\right) =\dfrac{2}{5}$ ( note:- one ball is taken out already from box

$-2$ )

probabilty of drawing a white ball from box

$-3 = p\left(W|{B}_{3}\right) =\dfrac{1}{2}$

probability of getting the ball from box

$-2$ if white ball is selected is given by

$P\left({B}_{2}|W\right)=\dfrac{P\left({B}_{2}\right)\times p\left(W|{B}_{2}\right)}{P\left({B}_{1}\right)\times p\left(W|{B}_{1}\right)+P\left({B}_{2}\right)\times p\left(W|{B}_{2}\right)+P\left({B}_{3}\right)\times p\left(W|{B}_{3}\right)}$

$=\dfrac{\dfrac{1}{3}\times\dfrac{2}{5}}{\dfrac{1}{3}\times\dfrac{1}{6}+\dfrac{1}{3}\times\dfrac{2}{5}+\dfrac{1}{3}\times\dfrac{1}{2}}$

$=\dfrac{\dfrac{2}{15}}{\dfrac{1}{18}+\dfrac{2}{15}+\dfrac{1}{6}}$

$=\dfrac{3}{8}$ ......

$(2)$

Hence probability of getting the balls from box

$-2$ if red ball selected first time and white ball second time

$=\dfrac{1}{2}\times \dfrac{3}{8}=\dfrac{3}{16} ...................$ (3)$$

Now we consider the event, first selected ball is white and the second one red

probabilty of drawing a white ball from box

$-1 = p\left(W|{B}_{1}\right) =\dfrac{1}{6}$

probabilty of drawing a white ball from box

$-2 = p\left(W|{B}_{2}\right) =\dfrac{1}{3}$

probabilty of drawing a white ball from box

$-3 = p\left(W|{B}_{3}\right) =\dfrac{1}{2}$

probability of getting the ball from box

$-2$ if white ball is selected is given by

$=\dfrac{\dfrac{1}{3}\times\dfrac{1}{3}}{\dfrac{1}{3}\times\dfrac{1}{6}+\dfrac{1}{3}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{2}}$

$=\dfrac{\dfrac{1}{9}}{\dfrac{1}{18}+\dfrac{1}{9}+\dfrac{1}{6}}$

$=\dfrac{1}{3}$ ......

$(4)$

Now let us consider red ball is drawn second time

probabilty of drawing a red ball from box

$-1 = p\left(R|{B}_{1}\right) =\dfrac{1}{3}$

probabilty of drawing a red ball from box

$-2 = p\left(R|{B}_{2}\right) =\dfrac{3}{5}$

probabilty of drawing a red ball from box

$-3 = p\left(R|{B}_{3}\right) =\dfrac{1}{6}$

probability of getting the ball from box

$-2$ if red ball is selected second time is given by

$=\dfrac{\dfrac{1}{3}\times\dfrac{3}{5}}{\dfrac{1}{3}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{5}+\dfrac{1}{3}\times\dfrac{1}{6}}$

$=\dfrac{\dfrac{3}{15}}{\dfrac{1}{9}+\dfrac{1}{5}+\dfrac{1}{18}}$

$=\dfrac{6}{11}$ ......

$(5)$

Hence probability of getting the balls from box

$-2$ if white ball is selected first time and red ball is selected second time

$=\dfrac{1}{3}\times \dfrac{6}{11}=\dfrac{2}{11}$ ...................

$(6)$

Using

$(3)$ and

$(6)$ , probability of getting two balls from box

$-2$ if one of then red and other one is white

$=\dfrac{3}{16}+\dfrac{2}{11}=\dfrac{65}{176}=0.37$

Two cards are selected at random from a box which contains five cards numbered 1,1,2,2, andLet X denote the sum and Y the maximum of the two numbers drawn. Find the probability distribution, mean and variance of X and Y.

2039713_1664210_ans_49df37ab2f9d49fdb799778f85880ddf.jpg

A die is tossed 180 times. Find the expected value $(\mu )$ , variance $(q^{2})$ and S.D $(\sigma )$ of number of times the face 3 will appear.

Let E be event of getting face

$3$ in a single throw of dice

Let

$p=p(E)$

Since these are total

$6$ possible outcomes

$(1, 2, 3, 4, 5, 6)$ and

$1$ favourable outcomes i.e.,

$3$ .

$p=p(t)=\dfrac{1}{6}$

$p(\bar{E})=1-p=1-\dfrac{1}{6}=\dfrac{5}{6}$

Mean

$=$ np

Variance

$=np(1-p)$

Standard deviation

$=\sqrt{np(1-p)}$

Mean

$=180\times \dfrac{1}{6}30$

Variance

$=180\times \dfrac{1}{6}\left(1-\dfrac{1}{6}\right)=25$

Standard deviation

$=\sqrt{180\times \dfrac{1}{6}\left(1-\dfrac{1}{6}\right)}=\sqrt{25}=5$ .

1231163_1502237_ans_097841d6d44c4ef7a0e65ef9105ea86b.jpg

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, find the probability distribution of X.

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By examining the chest X-ray , the probability that TB is detected when a person is actually suffering from it is $0.99$ . The probability of an healthy person diagnosed to having TB is $0.001$ . In a certain city $1$ in $1000$ people suffers from TB . A person is selected at random and is diagnosed to have TB . What is the probability that he/she actually have TB ?

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Three numbers are selected at random (without replacement) from first six positive integers . If X denotes the smallest of the three numbers obtained , find the probability distribution of X . Also , find the mean and variance of the distribution.

First six positive integers are

$1 , 2 , 3 ,4 , 5$ and

$6$ .
If three numbers are selected at random from above six numbers then the number of elements in sample space

$S$ is given by

i.e.,

$n (s) = ^{6}C_{3} = \dfrac{6 !}{3 ! \, 3!} = \dfrac{6 \times 5 \times 4}{3 \times 2} = 20$

Here ,

$X$ smallest of the three numbers obtained , is random variable

$X$ may have value

$1 , 2 , 3$ and

$4$ .
Therefore , required probability distribution is given as

$P(X = 1)$ = Probability of event getting

$1$ as smallest number

$= \dfrac{^{5}C_{2}}{20} = \dfrac{5!}{2! 3! \times 20} = \dfrac{5 \times 4}{2 \times 20} = \dfrac{10}{20} = \dfrac{1}{2} [^{5}C_{2} \equiv$ selection of two numbers out of

$2 , 3 , 4 , 5 , 6 ]$

$P(X = 2)$ = Probability of events getting

$2$ as smallest number.

$= \dfrac{^{4}C_{2}}{20} = \dfrac{4!}{2!2! \times 20} = \dfrac{6}{20} = \dfrac{3}{10}$

$[^{4}C_{2} \equiv$ Selection of two numbers out of

$3 , 4 , 5 , 6 ]$

$P(X = 3)$ = Probability of events getting

$3$ as smallest number.

$= \dfrac{^{3}C_{2}}{20} = \dfrac{3!}{2!1! \times 20} = \dfrac{3}{20}$

$[^{3}C_{2} \equiv$ Selection of two numbers out of

$4 , 5 , 6 ]$

$P(X = 4)$ = Probability of events getting

$4$ as smallest number.

$= \dfrac{^{2}C_{2}}{20} = \dfrac{1}{20}$

$[^{2}C_{2} \equiv$ Selection of two numbers out of

$5 , 6 ]$

Required probability distribution table is

$X$ or $x_i$	$1$	$2$	$3$	$4$
$P(X)$ or $p_i$	$\dfrac{1}{2}$	$\dfrac{3}{10}$	$\dfrac{3}{20}$	$\dfrac{1}{20}$

Mean

$= S(X) = Sp_ix_i$

$= 1 \times \dfrac{1}{2} + 2 \times \dfrac{3}{10} + 3 \times \dfrac{3}{20} + 4 \times \dfrac{1}{20}$

$= \dfrac{1}{2} + \dfrac{6}{10} + \dfrac{9}{20} + \dfrac{4}{20} = \dfrac{10 + 12 + 9 + 4}{20} = \dfrac{35}{20} = \dfrac{7}{4}$

Variance

$= \sum x^{2}_{i}p_{i} - ( \sum X)^2$

$= \left \{ 1^2 \times \dfrac{1}{2} + 2^2 \times \dfrac{3}{10} + 3^2 \times \dfrac{3}{20} + 4^2 \times \dfrac{1}{20} \right \} - \left ( \dfrac{7}{4} \right )^2$

$= \dfrac{1}{2} + \dfrac{12}{10} + \dfrac{27}{20} + \dfrac{16}{20} - \dfrac{49}{16} = \dfrac{10 + 24 + 27 + 16}{20} - \dfrac{49}{16}$

$= \dfrac{77}{20} - \dfrac{49}{16} = \dfrac{308 - 245}{80} = \dfrac{63}{80}$

School	1	2	3	4	5	6	7	8	9	10
Number of planet survived	40	35	32	45	35	30	45	40	35	25

	White	Black	Red
${B}_{1}$	2	1	2
${B}_{2}$	3	2	4
${B}_{3}$	4	3	2

Outcome:	No head	One head	Two heads	Three heads
Frequency	14	38	36	12

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$2k$	$3k$	$k^{2}$	$2k^{2}$	$7k^{2}+k$

Probability - Class 12 Commerce Maths - Extra Questions

Given that EE and FF are events such that P(E)=0.6,P(F)=0.3P\left(E\right) = 0.6 , P\left(F\right) = 0.3 and P(E∩F)=0.2P\left(E \cap F\right) = 0.2, find 6P(F|E)6P\left(F | E\right).

Compute 100×P(A|B)100\times P\left(A | B\right), if P(B)=0.5P\left(B\right) = 0.5 and P(A∩B)=0.32P\left(A \cap B\right) = 0.32.

If P(A)=35P\left( A \right) = \displaystyle\frac { 3 }{ 5 } and P(B)=15P\left( B \right) = \displaystyle\frac { 1 }{ 5 } , find 100P(A∩B)100P\left( A\cap B \right) if AA and BB are independent events.

A card from a pack of 5252 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. If the probability of the lost card is a diamond is pp enter 100p100p

A fair coin is tossed 88 times. Find the probability that it shows heads at least once.

From a bag containing 2 one rupee and 3 two rupee coins a person is allowed to draw 2 coins indiscriminately; find the value of his expectation.

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as(i) number greater than 4(ii) six appears on at least one die

8p(\bar {A})8p(\bar {A})

Find P(B/A)P(B/A) when P(A)=\dfrac {5}{11},P(B)=\dfrac {6}{11}P(A)=\dfrac {5}{11},P(B)=\dfrac {6}{11} and P(A \cup B)=\dfrac {7}{11}P(A \cup B)=\dfrac {7}{11}.

The probability that AA hit a target is 1/41/4 and the probability that BB hits the target is 1/31/3. If each of them fired once, what is the probability that the target will be hit atleast once?

Find the probability of successes in toss of a die, where a success is defined as (i) Number greater than 44(ii) Six appears on die

A black and a red die are rolled together. Find the conditional probability of obtaining the sum 88, given that the red die resulted in a number less than 44.

For two events AA and BB with P(B)\ne 1P(B)\ne 1 prove that P(A|B')=\dfrac{P(A)-P(A\cap B)}{1-P(B)}P(A|B')=\dfrac{P(A)-P(A\cap B)}{1-P(B)}.

A card from pack of 52 cards is lost. From the remaining card of the pack, one card is drawn and is found to be heart, find the probability of missing card to be (I) heart (II) club.

A man is know to speak truth 44 out of 55 times. He tosses a coin and reports that it is head. Find the probability that it is actually head.

An urn contains 5 red and 2 black balls . Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ? Is X a random variable ?

A bag contains 1010 white and 1515 black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

An urn contains 55 red and 22 black balls. Two balls are randomly selected. If XX represents the number of black balls, what are the possible values of XX?

X=xX=x-2-2-1-100112233P(X=x)P(X=x)0.10.1KK0.20.22K2K0.30.3KKis the probability distribution of random variable XX. Find the value of KK and variance of XX.

A box contain 10 red balls , 20 yellow balls and 50 blue balls. If a ball is drwan at random from the box , find the probability that it will be (i) a blue ball (ii) neither yellow nor blue

If two different numbers are taken from the set \{0,1,2,3............,10\}\{0,1,2,3............,10\}; then the probability that their sum as well as absolute different are both multiple of 44 , is :

Compute P(A|B)P(A|B) if P(B)=0.5P(B)=0.5 and P(A \cap B)=0.32P(A \cap B)=0.32.

50 plants were planted each school out of 10 schools. After a month, the number of planets that survived are given below:Find Mean and VarianceSchool 12345678910Number of planet survived 40353245353045403525

IF P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{12}P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{12} and P(A \cap B)=\dfrac{1}{36}P(A \cap B)=\dfrac{1}{36}, are the events AA and BB independent ?

If 12 identical balls are to be placed in 3 different boxes, then the probability that one of the boxes contains excatly 3 balls, is :

A number is chosen at random from numbers -3,-2, -1,0,1,2,What is the probability that |x| < 2.

Probability of getting 2 when we roll a dice.

If p \in = 0.6p \in = 0.6 and p ( E \cap F ) = 0.2p ( E \cap F ) = 0.2 then find p ( F / E )p ( F / E )

Bag B_{1}B_{1} contains 44 white and 22 black balls. Bag B_{2}B_{2} cantains 33 white and 44 black balls. A bag is drawn at random and a ball is chosen at random from it. What is the probability that the ball drawn is white?

A bag contains 1010 red balls and 88 green balls. 22 balls are drawn at random one by one with replacement. Find the probability that both the balls are green that a second year student is chosen.

A bag contains 6 red balls and 8 green balls. A ball is drawn at random . What is the probability that the ball is green

Chance that Sheela tells truth is 35% and for Ramesh is 75%. In what percent they likely to contradict each other in the same question?

A die is thrown once. If A is the event "the number appearing is a multiple of 3"3" and B is the event "the number appearing is even" Are the event A and B independent ?

A bag contains 4 balls.Two balls drawn at random without replacement and are found to be white. What is the probability that all balls are white?

Compute P ( A | B ) ,P ( A | B ) , if P ( B ) = 0.25P ( B ) = 0.25 and P ( A \cap B ) = 0.18P ( A \cap B ) = 0.18.

Evaluate P(A \cup B)P(A \cup B), if 2P(A)=P(B)=\dfrac{5}{13}2P(A)=P(B)=\dfrac{5}{13} and P(A/B)=\dfrac{2}{5}P(A/B)=\dfrac{2}{5}.

Find the expected number of boys in a family with 8 children,assuming the sex distribution to be equally probable.

Compute P ( A | B ) ,P ( A | B ) , if P ( B ) = 0.5P ( B ) = 0.5 and P ( A \cap B ) = 0.32P ( A \cap B ) = 0.32.

The probability distribution function of a random variable XX is given byx_{i}x_{i}:012p_{i}p_{i}:3c^{}3c^{}10c10c5c-15c-1where c > 0c > 0. Find cc.

The probability of student AA passing an examination is \dfrac{2}{9}\dfrac{2}{9} and of students, BB passing is \dfrac{5}{9}\dfrac{5}{9}. Assuming the two events: AA passes'. BB passes' as independent, find the probability of both passing the examination.

An urn contains 1010 black and 55 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Two numbers are selected at random from integers 11 through 99. If the sum is even, find the probability that both the numbers are odd.

Kamal and Monika appeared for an interview for two vacancies. The probability of Kamal's selection is 1/31/3 and that of Monika's selection is 1/51/5. Find the probability that at least one of them will be selected.

Find the value of kk in the given probability distribution:x_{i}x_{i}:234p_{i}p_{i}:0.2kk0.3

The probability that it will rain tomorrow is 0.850.85. What is the probability that it will not rain tomorrow ?

To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:Opinion:LikeDislikeNumber of students:13565Find the probability that a student chosen at random likes Mathematics

A school gives awards to the student each class 55 for bravery, 33 for punctuality, 33 for full attendance, 44 for social services 55 self confidence. An awarded student is selected at random. What is the probability that he /she is being awarded for self confidence.

A school gives awards to the student each class 55 for bravery, 33 for punctuality, 33 for full attendance, 44 for social services 55 self confidence. An awarded student is selected at random. What is the probability that he /she is being awarded for punctuality

Match column I with column II

In a class 3030% students fail in English; 2020% students fail in Hindi and 1010% students fail in English and Hindi both. A student is chosen at random, then the probability that he will fail English if he has failed in Hindi is \dfrac { k }{ 2 } \dfrac { k }{ 2 } .The value of kk is

An urn contains 2 white and 2 black balls. A ball is drawn at random. If it is white it is not replaced into the urn. Otherwise it is along with another ball of same colour. The process is repeated. the probability that the third ball drawn is black is a/b (no common divisor), then b-a =

\displaystyle P\left ( A\cup B \right )=P\left ( A\cap B \right )\displaystyle P\left ( A\cup B \right )=P\left ( A\cap B \right ) if and only if the ralation between P(A)P(A) and P(B)P(B) is P(A)+P(B)=2P(AB)P(A)+P(B)=2P(AB). If this is true enter 1, else enter 0.

One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 98,If X and Y denote the sum and the product of the digits on the tickets, What is the value of 57P(X=9|Y=0)57P(X=9|Y=0)?

A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. The probability that the second selected ball happens to be red if the first selected ball is red, is \dfrac{2}{k}.\dfrac{2}{k}. Find k.

A bag contains n+1 coins. It is known that one of these coins has heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that the toss results in heads is 7/12, find n

A speaks truth 33 out of 44 times. He reported that Mohan Bagan has won the match. The probability that his report was correct is \displaystyle \frac { k }{ 5 } \displaystyle \frac { k }{ 5 } , then k=k=

The probability that an archer hits the target when it is windy is 0.4; when it is not windy, her probability of hitting the target is 0.On any shot, the probability of a gust of wind is 0.The probability that she hits the target on first shot is 0.x10.x1. Find xx.

If three coins are tossed, represent the sample space and the event of getting atleast two heads, then find the number of elements in them.

Box-I contains 55 red and 44 white balls while box-II contains 44 red and 22 white balls. A fair die is thrown. If it turns up a multiple of 3,3, a ball is drawn, from box-I else a ball is drawn from box-II. The probability that the ball drawn is white is \dfrac{a}{27}\dfrac{a}{27}. Find aa.

AA speaks the truth '33 times out of 44' and BB speaks the truth '22 times out of 33', AA die is thrown. Both assert that the number turned up is 2.2. Find the probability of the truth of their assertion.

Cards of an ordinary deck of playing cards are placed into two heaps. Heap-I consists of all the red cards and heap-II consists of all the black cards. A heap is chosen at random and a card is drawn, find the probability that the card drawn is a king.

Mother, father and son line up at random for a family pictureE : E : son on one end, F : F : father in middle.Determine P\left( E|F \right) P\left( E|F \right)

A die is thrown three times,E\ :\ 4E\ :\ 4 appears on the third toss, F\ :\ 6F\ :\ 6 and 55 appears respectively on first two tosses.Determine 36P36P where P\left( E|F \right) P\left( E|F \right)

If P\left(A\right) = 0.8, P\left(B\right) = 0.5P\left(A\right) = 0.8, P\left(B\right) = 0.5 and P\left(B | A\right) = 0.4P\left(B | A\right) = 0.4, find(i) P\left(A \cap B\right)P\left(A \cap B\right)(ii) P\left(A | B\right)P\left(A | B\right)(iii) P\left(A \cup B\right)P\left(A \cup B\right)

A bag contains 44 red and 44 black balls, another bag contains 22 red and 66 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Consider the experiment of throwing a die, if a multiple of 33 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a 33'.

Two cards are drawn at random and without replacement from a pack of 5252 playing cards. Find the probability that both the cards are black.

State which of the following are not the probability distributions of a random variable. Give reasons for your answer.

Find the probability distribution of(i) number of heads in two tosses of a coin.

There are three coins. One is a two-headed coin another is a biased coin that comes up heads 7575% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?

An insurance company insured 20002000 scooter drivers, 40004000 car drivers and 60006000 truck drivers. The probability of an accidents are 0.01, 0.030.01, 0.03 and 0.150.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ?

Two numbers are selected at random (without replacement) the first six positive integers. Let XX denote the larger of the two numbers obtained. Find E\left(X\right)E\left(X\right).

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

Two dice are thrown simultaneously. If XX denotes the number of sixes, find the expectation of XX.

What is the minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than 90\%90\% ?

Bag II contains 33 red and 44 black balls and Bag IIII contains 44 red and 55 black balls. One ball is transferred from Bag II and Bag IIII and then a ball is drawn from Bag IIII. The ball so drawn is found to be red in colour. Find 310310 times the probability that the transferred ball is black.

In a hurdle race, a player has to cross 1010 hurdles. The probability that he will clear each hurdle is \displaystyle\frac { 5 }{ 6 } \displaystyle\frac { 5 }{ 6 } . What is the probability that he will knock down fewer than 22 hurdles ?

In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he win / loses.

Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is 66.

Given that $E$ and $F$ are events such that $P\left(E\right) = 0.6 , P\left(F\right) = 0.3$ and $P\left(E \cap F\right) = 0.2$ , find $6P\left(F | E\right)$ .

Compute $100\times P\left(A | B\right)$ , if $P\left(B\right) = 0.5$ and $P\left(A \cap B\right) = 0.32$ .

If $P\left( A \right) = \displaystyle\frac { 3 }{ 5 }$ and $P\left( B \right) = \displaystyle\frac { 1 }{ 5 }$ , find $100P\left( A\cap B \right)$ if $A$ and $B$ are independent events.

A card from a pack of $52$ cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. If the probability of the lost card is a diamond is $p$ enter $100p$

A fair coin is tossed $8$ times. Find the probability that it shows heads at least once.

Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as
(i) number greater than 4
(ii) six appears on at least one die

$8p(\bar {A})$

Find $P(B/A)$ when $P(A)=\dfrac {5}{11},P(B)=\dfrac {6}{11}$ and $P(A \cup B)=\dfrac {7}{11}$ .

The probability that $A$ hit a target is $1/4$ and the probability that $B$ hits the target is $1/3$ . If each of them fired once, what is the probability that the target will be hit atleast once?

Find the probability of successes in toss of a die, where a success is defined as
(i) Number greater than $4$
(ii) Six appears on die

A black and a red die are rolled together. Find the conditional probability of obtaining the sum $8$ , given that the red die resulted in a number less than $4$ .

For two events $A$ and $B$ with $P(B)\ne 1$ prove that $P(A|B')=\dfrac{P(A)-P(A\cap B)}{1-P(B)}$ .

A man is know to speak truth $4$ out of $5$ times. He tosses a coin and reports that it is head. Find the probability that it is actually head.

A bag contains $10$ white and $15$ black balls. Two balls are drawn in succession without replacement. What is the probability that first is white and second is black?

An urn contains $5$ red and $2$ black balls. Two balls are randomly selected. If $X$ represents the number of black balls, what are the possible values of $X$ ?

$X=x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X=x)$ $0.1$ $K$ $0.2$ $2K$ $0.3$ $K$
is the probability distribution of random variable $X$ . Find the value of $K$ and variance of $X$ .

A box contain 10 red balls , 20 yellow balls and 50 blue balls. If a ball is drwan at random from the box , find the probability that it will be
(i) a blue ball (ii) neither yellow nor blue

If two different numbers are taken from the set $\{0,1,2,3............,10\}$ ; then the probability that their sum as well as absolute different are both multiple of $4$ , is :

Compute $P(A|B)$ if $P(B)=0.5$ and $P(A \cap B)=0.32$ .

50 plants were planted each school out of 10 schools. After a month, the number of planets that survived are given below:Find Mean and Variance
School 1 2 3 4 5 6 7 8 9 10
Number of planet survived 40 35 32 45 35 30 45 40 35 25

IF $P(A)=\dfrac{1}{3}, P(B)=\dfrac{1}{12}$ and $P(A \cap B)=\dfrac{1}{36}$ , are the events $A$ and $B$ independent ?

If $p \in = 0.6$ and $p ( E \cap F ) = 0.2$ then find $p ( F / E )$

Bag $B_{1}$ contains $4$ white and $2$ black balls. Bag $B_{2}$ cantains $3$ white and $4$ black balls. A bag is drawn at random and a ball is chosen at random from it. What is the probability that the ball drawn is white?

A bag contains $10$ red balls and $8$ green balls. $2$ balls are drawn at random one by one with replacement. Find the probability that both the balls are green that a second year student is chosen.

A die is thrown once. If A is the event "the number appearing is a multiple of $3"$ and B is the event "the number appearing is even" Are the event A and B independent ?

Compute $P ( A | B ) ,$ if $P ( B ) = 0.25$ and $P ( A \cap B ) = 0.18$ .

Evaluate $P(A \cup B)$ , if $2P(A)=P(B)=\dfrac{5}{13}$ and $P(A/B)=\dfrac{2}{5}$ .

Compute $P ( A | B ) ,$ if $P ( B ) = 0.5$ and $P ( A \cap B ) = 0.32$ .

The probability distribution function of a random variable $X$ is given by
$x_{i}$ : 0 1 2
$p_{i}$ : $3c^{}$ $10c$ $5c-1$
Find $c$ .

The probability of student $A$ passing an examination is $\dfrac{2}{9}$ and of students, $B$ passing is $\dfrac{5}{9}$ . Assuming the two events: $A$ passes'. $B$ passes' as independent, find the probability of both passing the examination.

An urn contains $10$ black and $5$ white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Two numbers are selected at random from integers $1$ through $9$ . If the sum is even, find the probability that both the numbers are odd.

Kamal and Monika appeared for an interview for two vacancies. The probability of Kamal's selection is $1/3$ and that of Monika's selection is $1/5$ . Find the probability that at least one of them will be selected.

Find the value of $k$ in the given probability distribution:
$x_{i}$ : 2 3 4
$p_{i}$ : 0.2 $k$ 0.3

The probability that it will rain tomorrow is $0.85$ . What is the probability that it will not rain tomorrow ?

To know the opinion of the students about Mathematics, a survey of 200 students was conducted. The data is recorded in the following table:
Opinion: Like Dislike
Number of students: 135 65
Find the probability that a student chosen at random likes Mathematics

A school gives awards to the student each class $5$ for bravery, $3$ for punctuality, $3$ for full attendance, $4$ for social services $5$ self confidence. An awarded student is selected at random. What is the probability that he /she is being awarded for self confidence.

A school gives awards to the student each class $5$ for bravery, $3$ for punctuality, $3$ for full attendance, $4$ for social services $5$ self confidence. An awarded student is selected at random. What is the probability that he /she is being awarded for punctuality

In a class $30$ % students fail in English; $20$ % students fail in Hindi and $10$ % students fail in English and Hindi both. A student is chosen at random, then the probability that he will fail English if he has failed in Hindi is $\dfrac { k }{ 2 }$ .The value of $k$ is

$\displaystyle P\left ( A\cup B \right )=P\left ( A\cap B \right )$ if and only if the ralation between $P(A)$ and $P(B)$ is $P(A)+P(B)=2P(AB)$ . If this is true enter 1, else enter 0.

One ticket is selected at random from 100 tickets numbered 00, 01, 02, ..., 98,If X and Y denote the sum and the product of the digits on the tickets, What is the value of $57P(X=9|Y=0)$ ?

A bag contains 3 red and 7 black balls. Two balls are selected at random one-by-one without replacement. The probability that the second selected ball happens to be red if the first selected ball is red, is $\dfrac{2}{k}.$ Find k.

A speaks truth $3$ out of $4$ times. He reported that Mohan Bagan has won the match. The probability that his report was correct is $\displaystyle \frac { k }{ 5 }$ , then $k=$

he hits the target on first shot is $0.x1$ . Find $x$ .

Box-I contains $5$ red and $4$ white balls while box-II contains $4$ red and $2$ white balls. A fair die is thrown. If it turns up a multiple of $3,$ a ball is drawn, from box-I else a ball is drawn from box-II. The probability that the ball drawn is white is $\dfrac{a}{27}$ . Find $a$ .

$A$ speaks the truth ' $3$ times out of $4$ ' and $B$ speaks the truth ' $2$ times out of $3$ ', $A$ die is thrown. Both assert that the number turned up is $2.$ Find the probability of the truth of their assertion.

Mother, father and son line up at random for a family picture
$E :$ son on one end,
$F :$ father in middle.
Determine $P\left( E|F \right)$

A die is thrown three times,
$E\ :\ 4$ appears on the third toss, $F\ :\ 6$ and $5$ appears respectively on first two tosses.
Determine $36P$ where $P\left( E|F \right)$

If $P\left(A\right) = 0.8, P\left(B\right) = 0.5$ and $P\left(B | A\right) = 0.4$ , find
(i) $P\left(A \cap B\right)$
(ii) $P\left(A | B\right)$
(iii) $P\left(A \cup B\right)$

A bag contains $4$ red and $4$ black balls, another bag contains $2$ red and $6$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Consider the experiment of throwing a die, if a multiple of $3$ comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event 'the coin shows a tail', given that 'at least one die shows a $3$ '.

Two cards are drawn at random and without replacement from a pack of $52$ playing cards. Find the probability that both the cards are black.

Find the probability distribution of
(i) number of heads in two tosses of a coin.

There are three coins. One is a two-headed coin another is a biased coin that comes up heads $75$ % of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two-headed coin?

An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents are $0.01, 0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver ?

Two numbers are selected at random (without replacement) the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E\left(X\right)$ .

Two dice are thrown simultaneously. If $X$ denotes the number of sixes, find the expectation of $X$ .

What is the minimum number of times must a man toss a fair coin so that the probability of having at least one head is more than $90\%$ ?

In a hurdle race, a player has to cross $10$ hurdles. The probability that he will clear each hurdle is $\displaystyle\frac { 5 }{ 6 }$ . What is the probability that he will knock down fewer than $2$ hurdles ?

Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is $6$ .

Three cards are drawn successively with replacement from a well shuffled pack of $52$ cards. Find the probability distribution of the number of spades. Hence find the mean of the distribution.

$8p(\bar A \cup \bar B)$

$8p(\bar A \cap B)$

If A and B are both random events.
$p(A)=\dfrac{3}{8}$ ; $p(B)=\dfrac{1}{2}$ ; $p(A\ \cap\ B)=\dfrac{1}{4}$
Calculate:
$8p(\bar A \cap \bar B)$

$4p(\bar {B})$

$8p( A \cap \bar B)$

From a pack of $52$ cards, two cards are drawn randomly one by one without replacement. Find the probability that both of them are red.

A die is thrown twice and the sum of the numbers appearing is observed to be $7$ . Find the conditional probability that the number $3$ has appeared at least once.

If $P(A)=\dfrac{4}{5}$ and $P(B/A)=\dfrac{2}{5}$ , find $P(A\cap B).$

If $P(A)=0.8$ and $P(B|A)=0.4$ , then find $P(A\cap B)$ .

Three boxes numbered I, II, III contains the balls as follows:
White Black Red
I $1$ $2$ $3$
II $2$ $1$ $1$
III $4$ $5$ $3$
One box is randomly selected and a ball is drawn from it. If the ball is red, then find the probability that it is from box II

A person buys a lottery ticket in 50 lotteries in each of which his chance of winning a price is $\dfrac{1}{100}$ . What is the probability that he will win a price exactly once?

If $P(A)=0.6, P(B)=0.7$ and $P(A\cup B)=0.9$ , then find $P\left(\dfrac{A}{B}\right)$ and $P\left(\dfrac{B}{A}\right)$ .

If $mn$ coins have been distributed into na purses, n into each, find $(1)$ the chance that two specified coins will be found in the same purse; and $(2)$ what the chance becomes when $r$ purses have been examined and found not to contain either of the specified coins.

A certain stake is to be won by the first person who throws an ace with an octahedral die, if there are $4$ persons what is the chance of the last to win the stake?

From a lot of $6$ items containing $2$ defective items, a sample of $4$ items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find the probability distribution of X.

If A and B are events such that $P(A)=\displaystyle\frac{1}{2}$ , $P(B)=\displaystyle\frac{1}{3}$ and $P(A\cap B)=\displaystyle\frac{1}{4}$ , then find $P(B/A)$ .

If A and B are events such that $P(A)=\displaystyle\frac{1}{2}$ , $P(B)=\displaystyle\frac{1}{3}$ and $P(A\cap B)=\displaystyle\frac{1}{4}$ , then find $P(A/B)$ .

If $P(A) = \dfrac {7}{13}, P(B) = \dfrac {9}{13}$ and $P(A\cap B) = \dfrac {4}{13}$ , find $P(A/ B)$ .

A random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$
$P(X)$ $0.1$ $k$ $2k$ $2k$ $k$
Determine:
(i) $k$
(ii) $P(X \geq 2)$ .