MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 10

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least

$0.8$ , is

0%
$7$
0%
$6$
0%
$5$
0%
$3$

Explanation

$\textbf{Step -1: Finding the probability of getting head in each toss.}$

$\text{Let a coin is tossed }n\text{ times.}$

$\text{The probability of getting head in one trial, }p=\dfrac{1}{2}.$

$\textbf{Step -2: Finding the number of times a coin is tossed.}$

$\text{Probability of getting at least one head}=P(X\geq1)$

$=1-P(X=0)$

$=1-^{n}C_{0}(1-p)^n$

$=1-\left(1-\dfrac{1}{2}\right)^n$

$=1-\dfrac{1}{2^n}\geq0.8$ $\text{(Given)}$

$\Rightarrow \dfrac{1}{2^n}\leq1-0.8$

$\Rightarrow \dfrac{1}{2^n}\leq0.2$

$\text{When }n=1,$

$\dfrac{1}{2^n}=\dfrac{1}{2}=0.5>0.2$

$\text{When }n=2,$

$\dfrac{1}{2^n}=\dfrac{1}{2^2}=0.25>0.2$

$\text{When }n=3,$

$\dfrac{1}{2^n}=\dfrac{1}{2^3}=0.125<0.2$

$\therefore n\text{ should be }3.$

$\textbf{Hence, correct option is D.}$

A fair coin is tossed

$n$ times. Let

$X=$ the number of times head occurs

$P(X=4),P(X=5)$ and

$P(X=6)$ are in A.P., then the value of

$n$ can be

0%
$7$
0%
$10$
0%
$12$
0%
$14$

Explanation

P(X=4), P(X=5), P(X=6) is given by

$\dfrac { { C }_{ 4 }^{ n } }{ { 2 }^{ n } } ,\dfrac { { C }_{ 5 }^{ n } }{ { 2 }^{ n } } ,\dfrac { { C }_{ 6 }^{ n } }{ { 2 }^{ n } }$ respectively.

Since they are in AP, we have:

${ C }_{ 4 }^{ n },{ C }_{ 5 }^{ n },{ C }_{ 6 }^{ n }$ in AP.
Thus,

${ C }_{ 4 }^{ n }+{ C }_{ 6 }^{ n }=2*{ C }_{ 5 }^{ n }$
This is satisfied when

$n = 7$ or

$14.$

India plays two matches each with West Indies and Australia. In any match the probabilities of India getting points

$0, 1$ and

$2$ are

$0.45, 0.05,$ and

$0.50$ respectively. Assuming that the outcomes are independent, the probability of India getting at least

$7$ points is

0%
$0.8750$
0%
$0.0875$
0%
$0.0625$
0%
$0.0250$

Explanation

Total points should be

$\geq7$
Possible scores are

Points =7
Sequence of points will be

$1,2,2,2$

$P(7) = ^4C_1(0.5)^30.05 = 0.025$

Points =8
Sequence of points will be

$2,2,2,2$

$P(8) = (0.5)^4 = 0.0625$

Points >7

$P(\ge7) = 0.0625+0.025 = 0.0875$

$A$ is a set containing

$n$ elements. A subset

$P$ of

$A$ is chosen at random. The set

$A$ is reconstructed by replacing the elements of

$P$ . A subset

$Q$ of

$A$ is again chosen at random. Find the probability that

$P$ and

$Q$ have no common elements.

0%
${(3/4)}^{n}$
0%
${(1/4)}^{n}$
0%
${(2/3)}^{n}$
0%
${(1/3)}^{n}$

Explanation

Since, set

$A$ contains

$n$ elements. So, it has

${ 2 }^{ n }$ subsets.
Therefore set

$P$ can be chosen in

${ 2 }^{ n }$ ways, similarly set

$Q$ can be chosen in

${ 2 }^{ n }$ ways.
Therefore

$P$ and

$Q$ can be chosen in

$({ 2 }^{ n })({ 2 }^{ n })={ 4 }^{ n }$ ways.

Suppose,

$P$ contains

$r$ elements, where

$r$ varies from

$0$ to

$n$ . Then,

$P$ can be chosen in

$^{ n }{ { C }_{ r } }$ ways, for

$0$ to be disjoint from

$A$ , it should be chosen from set of all subsets of set consisting of remaining

$(n-r)$ elements.This can be done in

${ 2 }^{ n-r }$ ways.
Therefore

$P$ and

$Q$ can be chosen in

$_{ }^{ n }{ { C }_{ r } }.{ 2 }^{ n-r }$ ways.
But,

$r$ can be vary from

$0$ to

$n$ .
Therefore total number of disjoint sets

$P$ and

$Q$

$=\sum _{ r=0 }^{ n }{ _{ }^{ n }{ { C }_{ r } }.{ 2 }^{ n-r } } ={ \left( 1+2 \right) }^{ n }={ 3 }^{ n }$
Hence required probability

$\displaystyle =\frac { { 3 }^{ n } }{ { 4 }^{ n } } ={ \left( \frac { 3 }{ 4 } \right) }^{ n }$

Numbers are selected at random, one at a time, from the two digit numbers 00, 01, 02......., 99 with replacement. An even E occurs if and if the product of the two digits of a selected number isIf four numbers are selected, find the probability that the event E occurs at least 3 times.

0%
$\displaystyle 97/(5)^4$
0%
$\displaystyle 97/(25)^4$
0%
$\displaystyle 98/(25)^4$
0%
$\displaystyle 96/(25)^4$

Explanation

Out of the number 00,01,02...99 those number the product of whose digits is 18 ar 29,36,63,92

$\displaystyle \therefore P\left( E \right) =\frac { 4 }{ 100 } =\frac { 1 }{ 25 }$

$\displaystyle P\left( \overline { E } \right) =\frac { 24 }{ 25 }$
The probability that in selecting from number the event E occur at least three times

$\displaystyle =^{ 4 }{ { C }_{ 1 } }{ \left( \frac { 1 }{ 25 } \right) }^{ 3 }\frac { 24 }{ 25 } +^{ 4 }{ { C }_{ 4 } }{ \left( \frac { 1 }{ 25 } \right) }^{ 4 }=\frac { 4\times 24 }{ { \left( 25 \right) }^{ 4 } } +\frac { 1 }{ { 24 }^{ 4 } } =\frac { 97 }{ { \left( 25 \right) }^{ 4 } }$

One hundred identical coins, each with probability,

$p$ , of showing up heads are tossed once. If

$0<p<1$ and the probability of heads showing on fifty coins is equal to that of heads showing on

$51$ coins, then the value of

$p$ is :

0%
$1/2$
0%
$49/101$
0%
$50/101$
0%
$51/101$

Explanation

Applying binomial distribution for 100 trials we get

$\:^{100}C_{50}p^{50}(1-p)^{50}=\:^{100}C_{51}p^{51}(1-p)^{49}$

$\dfrac{(1-p)100!}{50!50!}=\dfrac{(p)100!}{51!49!}$

$\dfrac{1-p}{50}=\dfrac{p}{51}$

$51-51p=50p$

$101p=51$

$p=\dfrac{51}{101}$

Hence the required probability is

$\dfrac{51}{101}$

A card is drawn from a pack. The card is replaced and the pack is reshuffled. If this is done six times, the probability that

$2$ hearts,

$2$ diamonds and

$2$ black cards are drawn is

0%
$\displaystyle 90.\frac {1}{4}^6$
0%
$\displaystyle \frac {45}{2}.\frac {3}{4}^4$
0%
$\displaystyle \frac {90}{2^{10}}$
0%
none of these

Explanation

The probability of getting a heart in one draw

$\displaystyle =\frac { 13 }{ 52 } =\frac { 1 }{ 4 }$
Similarly, for a diamond the probability

$\displaystyle =\frac { 13 }{ 52 } =\frac { 1 }{ 4 }$
The probability of getting a black card in one draw

$\displaystyle =\frac { 26 }{ 52 } =\frac { 1 }{ 2 }$

$\therefore$ The required probability

$\displaystyle =^{ 6 }{ { C }_{ 2 } }\times ^{ 4 }{ { C }_{ 2 } }\times ^{ 2 }{ { C }_{ 2 } }\times { \left( \frac { 1 }{ 4 } \right) }^{ 2 }\times { \left( \frac { 1 }{ 4 } \right) }^{ 2 }\times { \left( \frac { 1 }{ 2 } \right) }^{ 2 }=90{ \left( \frac { 1 }{ 2 } \right) }^{ 10 }$

Suppose

$X$ follows a binomial distribution with parameters

$n$ and

$p$ , where

$0<p<1$ . If

$\displaystyle \frac { P\left( X=r \right) }{ P\left( X=n-r \right) }$ is independent of

$n$ and

$r$ , then

$p$ is equal to

0%
$\displaystyle \frac { 1 }{ 3 }$
0%
$\displaystyle \frac { 1 }{ 2 }$
0%
$\displaystyle \frac { 1 }{ 4 }$
0%
None of these

Explanation

$\displaystyle \frac { P\left( X=r \right) }{ P\left( X=n-r \right) } =\frac { _{ }^{ n }{ { C }_{ r } }{ p }^{ r }{ \left( 1-p \right) }^{ n-r } }{ _{ }^{ n }{ { C }_{ n-r }{ p }^{ n-r } }{ \left( 1-p \right) }^{ r } } =\frac { \left( 1-p \right) ^{ n-2r } }{ { p }^{ n-2r } } ={ \left( \frac { 1 }{ p } -1 \right) }^{ n-2r }$
For

$\displaystyle { \left( \frac { 1 }{ p } -1 \right) }^{ n-2r }$ to be independent of

$n$ and

$\displaystyle r,\frac { 1 }{ p } -1=1$

$\displaystyle \Rightarrow \frac { 1 }{ p } =1\Rightarrow p=\frac { 1 }{ 2 } .$

A coin is tossed

$n$ times. The probability of getting at least one head is greater than that of getting at least two tails by

$\displaystyle \frac {5}{32}$ . Then

$n$ is

0%
$5$
0%
$10$
0%
$15$
0%
none of these

Explanation

Probability of getting atleast one head

$=1-$ (Probability of 0 heads)

$=1-\:^{n}C_{0}.\dfrac{1}{2^{n}}$

$=1-\dfrac{1}{2^{n}}$ ...(i)
Probability of atleast two tails

$=1-$ (Probability of atmost 1 tails)

$=1-\dfrac{\:^{n}C_{0}}{2^{n}}-\dfrac{\:^{n}C_{1}}{2^{n}}$

$=1-\dfrac{1}{2^{n}}-\dfrac{n}{2^{n}}$ ...(ii)
Now subtracting ii from i, we get

$\dfrac{n}{2^{n}}$

$=\dfrac{5}{32}$
Hence

$n=5$ .

A coin is tossed 10 times, probability that on the

$10^{th}$ throw to observed

$5^{th}$ head is

0%
$\displaystyle ^{9}\textrm{C}_{5}\:\displaystyle \frac{1}{2^{5}}$
0%
$\displaystyle\frac{ ^{9}\textrm{C}_{4}}{2^{10}}$
0%
$\displaystyle\frac{ ^{9}\textrm{C}_{5}}{2^{10}}$
0%
None of these

Explanation

Since

$5^{th}$ head is needed on the tenth throw, 4 heads have already been appeared in the previous 9 throws.
Thus, the required probability equals

$^9C_4 (\dfrac{1}{2})^5 (\dfrac{1}{2})^4 (\dfrac{1}{2}) = ^9C_4 (\dfrac{1}{2})^{10}$

$^9C_4 = ^9C_5$ and so option c is also correct.

A dice is thrown

$2n + 1$ times,

$\displaystyle n \: \epsilon \: N$ . The probability that faces with even numbers show odd number of times is

0%
$\displaystyle \frac {2n + 1}{4n + 3}$
0%
less than $\displaystyle \frac {1}{2}$
0%
greater than $\displaystyle \frac {1}{2}$
0%
none of these

Explanation

Probability of choosing an even number or an odd number =

$\dfrac { 1 }{ 2 }$
The even number can come up either once or thrice or 5 times ... or 2n+1 times; the remaining throws should give odd numbers.
Therefore, probability =

$_{ 1 }^{ (2n+1) }{ C }*{ (\dfrac { 1 }{ 2 } ) }^{ (2n+1) }+_{ 3 }^{ (2n+1) }{ C }*{ (\dfrac { 1 }{ 2 } ) }^{ (2n+1) }+...+_{ (2n+1) }^{ (2n+1) }{ C }*{ (\dfrac { 1 }{ 2 } ) }^{ (2n+1) }=\dfrac { _{ 1 }^{ (2n+1) }{ C }+_{ 3 }^{ (2n+1) }{ C }+...+_{ (2n+1) }^{ (2n+1) }{ C } }{ { 2 }^{ (2n+1) } } =\dfrac { { 2 }^{ 2n } }{ { 2 }^{ 2n+1 } } =\dfrac { 1 }{ 2 }$

There is

$30%$ chance that it rains on any particular day. Given that there is at least one rainy day in a week, what is the probability that there are at least two rainy days in the week?

0%
$\displaystyle \frac { 1-4 (0.7) ^7}{ 1-(0.7)^7 }$
0%
$\displaystyle \frac { 4 (0.7) ^7}{ 1-(0.7)^7 }$
0%
$\displaystyle \frac { 1- (0.7) ^7}{ 1-4.(0.7)^7 }$
0%
$\displaystyle \frac { (0.7) ^7}{ 1-4.(0.7)^7 }$

Explanation

Let

$A$ be the event that there is at least one rainy day in a week.
Let

$B$ be the event that there are at least two rainy days in a week

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

$B\subseteq A$

$=\dfrac{(1-P(\text{one rainy day or no rainy day}))}{(1-P(\text{no rainiy day}))}$

$\displaystyle =\frac { 1-\left[ ^{ 7 }{ { C }_{ 1 }{ \left( 0.3 \right) }^{ 1 }{ \left( 0.7 \right) }^{ 6 }+^{ 7 }{ { C }_{ 0 }{ \left( 0.7 \right) }^{ 7 } } } \right] }{ 1-{ \left( 0.7 \right) }^{ 7 } } =\frac { 1-\left( 4 \right) { \left( 0.7 \right) }^{ 7 } }{ 1-{ \left( 0.7 \right) }^{ 7 } }$

A bag contains

$14$ balls of two colours, the number of balls of each colour being the same.

$7$ balls are drawn at random one by one. The ball in hand is returned to the bag before each new draw. If the probability that at least

$3$ balls of each colour are drawn is

$p$ then

0%
$\displaystyle p > \frac {1}{2}$
0%
$\displaystyle p = \frac {1}{2}$
0%
$\displaystyle p < 1$
0%
$\displaystyle p < \frac {1}{2}$

Explanation

There are two possible cases.
Case One,

$3$ balls are drawn of first color and

$4$ balls are drawn of second color.
Case Two,

$4$ balls are drawn of first color and

$3$ balls are drawn of second color.
The total probability is therefore,

$^7{ C }_{ 3 }{ \left(\dfrac { 7 }{ 14 } \right) }^{ 3 }{ \left(\dfrac { 7 }{ 14 } \right) }^{ 4 }+^7{ C }_{ 4 }{ \left(\dfrac { 7 }{ 14 } \right) }^{ 4 }{ \left(\dfrac { 7 }{ 14 } \right) }^{ 3 }=\dfrac { 35 }{ 64 } >\dfrac { 1 }{ 2 }$

An ordinary dice is rolled a certain number of times. If the probability of getting an odd number

$2$ times is equal to the probability of getting an even number

$3$ times. Then the probability of getting an odd number for odd number of times is:

0%
$\displaystyle \frac {1}{32}$
0%
$\displaystyle \frac {5}{16}$
0%
$\displaystyle \frac {1}{2}$
0%
$\dfrac{3}{16}$

Explanation

Let the no. of throws be n.
Probability of getting an odd no. twice =

$_{ 2 }^{ n }{ C }*{ (\dfrac { 1 }{ 2 } ) }^{ n }$
Probability of getting an even no. thrice =

$_{ 3 }^{ n }{ C }*{ (\dfrac { 1 }{ 2 } ) }^{ n }$
Since the above are equal, we get:

$_{ 2 }^{ n }{ C }=_{ 3 }^{ n }{ C } => n=5$
Probability of getting an odd number once or thrice or 5 times =

$(_{ 1 }^{ 5 }{ C }+_{ 3 }^{ 5 }{ C }+_{ 5 }^{ 5 }{ C })*{ (\dfrac { 1 }{ 2 } ) }^{ 5 }=\dfrac { 1 }{ 2 }$
Hence, (c) is correct.

A factory

$A$ produces

$10\%$ defective valves and another factory

$B$ produces

$20\%$ defective. A bag contains

$4$ valves of factory

$A$ and

$5$ valves of factory

$B$ . If two valves are drawn at random from the bag. Find the probability that at least one valves is defective. Give your answer upto two places of decimals.

0%
$0.19$
0%
$0.39$
0%
$0.29$
0%
$0.25$

Explanation

Factory

$A'$ s contribution

$=4$ valves (

$10\%$ defective)
Factory

$B'$ s contribution

$= 5$ valves (

$20\%$ defective)
Two valves can be selected in

$^{ 9 }{ { C }_{ 2 } }=36$ ways
Now probability ( at least one valve defective)

$= 1-$ P(no defective valve)
The two selected valves may be
A) both from factory

$A$
b) both from factory

$B$
c) one from each
Therefore required probability

$\displaystyle 1-\frac { 1 }{ ^{ 9 }{ { C }_{ 2 } } } \left\{ ^{ 4 }{ { C }_{ 2 }{ \left( \frac { 9 }{ 10 } \right) }^{ 2 }+^{ 5 }{ { C }_{ 2 }{ \left( \frac { 4 }{ 5 } \right) }^{ 2 } }+^{ 4 }{ { C }_{ 1 } }^{ 5 }{ { C }_{ 1 } }\frac { 9.4 }{ 10.5 } } \right\}$

$\displaystyle =1-\frac { 1 }{ 36 } \left\{ \frac { 486 }{ 100 } +\frac { 160 }{ 25 } +\frac { 720 }{ 50 } \right\} =1-\frac { 2566 }{ 3600 } =\frac { 517 }{ 1800 }$

A fair coin is tossed

$99$ times. If

$X$ is the number of times heads occurs

$P(X=r)$ is maximum when

$r$ , is

0%
$49$
0%
$50$
0%
$51$
0%
None of these

Explanation

Putting

$n=99$ and

$\displaystyle p=\frac { 1 }{ 2 } ,$ we have

$\displaystyle \left( n+1 \right) p=\left( 100 \right) \left( \frac { 1 }{ 2 } \right) =50,$ and

$\left( n+1 \right) p-1=49$
For maximum value of

$P(X=r),(n+1)p-1\le r\le \left( n+1 \right) p$

$\Rightarrow 49\le r\le 50 \therefore r=49$ and

$50$
Hence the maximum value of

$P(X=r)$ occurs at

$r=50$ and

$49$

The sum and product of mean and variance of a Binomial distribution are

$24$ and

$128$ respectively then the value of

$n$ is

0%
$16$
0%
$32$
0%
$24$
0%
None of these

Explanation

Given

$np+npq=24$ and

$np (npq)=128$

$np(1+q)=24$ and

$n^{2}p^{2}q=128$

$\therefore \displaystyle \frac{n^{2}p^{2}(1+q)^{2}}{n^{2}p^{2}q}=\frac{24\times 24}{128}$

$\Rightarrow 2q^{2}-5q+2=0$

$\therefore p=q=1/2$ now

$np(1+q)=24$

$\therefore n\left ( \displaystyle\frac{1}{2} \right )\left (\displaystyle \frac{3}{2} \right )=24$

$\Rightarrow n = \displaystyle\frac{24\times 4}{3}=32$

0%
Both Assertion & Reason are individually true & Reason is correct explanation of Assertion
0%
Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true

Explanation

Let P(getting a head) =

$p = \dfrac 12$

and let P(getting a tail)=

$q = 1-p = 1-\dfrac 12 = \dfrac 12$

As n = 8

The general form of binomial distribution for determining the probability of r successes is

$P(r) = ^nC_r\times q^{(n-r)}\times p^r$

Probability of getting exactly 2 tails=

$P(r = 2)$

$\implies P(2)=^{8}C_2\times \left(\dfrac 12\right)^{(8-2)}\times\left(\dfrac 12\right)^2=\dfrac {8\times7}{2\times1} \left(\dfrac 12\right)^{8}=28\times\left(\dfrac 12\right)^{8} ..............(i)$

Probabibility of getting exactly 6 tails =

$P(r = 6)$

$\implies P(6)=^{8}C_6\times \left(\dfrac 12\right)^{(8-6)}\times\left(\dfrac 12\right)^6=\dfrac {8\times7\times6\times5\times4\times3}{6\times5\times4\times3\times2\times1}\times \left(\dfrac 12\right)^{8}=28\times\left(\dfrac 12\right)^{8}.....(ii)$

From equation (i) and (ii)

We can conclude that if 8 coins are thrown simultaneously, then probability of occurence of exactly 2 tails is equal to the probability of occurence of exactly 6 tails.

A coin is tossed

$\left (2n + 1 \right )$ times, the probability that head appear odd number of times is

0%
$\displaystyle \frac{n}{2n+1}$
0%
$\displaystyle \frac{n+1}{2n+1}$
0%
$\displaystyle \frac{1}{2}$
0%
None of these

Explanation

$P(H)=P(T)=\dfrac{1}{2}$ (with a single coin)
The required probability

$=p \$ (that head occurs

$1, 3, ... or\ (2n + 1)$ times)
We know

$P(X=r) =\displaystyle^{2n+1}\textrm{C}_{r}\left (\frac{1}{2}\right )^{r}\left (\frac{1}{2}\right )^{2n+1-r}$

$\therefore P(1)+P(3)+P(5)+\cdots+P(2n+1)$

$\displaystyle={ }^{2n+1}\textrm{C}_{1}\left(\frac{1}{2}\right )^{1}\left (\frac{1}{2}\right)^{2n}+{ }^{2n+1}\textrm{C}_{3}\left (\frac{1}{2}\right )^{3}\left (\frac{1}{2}\right )^{2n-2}+\cdots+{ }^{2n+1}\textrm{C}_{2n+1}\left (\frac{1}{2}\right )^{2n}$

$=\displaystyle\dfrac{1}{2^{2n+1}}\left [ ^{2n+1} \textrm{C}_{1} + { }^{2n+1} \textrm{C}_{3}+\cdots + { }^{2n+1}\textrm{C}_{2n+1} \right ]$

$=\displaystyle\frac{1}{2^{2n+1}}\times 2^{2n}=\frac{1}{2}$

Hence, option C is correct.

$x$ is a random poission variable such that

$B= P\left ( X= 2 \right )= P\:\left ( X= 3 \right )$ , then

$P\left ( X= 4 \right )$ equals

0%
$2B$
0%
$\displaystyle \frac{3B}{4}$
0%
$\displaystyle \frac{3B^{2}}{4}$
0%
None of these

Explanation

Say m be the mean of the P.D. then

$\displaystyle p\left ( X=r \right )=\frac{m^{r}e^{-m}}{r!},r=0, 1, 2, \cdots$

Now

$\displaystyle \Rightarrow B=\frac{m^{2}e^{-m}}{2!}=\frac{m^{3}e^{-m}}{3!}\Rightarrow m = 3$

now

$B =P(X=2) \displaystyle =B\frac{3^{2}e^{-3}}{2!}=\frac{9}{2e^{3}}$

$\displaystyle \therefore P\left ( X=4 \right )=\frac{3^{4}e^{-3}}{4!}=\frac{27}{8e^3}=\frac{3B}{4}$

In a binomial distribution

$\displaystyle B\left(n,p=\frac{1}{4}\right)$ if the probability of at least one success is greater than or equal to

$\displaystyle \frac{9}{10}$ , then

$n$ is greater than

0%
$\displaystyle \frac{1}{\log_{10}4+\log_{10}3}$
0%
$\displaystyle \frac{9}{\log_{10}4-\log_{10}3}$
0%
$\displaystyle \frac{4}{\log_{10}4-\log_{10}3}$
0%
$\displaystyle \frac{1}{\log_{10}4-\log_{10}3}$

Explanation

Probability of at least one success

$\displaystyle =1-$ No success

$\displaystyle =1- ^{n}C_{n} q^{n}$
where

$\displaystyle q=1-p=3/4$
we want

$\displaystyle =1- \left(\frac{3}{4}\right)^{n}\geq \frac{9}{10}$

$\displaystyle \Rightarrow \frac{1}{10}\geq \left(\frac{3}{4}\right)^{n} \Rightarrow \left(\frac{3}{4}\right)^{n}\leq \frac{1}{10}$
Taking logarithm on base 10 we have

$\displaystyle n\log _{10}(3/4)\leq \log_{10}10^{-1}$

$\displaystyle \Rightarrow n(\log _{10}3 -\log _{10}4 )\leq -1$

$\displaystyle \Rightarrow n(\log _{10}4 -\log _{10}3 ) \geq 1$

$\displaystyle \Rightarrow n \geq \frac{1}{\log _{10}4 -\log _{10}3 }$

$3 percent$ bulb manufactured by a company are defective; the probability that in a sample of

$100$ bulbs exactly five defective is

0%
$\dfrac { { { e }^{ -0.003 } }\left( 0.03 \right) ^{ 5 } }{ 5! }$
0%
$\dfrac { { { e }^{ -0.3 } }0.03^{ 5 } }{ 5!}$
0%
$\dfrac { { { e }^{ -3 } }3^{ 5 } }{5! }$
0%
$\dfrac { { e }^{ -0.3 }{ 3 }^{ -5 } }{5! }$

Explanation

$In\quad poison\quad distribution\quad \frac { { e }^{ -u }{ u }^{ x } }{ x! } \\ Here\quad 3\quad are\quad defective\quad in\quad 100\quad so\quad u=3\\ Probaility\quad that\quad exactly\quad 5\quad are\quad defective\quad (x=5)=\quad \frac { { e }^{ -3 }{ (3) }^{ 5 } }{ 5! }$

$n$ independent trials (finite) of a random experiment, let

$X$ be the number of times an event

$A$ occurs. If the probability of success of one trial say

$p$ and we get the probability of failure of event

$A$ i.e.

$P\left ( \bar{A} \right )=1-p=q$ . The probability of

$r$ success in

$n$ trials is denoted by

$P(X = r)$ such that

$P(X = r) = ^{n}C_{r}p^{r}q^{n-r}$ , known as binomial distribution of random variable

$X$ . We also have the following result-

(i) Probability of getting at least

$k$ successes is

$P\left ( X\geq K \right )=\sum_{r=k}^{n}\ ^{n}C_{r}p^{r}q^{n-r}$

(ii) Probability of getting at the most

$k$ successes is

$P\left ( X\leq K \right )=\sum_{r=0}^{n}\ ^{n}C_{r}p^{r}q^{n-r}$

(iii)

$\sum_{r=0}^{n}=^nC_{r}p^{r}q^{n-r}=\left ( p+q \right )^{n}=1$

On the basis of above information answer the following question.

In a hurdle race, a player has to cross

$10$ hurdles.The probability that he will cross each hurdle is

$\displaystyle \frac{5}{6}$ , then the probability that he will knock down fewer than two hurdle is

0%
$\displaystyle 7\left (\frac{5}{6} \right )^{10}$
0%
$\displaystyle 5\left (\frac{5}{6} \right )^{10}$
0%
$\displaystyle 3\left (\frac{5}{6} \right )^{10}$
0%
$\displaystyle \frac{3}{5}\left (\frac{5}{6} \right )^{10}$

Explanation

Here ,

$\displaystyle n=10,p=$ knocking down the hurdle

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

and

$q=$ Clearing the hurdle

$=\displaystyle \frac{5}{6}$

Using

$\displaystyle P(X=r)= \, ^{10}C_{r}\left(\frac{1}{6}\right)^{r}\left(\frac{5}{6}\right)^{10-r}$

$\therefore$ Required probability

$=\displaystyle P(X=r<2)=P(0)+P(1)=\left(\dfrac{5}{6}\right)^{10}+(10)\dfrac{1}{6}\left(\dfrac{5}{6}\right)^{9}=\left(\dfrac{5}{6}+\dfrac{10}{6} \right)\left(\dfrac{5}{6}\right)^{9}=3\left(\dfrac{5}{6}\right)^{10}$

The probability of a man hitting a target in one trial is

$\displaystyle \frac{1}{4}$ .The chances of hitting the target at least once in

$n$ trials exceeds

$\displaystyle \frac{2}{3}$ , then the value of

$n$ equals

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

$\displaystyle p=\dfrac{1}{2},q=\dfrac{3}{4},n=?$

Now using,

$p(X=r)=^{n}C_{r}p^{r}q^{n-r}$

Given

$\displaystyle P(X\geq 1)>\dfrac{2}{3}$

$\displaystyle \Rightarrow 1-P(0)>\dfrac{2}{3}\Rightarrow 1-\left(\dfrac{3}{4}\right)^{n}>\dfrac{2}{3}\Rightarrow \left(\dfrac{1}{3}\right)>\left(\dfrac{3}{4}\right)^{n}$

For

$\displaystyle n=1,\dfrac{1}{3}\ngtr \dfrac{3}{4}$

For

$\displaystyle n=2,\dfrac{1}{3}\ngtr\dfrac{9}{16}$

For

$n=3,\dfrac{1}{3}\ngtr \dfrac{27}{64}$

For

$n=4,\dfrac{1}{3}> \dfrac{81}{256}$

$\displaystyle \therefore$ Minimum value of

$\displaystyle n=4$

A pair of fair dice is thrown independently three times. The probability of getting a score of 9 exactly twice is

0%
$8/729$
0%
$8/243$
0%
$1/729$
0%
$8/9$

Explanation

$\textbf{Step 1: Given conditions & probability determination.}$

$\text{Given:}$

$\text{A pair of dice are thrown independently thrice}$

$\text{The sample size for the given condition is 36}$

$\text{The possibility of getting 9 in a single throw is (5,4),(4,5),(3,6),(6,3)}$

$\text{Hence the probability of getting 9 in a single throw is 4/36 = 1/9}$

$\textbf{Step 2: Calculation of the required probability.}$

$\text{The probability of getting 9 twice is calculated as follows:}$

${^3C_2 (1/9)^2 (8/9) = 8/243}$

$\text{Therefore the required probability is 8/243.}$

$\textbf{The correct option is B}$

The problem of points. Two player

$A$ and

$B$ want respectively

$m$ and

$n$ points of winning a single game are

$p$ and

$q$ respectively where

$p+q =1$ . The stake is to belong to the player who first makes up his set; find the probabilities in favour of each player.

0%
$\displaystyle P^{m}\left [ 1+mq+\frac{m(m+1)}{1.2} \right ]q^{2}+...\frac{(m+n-2)!}{(m-1!)(n-1)!}q^{n-1}$ , $\displaystyle q^{n}\left [ 1+np+\frac{n(n+1)}{1.2}p^{2}+......+\frac{(m+n-2)}{(m-1)!(n-1)!}p^{m-1} \right ]$
0%
$\displaystyle P^{m}\left [ 1+mq+\frac{m(m+1)}{1.2} \right ]q^{2}+...\frac{(m+n-1)!}{(m-1!)(n-1)!}q^{n-1}$ , $\displaystyle q^{n}\left [ 1+np+\frac{n(n+1)}{1.2}p^{2}+......+\frac{(m+n-2)}{(m-1)!(n-1)!}p^{m-1} \right ]$
0%
$\displaystyle P^{m}\left [ 1+mq+\frac{m(m+1)}{1.2} \right ]q^{2}+...\frac{(m+n-2)!}{(m-1!)(n-1)!}q^{n-1}$ , $\displaystyle q^{n}\left [ 1+np+\frac{n(n+1)}{1.2}p^{2}+......+\frac{(m+n-1)}{(m-1)!(n-1)!}p^{m-1} \right ]$
0%
$\displaystyle P^{m}\left [ 1+mq+\frac{m(m+1)}{1.2} \right ]q^{2}+...\frac{(m+n-1)!}{(m-1!)(n-1)!}q^{n-1}$ , $\displaystyle q^{n}\left [ 1+np+\frac{n(n+1)}{1.2}p^{2}+......+\frac{(m+n-1)}{(m-1)!(n-1)!}p^{m-1} \right ]$

Explanation

Suppose

$A$ wins in exactly

$m+r$ games.
To do so he must win the game and

$m-1$ out of the preceding

$\displaystyle ^{m+r-1}C_{m-1}P^{m-1}P^{m-1}q^{r}p,i.e. ^{m+r-1}C_{m-1}P^{m}q^{r}.$
Now the set will definitely be decided in

$m+n-1$ games; and

$A$ may win these

$m$ games in exactly

$m$ games, or

$m+1$ games,...., or

$m+n-1$ games.
Hence we shall obtain the chance of

$A$ 's winning the set by giving to

$r$ values

$0,1,2,.....,n-1$ in succession in the expression

$\displaystyle ^{m+r-1}C_{m-1}P^{m}q^{r}$ .
Hence

$A$ 's probability to win the set is

$\displaystyle P^{m}\left [ 1+mq+\frac{m(m+1)}{1.2} \right ]q^{2}+...\frac{(m+n-2)!}{(m-1!)(n-1)!}q^{n-1}$ ;
similarly

$B$ 's probability to win the set is

$\displaystyle q^{n}\left [ 1+np+\frac{n(n+1)}{1.2}p^{2}+......+\frac{(m+1-2)}{(m-1)!(n-1)!}p^{m-1} \right ]$

One hundred identical coins, each with probability

$p$ , of showing up heads are tossed. If

$\displaystyle 0< p< 1$ and the probability of heads showing on

$50$ coins is equal to that of the heads showing in

$51$ coins, then the value of

$p$ is

0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{49}{101}$
0%
$\displaystyle \frac{50}{101}$
0%
$\displaystyle \frac{51}{101}$

Explanation

Let

$X$ be the number of coins showing heads.

Then

$X$ follows a binomial distribution with parameters

$n=100$ and

$p$ .
Since,

$P(X=50)=P(X=51)$ ,

We get,

$^{100}C_{50}p^{50}(1-p)^{50}=\ ^{100}C_{51}p^{51}(1-p)^{49}$

$\Rightarrow \displaystyle \frac {100!}{50!50!}\cdot \frac {51!49!}{100!}=\frac {p}{1-p}$

$\Rightarrow \dfrac {51}{50}=\dfrac {p}{1-p}$

$\Rightarrow 51-51p=50p$

$\Rightarrow p=\dfrac{51}{101}$

In a certain experiment the probability of success is twice the probability of faliure. Find the probaility of at least four successes in six trails.

0%
$\displaystyle \frac{297}{729}$
0%
$\displaystyle \frac{233}{729}$
0%
$\displaystyle \frac{432}{729}$
0%
$\displaystyle \frac{496}{729}$

Explanation

Let p is the probability of success and q the probability of failure .
Then

$p+q=1$
Given that

$p=2q$

$\displaystyle \Rightarrow 2q+q=1$

$q=\dfrac{1}{3}$

$\Rightarrow p=\dfrac{2}{3}$

Now the probability of at least four successes in six trials

$=\displaystyle^{6}C_{4}p^{4}q^{2}+^{6}C_{5}p^{5}q+^{6}C_{6}p^{6}$

$\displaystyle = 15 \left ( \frac{2}{3} \right )^{4}\left ( \frac{1}{3} \right )^{2}+6\left ( \frac{2}{3} \right )^{5}\left ( \frac{1}{3} \right )+\left ( \frac{2}{3} \right )^{6}$

$\displaystyle =\frac{1}{3^{6}}[240+192+64]=\frac{496}{729}$

The distribution of a random variable

$X$ whose range is

$\{ 1,2,3,4 \}$ is given in the following table. Find the mean and variance of

$X$ respectively are:

$X$	$1$	$2$	$3$	$4$
$P(X = x)$	$K$	$2K$	$3K$	$4K$

0%
$2,3$
0%
$3,1$
0%
$3,2$
0%
$2,4$

the sample has

$3$ defective and

$5$ non-defective bulbs.

0%
$\displaystyle \frac{4.89.87.86}{7.97.95.47.31}$
0%
$\displaystyle \frac{5103}{12500000}$
0%
$\displaystyle \frac{45927}{12500000}$
0%
$\displaystyle \frac{413343}{25000000}$

Explanation

Total number of bulbs in the lot

$=10 \:defective +90 \:non-defective=100$

Number of bulbs selected is

$8$ .

Therefore total number of ways in selecting

$8$ bulbs from total of

$100$ bulbs is

$100C_8 =\dfrac{100\times 99\times 98\times 97\times 96\times 95\times 94\times 93}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}$

$\therefore n(S)=10\times 99\times 14\times 97\times 95\times 47\times 31$

Let

$A$ be the event of the sample having

$3$ defectives and

$5$ non-defectives.

Thus

$n(A)=$ (number of ways in which 3 defective bulbs can be selected from 10)

$\times$ (number of ways in which 5 non-defective bulbs can be selected from 90)

$\therefore n(A)=10C_3 \times 90C_5$

Now probabilty that the sample has

$3$ defective and

$5$ non-defective bulb is

$P(A)$ .

$P(A)=\dfrac{n(A)}{n(S)}$

$=\dfrac{10C_3 \times 90C_5}{10\times 99\times 14\times 97\times 95\times 47\times 31}$

$=\dfrac{\dfrac{10\times 9\times 8}{3\times 2\times 1} \times \dfrac{90\times 89\times 88\times 87\times 86}{5\times 4\times 3\times 2\times 1}}{10\times 99\times 14\times 97\times 95\times 47\times 31}$

$=\dfrac{4\times 90\times 89\times 22\times 87\times 86}{10\times 99\times 14\times 97\times 95\times 47\times 31}$

$P(A)=\dfrac{4\times 89\times 87\times 86}{7\times 97\times 95\times 47\times 31}$

Hence probabilty that the sample has

$3$ defective and

$5$ non-defective bulb is

$\dfrac{4\cdot 89\cdot 87\cdot 86}{7\cdot 97\cdot 95\cdot 47\cdot 31}$ .

$A$ and

$B$ play a match in which the chances of their winning a game is

$\frac{1}{4}$ for both of them and

$\frac{1}{2}$ is the probability that match is being drawn. Match is finished as soon as either player wins two games. Find the probability that the match will be finished in

$4$ or less games.

0%
$\dfrac{9}{16}$
0%
$\dfrac{7}{16}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$

Explanation

We have to find the probability that match will be finished in

$4$ or less games.
We will find the ways in which match is not finished in

$4$ games.
There are following

$4$ mutually exclusive ways:
(1) All the games are drawn.
(2)

$A$ and

$B$ both win one game each and remaining two are drawn.
(3)

$A$ wins one game and remaining three are drawn.
(4)

$B$ wins one game and remaining three are drawn.
So the probability that match is not finished in 4 games is equal to the sum of probabilities of

$4$ ways.

$=\displaystyle \left ( \frac{1}{2} \right )^{4}+^{4}C_{2}\left ( \frac{1}{4} \right )^{2}\left ( \frac{1}{2} \right )^{2}+^{4}C_{1}\left ( \frac{1}{4} \right )\left ( \frac{1}{2} \right )^{3}+^{4}C_{1}\left ( \frac{1}{4} \right )\left ( \frac{1}{2} \right )^{3}$

$\displaystyle =\frac{1}{16}+\frac{3}{16}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$
Hence the required probability

$=\displaystyle 1-\frac{1}{2}=\frac{1}{2}$

$A$ and

$B$ play a match to be decided as soon as either has won two games. The chance of either winning a game is

$\displaystyle \dfrac{1}{20}$ and of its being drawn

$\displaystyle \dfrac{9}{10}.$ What is the the chance that the match is finished in

$10$ or less games?

0%
$\displaystyle 0.18$ approx.
0%
$\displaystyle 0.16$ approx.
0%
$\displaystyle 0.17$ approx.
0%
$\displaystyle 0.15$ approx.

Explanation

If the match is not finished in

$10$ games, either of the following events. will possibly occur :

(i) All the games are drawn

(ii)

$A$ and

$B$ each win one game and the rest

$8$ games are drawn.

(iii)

$A$ or

$B$ wins one game and the rest

$9$ games are drawn. The corresponding probabilities of the above events are

(i)

$\displaystyle \left ( \frac{9}{10} \right )^{10},$ (ii) 2.

$\displaystyle ^{10}C_{2}\left ( \frac{1}{20} \right )^{2}\left ( \frac{9}{10} \right )^{8},$ (iii) 2.

$\displaystyle ^{10}C_{1}\left ( \frac{1}{20} \right )\left ( \frac{9}{10} \right )^{9},$
Since these events are mutually exclusive, the probability of
the game not being finished in 10 games

$\displaystyle =\left ( \frac{9}{10} \right )^{10}+2^{10}C_{2}\left ( \frac{1}{20} \right )^{2}\left ( \frac{9}{10} \right )^{8}+2^{10}C_{1}\left ( \frac{1}{20} \right )\left ( \frac{9}{10} \right )^{9}$

$\displaystyle =\frac{9^{8}\times 387}{2\times 10^{10}}$
Hence the probability that the match is finished in

$10$ or less number of games

$\displaystyle =1-\frac{9^{8}\times 387}{2x\times 10^{10}}=0.17$ approx.

In five throws with a single die what is the chance of throwing (1) three aces exactly, (2) three aces at least?

0%
$\dfrac{125}{3888}$ , $\dfrac{625}{648}$ , Favourable way for event $E$ when one number is selected are $(2,9)$
0%
$\dfrac{125}{3888}$ , $\dfrac{23}{648}$ , Favourable way for event $E$ when one number is selected are $(2,9)$
0%
$\dfrac{3763}{3888}$ , $\dfrac{23}{648}$ , Favourable way for event $E$ when one number is selected are $(2,9)$
0%
$\dfrac{3763}{3888}$ , $\dfrac{625}{648}$ , Favourable way for event $E$ when one number is selected are $(2,9)$

Explanation

Total number of throws i.e.

$n=5$
Let

$p$ be the probability of throwing an ace and

$q$ be of not throwing an ace in a single throw of die.
Probability of getting an ace (1) is

$p=\dfrac{1}{6}$

$\Rightarrow q=1-\dfrac{1}{6}=\dfrac{5}{6}$
Probability of exactly

$3$ aces is

$P(X=3)=^5C_3 p^3q^2$

$P(X=3)=10(\dfrac{1}{6})^3(\dfrac{5}{6})^2$

$\Rightarrow P(X=3)=\dfrac{125}{3888}$
Probability of at least

$3$ aces is

$P(X=3)+P(X=4)+P(X=5)$

$=^5C_3 p^3q^2+^5C_4 p^4q^1+^5C_5 p^5q^0$

$=10(\dfrac{1}{6})^3(\dfrac{5}{6})^2+5(\dfrac{1}{6})^4(\dfrac{5}{6})+(\dfrac{1}{6})^5$

$=\dfrac{1}{6^{5}}[250+25+1]=\dfrac{276}{7776}=\dfrac{23}{648}$

One hundred identical coins, each with probability

$p$ of showing heads are tossed once. If

$0 < p < 1$ and the probability of heads showing on

$50$ coins is equal to that of heads showing on

$51$ coins, the value of

$p$ is

0%
$1/2$
0%
$51/101$
0%
$49/101$
0%
$3/101$

Explanation

Let

$X$ be the number of coins showing heads. Then

$X$ follows a binomial distribution with parameters

$n=100$ and

$p$ .
Since

$P(X=50)=P(X=51)$ ,

we get,

$^{100}C_{50}p^{50}(1-p)^{50}=^{100}C_{51}l^{51}(1-p^{49}$

$\Rightarrow \displaystyle \frac {100!}{50!50!}\cdot \frac {51!49!}{100!}=\frac {p}{1-p}\Rightarrow \frac {51}{50}=\frac {p}{1-p}$

$\Rightarrow 51-51p=50p\Rightarrow p=51/101$ .

Suppose the probability for

$A$ to win a game against

$B$ is

$0.4, A$ has two options of playing matches against

$B$ , One 'the best of

$3$ games' and the other 'the best of

$5$ games'. Which of the options,

$A$ should choose?(No game ends in a draw).

0%
best of $3$ games
0%
best of $5$ games
0%
both options are equally probable
0%
can't say

Explanation

Case

$I$ when

$A$ plays

$3$ games against

$B$ . In this case, we have

$n=3,p=0.4$ and

$q=0.6$

Let

$X$ denote the number of wins. Then,

$PP\left( X=r \right) =_{ }^{ 3 }{ { C }_{ r } }{ \left( 0.4 \right) }^{ r }{ \left( 0.6 \right) }^{ 3-r };r=0,1,2,3$

${ P }_{ 1 }=$ Probability of winning he best of three games

$=P\left( X\ge 2 \right) =P\left( X=2 \right) +P\left( X=3 \right)$

$={ }^{ 3 }{ { { C }_{ 2 } } }{ \left( 0.4 \right) }^{ 2 }{ \left( 0.6 \right) }^{ 1 }+_{ }^{ 3 }{ { { C }_{ 3 } } }{ \left( 0.4 \right) }^{ 3 }{ \left( 0.6 \right) }^{ 0 }$

$=0.288+0.064=0.352$ Case

$II$ when

$A$ plays

$5$ games against

$B$ . In this case, we have

$n=5,p=0.4$ and

$q=0.6$

Let

$X$ denote the number of wins in

$5$ games. Then,

$PP\left( X=r \right) ={ }^{ 5 }{ { C }_{ r } }{ \left( 0.4 \right) }^{ r }{ \left( 0.6 \right) }^{ 5-r };r=0,1,2,3,...,5$

${ P }_{ 12 }=$ Probability of winning he best of five games

$=P\left( X\ge 3 \right) =P\left( X=3 \right) +P\left( X=4 \right)+P\left( X=5 \right)$

$={ }^{ 5 }{ { { C }_{ 3 } } }{ \left( 0.4 \right) }^{ 3 }{ \left( 0.6 \right) }^{ 2 }+_{ }^{ 5 }{ { { C }_{ 4 } } }{ \left( 0.4 \right) }^{ 4 }(0.6)+_{ }^{ 5 }{ { { C }_{ 5 } } }{ \left( 0.4 \right) }^{ 5 }{ \left( 0.6 \right) }^{ 0 }$

$=0.2304+0.0768+0.1024=0.31744$
Clearly,

${ P }_{ 1 }\ge { P }_{ 2 }$ .

Three fair coins are tarred simultaneously. Let x denote the number of heads of experiment.

0%
Write the sample space of experiment.
0%
Write the rang set of x
0%
contract the probability distribution of x
0%
Hence, find mean variance and standard deviation (s.d) of x

A man takes a step forward with probability 0.4 and backward with probability 0.Find the probability that at the end of eleven steps he is one step away from the starting point.

0%
$\displaystyle 0.368$
0%
$\displaystyle 0.147$
0%
$\displaystyle 0.22$
0%
$\displaystyle 0.073$

Explanation

${\textbf{Step -1: Finding the Probability of success and failure}}$

${\text{Let a step forward be a success and a step backward be a failure}}{\text{.}}$

${\text{Then, the probability of success in one step}}$

$\Rightarrow {\text{P = 0}}{\text{.4 = }}\dfrac{2}{5}$

${\text{The probability of failure in one step = Q = 0}}{\text{.6 = }}\dfrac{3}{5}$

${\text{In 11 steps he will be one step away from the staring point if the numbers of successes }}$

${\text{ and failure differ by 1}}{\text{.}}$

${\text{So, the number of successes = 6}}$

${\text{The number of failures = 5}}$

${\text{or the number of successes = 5,}}$

${\text{The number of failures = 6}}$

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

$\therefore {\text{The required probability = }}{}^{{\text{11}}}{{\text{C}}_{\text{6}}}{{\text{P}}^{\text{6}}}{{\text{Q}}^{\text{5}}} + {}^{{\text{11}}}{{\text{C}}_5}{{\text{P}}^5}{{\text{Q}}^6}$

$= {}^{{\text{11}}}{{\text{C}}_{\text{6}}}{\left( {\dfrac{2}{5}} \right)^6}.{\left( {\dfrac{3}{5}} \right)^5} + {}^{{\text{11}}}{{\text{C}}_5}{\left( {\dfrac{2}{5}} \right)^5}.{\left( {\dfrac{3}{5}} \right)^6}$

$= \dfrac{{11!}}{{6!.5!}}.{\left( {\dfrac{2}{5}} \right)^5}.{\left( {\dfrac{3}{5}} \right)^5}.\left\{ {\dfrac{2}{5} + \dfrac{3}{5}} \right\}$

$= \dfrac{{11.10.9.8.7}}{{120}}.\dfrac{{{2^5}{{.3}^5}}}{{{5^{10}}}}$

$= 462 \times {\left( {\dfrac{6}{{25}}} \right)^5}$

$= 0.368$

${\textbf{Hence, correct answer is option A}}$

The probability that an event

$A$ occurs in a single trial of an experiment is

$0.6$ . Three independent trials of the experiment are performed. The probability that the event

$A$ occurs at least twice is

0%
$0.636$
0%
$0.632$
0%
$0.648$
0%
$0.946$

Explanation

Given , number of independent trials

$n=3$
Let X be number of times event A occurs.
Probability that event A occurs in single trial

$p=0.6$

$\Rightarrow q=1-0.6=0.4$

Required probability

$=P(X\ge 2)$

$=P(X=2)+P(X=3)$

$=^3C_2(0.6)^2(0.4)+^3C_3(0.6)^3$

$=0.432+0.216=0.648$

Each of two persons

$A$ and

$B$ toss three fair coins. The probability that both get the same number of heads is

0%
$\dfrac {3}{8}$
0%
$\dfrac {1}{9}$
0%
$\dfrac {5}{16}$
0%
$\dfrac {7}{16}$

Explanation

Let X be the number of heads obtained by A and Y be the number of heads
obtained by B. Note that both X and Y are binomial variate with
parameters

$n=3$ and

$p=1/2$ .
Probability that both A and B obtain the same number of heads is

$=P(X=0) P(Y=0)+P(X=1) P(Y=1)+P(X=2)P(Y=2)+P(X=3)P(Y=3)$

$=\left[^3C_0\left (\frac {1}{2}\right )^3\right ]^2+\left [^3C_1\left (\frac {1}{2}\right )^3\right ]^2+\left [^3C_2\left (\frac {1}{2}\right )^3\right ]^2+\left [^3C_3\left (\frac {1}{2}\right )^3\right ]^2$

$\displaystyle =\left (\frac {1}{2}\right )^6[1+9+9+1]=\frac {20}{64}=\frac {5}{16}$ .

A fair coin is tossed

$100$ times. The probability of getting tails

$1, 3, ..., 49$ times is

0%
$1/2$
0%
$1/4$
0%
$1/8$
0%
$1/16$

Explanation

Let

$p=$ probability of getting a tail in a single trial

$=1/2$

$n=$ number of trails

$=100$
and

$X=$ number of tails in 100 trials.
We have

$P(X=r)=^{100}C_rp^rq^{n-r}$

$=^{100}C_r\left (\frac {1}{2}\right )^r\left (\frac {1}{2}\right )^{100-r}=^{100}C_r\left (\frac {1}{2}\right )^{100}$
Now,

$P(X=1)+P(X=3)+....+P(X=49)$

$=^{100}C_1\left (\frac {1}{2}\right )^{100}+{\;}^{100}C_3\left (\frac {1}{2}\right )^{100}+....+^{100}C_{49}\left (\frac {1}{2}\right )^{100}$

$=(^{100}C_1+^{100}C_3+...+^{100}C_{49})\left (\frac {1}{2}\right )^{100}$
But

$^{100}C_1+^{100}C_3+...+^{100}C_{99}=2^{99}$
Also,

$^{100}C_{99}=^{100}C_1$ ,

$^{100}C_{97}=^{100}C_3, ...^{100}C_{51}=^{100}C_{49}$
Thus,

$2(^{100}C_1+^{100}C_3+...+^{100}C_{49})=2^{99}$

$\Rightarrow ^{100}C_1+^{100}C_3+...+^{100}C_{49}=2^{98}$

$\therefore$ Probability of required event

$=\cfrac {2^{98}}{2^{100}}=\cfrac {1}{4}$ .

In a hurdle race, a runner has probability

$p$ of jumping over a specific hurdle. Given that in

$5$ trials, the rummer succeeded

$3$ times, the conditional probability that the runner had succeeded in the first trial is

0%
$\dfrac {3}{5}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {1}{5}$
0%
$\dfrac {4}{5}$

Explanation

Let

$A$ denote the event that the runner succeeds exactly

$3$ times out of five and

$B$ denote the event that the number succeeds on the first trial.

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

$P(B\cap A)=P$ (succeeding in the first trial and exactly once in two other trials)

$=p(^4C_2p^2(1-p)^2)=6p^3(1-p)^2$
and

$P(A)=^5C_3p^3(1-p)^2=10p^3(1-p)^2$
Thus,

$\displaystyle P(B|A)=\frac {6p^3(1-p)^2}{10p^3(1-p)^2}=\frac {3}{5}$ .

The minimum number of times a fair coin must be tossed so that the probability of getting at least one head is at least

$0.8$ is

0%
$7$
0%
$6$
0%
$5$
0%
$3$

Explanation

$\textbf{Step 1: Calculating Probability of getting atleast 1 Head}$

$\text{The probability of getting atleast one head should be atleast 0.8 }$

$\implies P(H)\geq 0.8$

$\implies \text{Choose the required terms from the binomial expansion}$

$\implies \dfrac{^nC_0}{2^n}+\dfrac{^nC_1}{2^n}+..+\dfrac{^nC_{n-1}}{2^n}\geq 0.8$

$\implies \dfrac{2^n-^nC_0}{2^n}\geq 0.8$

$\textbf{Step 2: Solving for n}$

$\implies ^nC_0\leq 0.2\times2^n$

$\implies 1\leq0.2(2^n)$

$\implies \text{The minimum value would be 3}$

$\textbf{Hence, number of times the coin must be tossed is 3}$

The probability that

$X$ wins the match after

$(n+1)$ games,

$(n\geq 1)$ is

0%
$na^2b^{n-1}$
0%
$a^2(nb^{n-1}+n(n-1)b^{n-2}c)$
0%
$na^2bc^{n-1}$
0%
none of these

Explanation

$X$ can win after the

$(n+1)^{th}$ game in the following two mutually exclusive ways
(i)

$X$ wins exactly one of the first

$n$ games draws

$(n-1)$ games and wins the

$(n+1)^{th}$ the games
(ii)

$X$ loses exactly one of the first

$n$ games, wins exactly one of the first

$n$ games and draws

$(n-2)$ games and wins the

$(n+1)^{th}$ game.
For (i) the probability is

$(^nP_1 ab^{n-1})a$ and for (ii) the probability is

$(^nP_2)(ac)b^{n-2}a$

$\therefore$ The probability

$X$ wins after the

$(n+1)^{th}$ game

$P{(X)}=na^2b^{n-1}+n(n-1)a^2b^{n-2}c$

$=a^2(nb^{n-1}+n(n-1)b^{n-2}c)$ .

A radar unit is used to measure speeds of cars on a motorway. The speeds are normally distributed with a mean of

$9$ km/hr and a standard deviation of

$10$ km/hr. What is the probability that a car picked at random is travelling at more than

$100$ km/hr?

0%
$0.1698$
0%
$0.1548$
0%
$0.1587$
0%
$0.1236$

Explanation

Let

$x$ be the random variable that represents the speed of cars.

$x$ has

$\mu=90,\sigma=10$ .

We have to find the probability that

$x$ is higher than

$100$ or

$P(x > 100)$

Given

$x, z=\dfrac{x-\mu}{\sigma}$ .

Thus for

$x=100, z=\dfrac{100-90}{10}=1$

$\Rightarrow P(x>100)=P(z=1)$

$=$ [total area]

$-$ [area to the left of

$z=1$ ]

$=1-0.8413$ (from normal distribution table)

$\therefore P(x>100)=0.1587$

Hence the probability that a car selected at a random has a speed greater than

$100$ km/hr is equal to

$0.1587$ .

$X$ follows a binomial distribution with parameter

$n=101$ and

$p=1/3$ , then

$P(X=r)$ is maximum if

$r$ equals

0%
$34$
0%
$33$
0%
$32$
0%
$35$

Explanation

Given

$n=101, p=\dfrac{1}{3}$

$\Rightarrow q=1-\dfrac{1}{3}=\dfrac{2}{3}$
We will check using options

$P(X=34)=^{101}C_{34}(\dfrac{1}{3})^{34} (\dfrac{2}{3})^{67}$

$\Rightarrow P(X=34)=\dfrac{101}{(67)!(34)!}\times (\dfrac{1}{3})^{101} 2^{67}$ .....(1)

$P(X=33)=^{101}C_{33}(\dfrac{1}{3})^{33} (\dfrac{2}{3})^{68}$

$\Rightarrow P(X=33)=\dfrac{101}{(68)!(33)!}\times (\dfrac{1}{3})^{101} 2^{68}$ .....(2)

$P(X=32)=^{101}C_{32}(\dfrac{1}{3})^{32} (\dfrac{2}{3})^{69}$

$\Rightarrow P(X=32)=\dfrac{101}{(69)!(32)!}\times (\dfrac{1}{3})^{101} 2^{69}$ .....(3)

$P(X=35)=^{101}C_{35}(\dfrac{1}{3})^{35} (\dfrac{2}{3})^{66}$

$\Rightarrow P(X=35)=\dfrac{101}{(66)!(35)!}\times (\dfrac{1}{3})^{101} 2^{66}$ .....(4)
Now, dividing (1) by (2), we get

$\dfrac{P(X=34)}{P(X=33)}=\dfrac{34}{68} \times 2 =1$

$\Rightarrow P(X=34)=P(X=33)$
Now, we will divide (1) by (3)

$\dfrac{P(X=34)}{P(X=32)}=\dfrac{69}{64}>1$

$\Rightarrow P(X=34)>P(X=32)$
Now, we will divide (1) by (4), we get

$\dfrac{P(X=34)}{P(X=35)}=\dfrac{70}{67}>1$

$\Rightarrow P(X=34)>P(X=35)$
Hence,

$P(X=34), P(X=33)$ is maximum.

$X$ and

$Y$ are the independent random variable for

$\displaystyle A\left( 5,\frac { 1 }{ 2 } \right)$ and

$\displaystyle B\left( 7,\frac { 1 }{ 2 } \right)$ , then

$P\left( X+Y\ge 1 \right) =$

0%
$\displaystyle \frac { 4095 }{ 4096 }$
0%
$\displaystyle \frac { 309 }{ 4096 }$
0%
$\displaystyle \frac { 4031 }{ 4096 }$
0%
none of these

Explanation

For random variable

$\displaystyle X,n=5,p=\frac { 1 }{ 2 }$ and for random variable

$\displaystyle Y,n=7,p=\frac { 1 }{ 2 }$

Now,

$\displaystyle P\left( X+Y\ge 1 \right) =1-P\left( X+Y<1 \right) =1-P\left( X+Y=0 \right)$

$\displaystyle =1-P\left( X=0,Y=0 \right)$

$\displaystyle =1-P\left( X=0 \right) P\left( Y=0 \right)$

$[\because X$ and

$Y$ are independent

$]$

$\displaystyle =1-^{ 5 }{ { C }_{ 0 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }.^{ 7 }{ { C }_{ 0 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 7 }$

$\displaystyle =1-{ \left( \frac { 1 }{ 2 } \right) }^{ 12 }=\frac { 4095 }{ 4096 } .$

Let, a binomial distribution

$\displaystyle B\left ( n,p=\frac{1}{4} \right )$ . If the probability of at least one success is greater than or equal to

$\displaystyle \frac{9}{10}$ , then

$n$ is greater than

0%
$\displaystyle \frac{1}{\log _{10}4+\log _{10}3}$
0%
$\displaystyle \frac{9}{\log _{10}4-\log _{10}3}$
0%
$\displaystyle \frac{4}{\log _{10}4-\log _{10}3}$
0%
$\displaystyle \frac{1}{\log _{10}4-\log _{10}3}$

Explanation

Probability of at least one success

$= 1- No \ success = 1- ^nC_nq^n$

where

$q=1-p=\dfrac 34$

we want

$1-\left(\dfrac 34 \right)^n\ge \dfrac 9{10}$

$\implies \dfrac 1{10}\ge \left(\dfrac 34 \right)^n\\\implies \left(\dfrac 34\right)^n\le \dfrac 1{10}$

Taking logarithm on base 10 we have

$n\log _{10}\left(\dfrac 34\right)\le \log _{10}\left(10\right)^{-1}\\\implies n(\log_{10} 3 -\log_{10} 4)\le -1\\\implies n(\log _{10}4-\log_{10}3)\ge 1\\\implies n\ge \dfrac 1{\log_{10}4-\log_{10}3}$

A fair coin is tossed

$100$ times. The probability of getting tails an odd number of times is

0%
$\dfrac {1}{2}$
0%
$\dfrac {1}{4}$
0%
$\dfrac {1}{8}$
0%
$\dfrac {3}{8}$

Explanation

Let

$X$ be the event of getting tail.

Required probability

$=P\left( X=1 \right) +P\left( X=3 \right) +\cdots +P\left( X=99 \right)$

$=_{ }^{ 100 }{ { C }_{ 1 } }\left( \dfrac { 1 }{ 3 } \right) { \left( \dfrac { 1 }{ 2 } \right) }^{ 99 }+^{ 100 }{ { C }_{ 3 } }{ \left( \dfrac { 1 }{ 2 } \right) }^{ 100 }+\cdots +^{ 100 }{ { C }_{ 99 } }{ \left( \dfrac { 1 }{ 2 } \right) }^{ 100 }$

$=\dfrac { _{ }^{ 100 }{ { C }_{ 1 } }+^{ 100 }{ { C }_{ 3 } }+\cdots +^{ 100 }{ { C }_{ 99 } } }{ { 2 }^{ 100 } }$

$=\dfrac { \dfrac { { 2 }^{ 100 } }{ 2 } }{ { 2 }^{ 100 } }$

$=\dfrac { { 2 }^{ 99 } }{ { 2 }^{ 100 } }$

$=\dfrac { 1 }{ 2 }$

An experiment succeeds twice as often as it fails. Find the probability that in next 6 trials, there will be more than 3 successes

0%
$\displaystyle \frac {496}{729}$
0%
$\displaystyle \frac {31}{729}$
0%
$\displaystyle \frac {16}{27}$
0%
$\displaystyle \frac {31}{54}$

Explanation

Given, An experiment succeeds twice often as it fails.

$\Rightarrow$ Probability of success,

$P(S)=\dfrac{2}{3}$ ,

Probability of Failure,

$P(F)=\dfrac{1}{3}$
An experiment is conducted

$6$ trials,

This is similar to binomial probability disribution

$(q+p)^n$ (where

$q=$ probability of success,

$p=$ proability of failure)

$\Rightarrow p=\dfrac{2}{3},q=\dfrac{1}{3},n=6$

$\therefore$ The probability to get more than 3 success=

$P(X=4)+P(X=5)+P(X=6)$

$={6}C_{4}q^2p^4+{6}C_{5}qp^5+{6}C_{6}p^6$

$={6}C_{4}(\dfrac{1}{3})^2(\dfrac{2}{3})^4+{6}C_{5}(\dfrac{1}{3})(\dfrac{2}{3})^5+{6}C_{6}(\dfrac{2}{3})^6$

$=\displaystyle\frac{(240+192+64)}{729}=\displaystyle\frac{496}{729}$

If the mean and variance of a binomial variate

$X$ are

$8$ and

$4$ respectively, then

$P(X < 3) =$

0%
$\dfrac{265}{2^{15}}$
0%
$\dfrac{137}{2^{16}}$
0%
$\dfrac{267}{2^{16}}$
0%
$\dfrac{265}{2^{16}}$

Explanation

Let '

$p$ ' denote the probability of success and '

$n$ ' denote the number of trials.

$E(x)=np$

$=8$ .
And

$Var(x)=np(1-p)$

$=4$
Therefore

$np(1-p)=4$

$8(1-p)=4$

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

$p=\dfrac{1}{2}$ ...(i)
Hence

$n(\dfrac{1}{2})=8$

$n=16$
Therefore

$P(X<3)$

$=P(0)+P(1)+P(2)$

$=\dfrac{1}{2^{16}}(\:^{16}C_{0}+\:^{16}C_{1}+\:^{16}C_{2})$

$=\dfrac{1+16+120}{2^{16}}$

$=\dfrac{137}{2^{16}}$

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 10 - MCQExams.com

0:0:3

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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