MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 6

A distributor of bean seeds determines from extensive tests that

$5$ % of seeds will not germinate. He sells the seeds in packets of

$200$ and guarantees

$90$ % germination. The probability that a particular packet will violate the guarantee is

0%
$\displaystyle \dfrac{e^{-10}(10)^{10}}{ 10!}$
0%
$\displaystyle \frac{e^{-10}(10)^{20}}{10!}$
0%
$1-\displaystyle \sum_{x=0}^{20}\frac{e^{-10}(10)^{x}}{ x!}$
0%
$1-\displaystyle \sum_{x=0}^{\infty }\frac{e^{-10}(10)^{x}}{ x!}$

Explanation

The probabilityof a seed not germinating

$\displaystyle=p=\frac { 5 }{ 100 } =0.05$

$\lambda =$ mean number of seeds in a sample of 200. which do not germinate

$=n p=200\times 0.05=10$
Let X = R.V. = number of seeds that do not germinate
A packet will violate guarantee if it contains more than 20 non-germinating seeds
Probability that the guarantee is violated

$\displaystyle =P\left( X>20 \right) =1-P\left( X\le 20 \right) =1-\sum _{ x=0 }^{ 20 }{ \frac { { e }^{ -10 }{ 10 }^{ x } }{ x! } }$

On the average a submarine on patrol sights

$6$ enemy ships per hour. Assuming the number of ships sighted in a given length of time is a poisson variate, the probability of sighting

$6$ ships in the next half an hour is

0%
$\displaystyle \frac{e^{-6}6^{6}}{ 6!}$
0%
$\displaystyle \frac{e^{-3}3^{3}}{ 3!}$
0%
$\displaystyle \frac{e^{-3}3^{6}}{ 6!}$
0%
$\displaystyle \frac{e^{-6}3^{3}}{ 6!}$

Explanation

Let

$\lambda$ be the Poisson Variate.
The

$P(k)=\dfrac{e^{-\lambda}.\lambda^{k}}{k!}$
Now here

$\lambda=6.t$ where

$t=$ time (hours).
Hence for half an hour

$\lambda=\dfrac{6}{2}$

$=3$ .

$P(k)=\dfrac{3^{k}.e^{-3}}{k!}$
Hence the probability of sighting

$6$ ships in half an hour, will be

$P(6)$

$=\dfrac{3^{6}.e^{-3}}{6!}$ .

A coin is tossed

$n$ times. The probability that head will turn up an odd number of times, is

0%
$\displaystyle \frac { 1 }{ 2 }$
0%
$\displaystyle \frac { n+1 }{ 2n }$
0%
$\displaystyle \frac { n-1 }{ 2n }$
0%
$\displaystyle \frac { { 2 }^{ n-1 }-1 }{ { 2 }^{ n } }$

Explanation

Let

$X$ denote the number of heads in

$n$ trails. Then,

$\displaystyle P\left( X=r \right) =_{ }^{ n }{ { C }_{ r } }{ \left( \frac { 1 }{ 2 } \right) }^{ r }{ \left( \frac { 1 }{ 2 } \right) }^{ n-r }=_{ }^{ n }{ { C }_{ r } }{ \left( \frac { 1 }{ 2 } \right) }^{ n }$
Thus, the required probability

$=P\left( X=1 \right) +P\left( x=3 \right) +P\left( X=5 \right) +...$

$\displaystyle =_{ }^{ n }{ { C }_{ 1 } }{ \left( \frac { 1 }{ 2 } \right) }^{ n }+^{ n }{ { C }_{ 3 } }{ \left( \frac { 1 }{ 2 } \right) }^{ n }+^{ n }{ { C }_{ 5 } }{ \left( \frac { 1 }{ 2 } \right) }^{ n }+...$

$\displaystyle ={ \left( \frac { 1 }{ 2 } \right) }^{ n }\left( ^{ n }{ { C }_{ 1 } }+^{ n }{ { C }_{ 3 } }+^{ n }{ { C }_{ 5 } }+... \right) =\frac { 1 }{ { 2 }^{ n } } \left( { 2 }^{ n-1 } \right) =\frac { 1 }{ 2 }$

A company knows on the basis of past experience that

$2$ % of its blades are defective. The probability of having

$3$ defective blades in a sample of

$100$ blades if

$e^{-2}=0.1353$ is

0%
$0.1353$
0%
$0.1804$
0%
$0.2706$
0%
$0.3606$

Explanation

Here P.D parameter

$\lambda = \cfrac{2}{100}\times 100=2$
Hence number of probability that 3 blade are defective is

$=P(X=3)=\cfrac{e^{-2}(2)^3}{3!}=\cfrac{4}{3}e^{-2}=\cfrac{4}{3}\times .1353=.1804$

Patients arrive randomly and independently at a Doctor's room from 8 AM at an average rate of one in 5 minutes. The waiting room can accommodate 12 persons. The probability that the room will be full when the doctor arrives at 9AM is

0%
$\displaystyle \frac{{e}^{-12}(12)^{12}}{ 12!}$
0%
$\displaystyle \sum_{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$
0%
$1-\displaystyle \sum_{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$
0%
$1-\displaystyle \sum_{{x}=0}^{\infty }\frac{{e}^{-12}(12)^{{x}}}{ x!}$

Explanation

Random Arrival process is a Poisson process
Given by

$P(k) = \dfrac{e^{-\lambda} \lambda^x}{x!}$

$given, \lambda = 1/5 min^{-1}= 12 s^{-1}$
Room is not full, if No.of patients

$< 12$
Hence

$P$ (room is full)

$= 1 - (P(1)+ . . . . +P(11))$

$=1-\displaystyle \sum_{{x}=0}^{11}\frac{{e}^{-12}(12)^{{x}}}{ x!}$

A box contains

$100$ bulbs out of which

$10$ are defective. A sample of

$5$ bulbs is drawn. The probability that none is defective, is

0%
$\displaystyle { \left( \frac { 1 }{ 10 } \right) }^{ 5 }$
0%
$\displaystyle { \left( \frac { 1 }{ 2 } \right) }^{ 5 }$
0%
$\displaystyle { \left( \frac { 9 }{ 10 } \right) }^{ 5 }$
0%
$\displaystyle \frac { 9 }{ 10 }$

Explanation

Let

$X$ be the number of defective bulbs in a sample of

$5$ bulbs.
Probability that a bulb is defective

$\displaystyle =p=\frac { 10 }{ 100 } =\frac { 1 }{ 10 }$
Then,

$\displaystyle P\left( X=r \right) =_{ }^{ 5 }{ { C }_{ r } }{ \left( \frac { 1 }{ 10 } \right) }^{ r }{ \left( \frac { 9 }{ 10 } \right) }^{ 5-r }$
The required probability

$\displaystyle P\left( X=0 \right) =_{ }^{ 5 }{ { C }_{ 0 } }{ \left( \frac { 1 }{ 10 } \right) }^{ 0 }{ \left( \frac { 9 }{ 10 } \right) }^{ 5 }={ \left( \frac { 9 }{ 10 } \right) }^{ 5 }$

What is the probability of guessing correctly at least 8 out of 10 answers on a true -false examination?

0%
$\displaystyle\frac{11}{128}$
0%
$\displaystyle\frac{5}{128}$
0%
$\displaystyle\frac{13}{128}$
0%
$\displaystyle\frac{7}{128}$

Explanation

$p(X \geq 8) = _{8}^{10}\textrm{C} (\displaystyle\frac{1}{2})^8(\displaystyle\frac{1}{2})^2+ _{9}^{10}\textrm{C} ( \displaystyle\frac{1}{2})^9 (\displaystyle\frac{1}{2})^1 + _{10}^{10}\textrm{C} {(\displaystyle\frac{1}{2})}^{(10)}$

$= {(\displaystyle\frac{1}{2})}^{(10)}[_{2}^{10}\textrm{C} +_{1}^{10}\textrm{C} +_{0}^{10}\textrm{C} ]$

$= {(\displaystyle\frac{1}{2})}^{10}[45+10+1]$ =

$\displaystyle\frac{56}{{8}\times{2^7}}$

$=\displaystyle\frac{7}{128}$

The constant term of the expansion of

$\left (3x-\dfrac{5}{x^{2}}\right)^{9}$ is.

0%
$^{9}\textrm{C}_{2}3^{6}5^{3}$
0%
$^{9}\textrm{C}_{4}3^{4}5^{5}$
0%
$^{9}\textrm{C}_{3}3^{6}5^{3}$
0%
$^{9}\textrm{C}_{3}$

A box contains

$24$ identical balls of which

$12$ are white and

$12$ are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is

0%
$\displaystyle \frac { 5 }{ 64 }$
0%
$\displaystyle \frac { 27 }{ 32 }$
0%
$\displaystyle \frac { 5 }{ 32 }$
0%
$\displaystyle \frac { 1 }{ 2 }$

Explanation

Probability of drawing a white ball 4th time at 7th draw

$=($ Probability of drawing

$3$ white balls in first

$6$ draws

$)($ Probability of drawing a white ball at the 7th draw

$)$

$\displaystyle =\left[ _{ }^{ 6 }{ { C }_{ 3 } }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 } \right] { \left( \frac { 1 }{ 2 } \right) }=\frac { 20 }{ { 2 }^{ 7 } } =\frac { 5 }{ 32 }$

$X$ follows a binomial distribution with parameters

$n=8$ and

$\displaystyle p=\frac { 1 }{ 2 }$ , then

$P\left( \left| X-4 \right| \le 2 \right) =$

0%
$\displaystyle \frac { 119 }{ 128 }$
0%
$\displaystyle \frac { 116 }{ 128 }$
0%
$\displaystyle \frac { 29 }{ 128 }$
0%
None of these

Explanation

We have

$P\left( \left| X-4 \right| \le 2 \right) =P\left( -2\le X-4\le 2 \right) =P\left( 2\le X\le 6 \right)$

$=1-\left[ P\left( X=0 \right) +P\left( X=1 \right) +P\left( X=7 \right) +P\left( X=8 \right) \right]$

$\displaystyle =1-\left[ _{ }^{ 8 }{ { C }_{ 0 }{ \left( \frac { 1 }{ 2 } \right) }^{ 8 } }+^{ 8 }{ { C }_{ 1 }{ \left( \frac { 1 }{ 2 } \right) }^{ 8 }+^{ 8 }{ { C }_{ 7 }{ \left( \frac { 1 }{ 2 } \right) }^{ 8 } } }+^{ 8 }{ { C }_{ 8 }{ \left( \frac { 1 }{ 2 } \right) }^{ 8 } } \right]$

$\displaystyle =1-{ \left( \frac { 1 }{ 2 } \right) }^{ 8 }\left( 1+8+8+1 \right) =1-\frac { 18 }{ { 2 }^{ 8 } } =\frac { 119 }{ 128 }$

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 only if the product of the two digits of selected number is

$18$ . If four numbers are selected, the probability that the event

$E$ occurs at least

$3$ times, is

0%
$\displaystyle \frac { 99 }{ { \left( 25 \right) }^{ 4 } }$
0%
$\displaystyle \frac { 86 }{ { \left( 25 \right) }^{ 4 } }$
0%
$\displaystyle \frac { 74 }{ { \left( 25 \right) }^{ 4 } }$
0%
$\displaystyle \frac { 97 }{ { \left( 25 \right) }^{ 4 } }$

Explanation

Out of the numbers

$00,01,02,...,99$ , those numbers the product of whose digits is

$18$ are

$29,36,63,92$ .

$\displaystyle \therefore P\left( E \right) =\frac { 4 }{ 100 } =\frac { 1 }{ 25 } \Rightarrow P\left( \overline { E } \right) =\frac { 24 }{ 25 }$
The probability that in selecting form numbers the events

$E$ occurs at least three times

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

Suppose

$X$ is a binomial variate

$B(5,p)$ and

$P(X=2)=P(X=3)$ , then

$p$ is equal to

0%
$1/2$
0%
$1/3$
0%
$1/4$
0%
$1/5$

Explanation

Here

$N = 5$
Given

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

$\Rightarrow ^5C_2 p^2(1-p)^3=^5C_3 p^3 (1-p)^2$

$\Rightarrow 1-p = p \therefore p =\cfrac{1}{2}$

A bag contains three white, two black and four red balls. If four balls are drawn at random with replacement, the probability that the sample contains just one white ball is

0%
$\displaystyle \frac { 32 }{ 81 }$
0%
$\displaystyle \frac { 11 }{ 2 }$
0%
$\displaystyle \frac { 32 }{ 93 }$
0%
None of these

Explanation

The probability

$p$ of getting w white ball in a single draw is

$\displaystyle \frac { 3 }{ 9 } =\frac { 1 }{ 3 }$ ,
If

$X$ is the number of white balls drawn, then

$X\sim B\left( n,p \right)$ , with

$n=4$ .
Therefore the probability of getting exactly one white ball is

$\displaystyle P\left( X=1 \right) =^{ 4 }{ { C }_{ 1 }^{ } }p{ q }^{ 3 }=4\left( \frac { 1 }{ 3 } \right) { \left( \frac { 2 }{ 3 } \right) }^{ 3 }=\frac { 32 }{ 81 }$

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

$0,1,2$ are

$0.45,0.05,0.50$ 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

Since there are just four matches to be played India can get a maximum of

$8$ points.

$\therefore P($ India gets at least

$7$ points

$) =P($ getting exactly

$7$ points

$)+P($ getting exactly

$8$ points

$)$

$=P($ getting

$2$ in each of the

$3$ matches and

$1$ in one match

$)+P($ getting

$2$ in each of the four matches

$)$

$=_{ }^{ 4 }{ { C }_{ 3 } }{ \left( 0.5 \right) }^{ 3 }\left( 0.05 \right) +_{ }^{ 4 }{ { C }_{ 4 } }{ \left( 0.5 \right) }^{ 4 }=0.025+0.0625=0.0875$

One hundred identical coins, each with probability

$p$ of showing up heads, are tossed. 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%
$\displaystyle \frac { 1 }{ 2 }$
0%
$\displaystyle \frac { 49 }{ 101 }$
0%
$\displaystyle \frac { 50 }{ 101 }$
0%
$\displaystyle \frac { 51 }{ 101 }$

Explanation

Let

$X\sim B\left( 100,p \right)$ be the number of coins showing heads and let

$q=1-p$ .
Then, since

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

$_{ }^{ 100 }{ { C }_{ 51 } }\left( { p }^{ 51 } \right) \left( { q }^{ 49 } \right) =_{ }^{ 100 }{ { C }_{ 50 } }\left( { p }^{ 50 } \right) \left( { q }^{ 50 } \right)$

$\displaystyle \Rightarrow \frac { p }{ q } =\left( \frac { 100! }{ 50!50! } \right) \left( \frac { 51!49! }{ 100! } \right)$

$\displaystyle \Rightarrow \frac { p }{ 1-p } =\frac { 51 }{ 50 } \Rightarrow 50p=51-51p\Rightarrow p=\frac { 51 }{ 101 }$

Six person try to swim across a wide river. It is known that on an average, only three persons out of the ten are successful in crossing the river. what is the probability that at most four of the six persons will cross river safely?

0%
$0.990523$
0%
$0.890523$
0%
$0.980523$
0%
None of these

Explanation

Let success be defined as crossing the river.

$\Rightarrow$ probability of success

$\displaystyle \left( p \right) =\frac { 3 }{ 10 }$ and that of failure

$\displaystyle \left( q \right) =\frac { 7 }{ 10 }$
Since number of trials

$(n)=6$

$\therefore P$ (at most

$4$ success in

$6$ trials)

$=1-^{ 6 }{ { C }_{ 5 }{ P }^{ 2 }q- }^{ 6 }{ { C }_{ 6 } }{ P }^{ 6 }$

$\displaystyle =1-6{ \left( \frac { 3 }{ 10 } \right) }^{ 5 }\left( \frac { 7 }{ 10 } \right) -{ \left( \frac { 3 }{ 10 } \right) }^{ 6 }=1-{ \left( \frac { 3 }{ 10 } \right) }^{ 6 }\left( 13 \right) =0.9905$

A die is thrown

$7$ times. What is the chance that an odd number turns up ?
(i) exactly

$4$ times
(ii) at least

$4$ times.

0%
(i) $\displaystyle \frac{35}{128}$ , (ii) $\displaystyle \frac{1}{2}$
0%
(i) $\displaystyle \frac{45}{128}$ , (ii) $\displaystyle \frac{1}{2}$
0%
(i) $\displaystyle \frac{35}{128}$ , (ii) $\displaystyle \frac{1}{4}$
0%
None of these

Explanation

Here number of trials,

$n = 7, p =$ probability of getting odd numbers

$=\cfrac{3}{6} = \cfrac{1}{2}$
(i)

$P(X=4) = ^7C_4 \left(\cfrac{1}{2}\right)^7 =\cfrac{35}{128}$
(ii)

$P(X\geq 4) = 1-P(X=0)-P(X=1)-P(X=2)-P(X=3)$

$= \quad 1-\left[^7C_0+^7C_1+^7C_2+^7C_3\right]\left(\cfrac{1}{2}\right)^7=\cfrac{1}{2}$

A die is thrown

$7$ times. What is the chance that an odd number turns up at least

$4$ times?

0%
$\cfrac{1}{3}$
0%
$\cfrac{1}{4}$
0%
$\cfrac{1}{2}$
0%
$\cfrac{35}{128}$

Explanation

For at least

$4$ successive odd number,
Probability

$\displaystyle =^{ 7 }{ C }_{ 4 }{ \left( \frac { 1 }{ 2 } \right) }^{ 4 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }+^{ 7 }{ C }_{ 5 }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }$

$\displaystyle +^{ 7 }{ C }_{ 6 }{ \left( \frac { 1 }{ 2 } \right) }^{ 6 }{ \left( \frac { 1 }{ 2 } \right) }^{ 1 }+^{ 7 }{ C }_{ 7 }{ \left( \frac { 1 }{ 2 } \right) }^{ 7 }=\frac { 1 }{ 2 }$

Suppose X is a binomial variate

$B ( 5, P )$ and

$P ( X=2 )$ and

$P ( X=3 )$ , then

$P$ is equal to

0%
$\displaystyle\frac{1}{2}$
0%
$\displaystyle\frac{1}{3}$
0%
$\displaystyle\frac{1}{4}$
0%
$\displaystyle\frac{1}{5}$

Explanation

Here

$N = 5$
Given

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

$\Rightarrow ^5C_2 P^2(1-P)^3=^5C_3 P^3 (1-P)^2$

$\Rightarrow 1-P = P \therefore P =\cfrac{1}{2}$

$X$ and

$Y$ are independent binomial variate of binomial distribution

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

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

$P(X+Y=3)$ is

0%
$\displaystyle \frac { 55 }{ 1024 }$
0%
$\displaystyle \frac { 55 }{ 4098 }$
0%
$\displaystyle \frac { 55 }{ 2048 }$
0%
none of these

Explanation

$\displaystyle P\left( X+Y=3 \right) =P\left( X=0,Y=3 \right) +P\left( X=1,Y=2 \right) +P\left( X=2,Y=1 \right) +P\left( X=3,Y=0 \right)$

$\displaystyle =P\left( X=0 \right) P\left( Y=3 \right) +P\left( X=1 \right) P\left( Y=2 \right) +P\left( X=2 \right) P\left( Y=1 \right) +P\left( X=3 \right) P\left( Y=0 \right)$

$(X$ and

$Y$ are independent

$)$

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

$\displaystyle ^{ 7 }{ C }_{ 3 }.{ \left( \frac { 1 }{ 2 } \right) }^{ 7 }$

$\displaystyle +^{ 5 }{ { C }_{ 1 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$

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

$\displaystyle +^{ 5 }{ { C }_{ 2 }.{ \left( \frac { 1 }{ 2 } \right) }^{ 5 } }$

$\displaystyle ^{ 7 }{ { C }_{ 1 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 7 }+$

$\displaystyle ^{ 5 }{ { C }_{ 3 } }.{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$

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

$\displaystyle ={ \left( \frac { 1 }{ 2 } \right) }^{ 12 }\left\{ \left( 1 \right) \left( 35 \right) +\left( 5 \right) \left( 21 \right) +\left( 10 \right) \left( 7 \right) +\left( 10 \right) \left( 1 \right) \right\}$

$\displaystyle =\frac { 220 }{ { 2 }^{ 12 } } =\frac { 55 }{ 1024 }$

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

$\textbf{Step 1: Calculating the value of minimum number of Tosses}$

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

$\implies P(H)\geq 0.8$

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

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

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

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

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

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

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

If X follows a binomial distribution with parameters n = 100 and

$P = \displaystyle\frac{1}{3}$ , then

$P (X = r)$ is maximum when

$r =$

0%
$33$
0%
$50$
0%
$25$
0%
none of these

Explanation

Here,

$n=100$ and

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

$\displaystyle\Rightarrow q=1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 }$

$\displaystyle\because np=100\times \frac { 1 }{ 3 } =33+\frac { 1 }{ 3 }$

$\therefore P(x=r)$ will be maximum.
when

$v=$ (an integer)

$=33$

If the mean and the variance of a binomial variate

$X$ are

$2$ and

$1$ respectively, then the probability that

$X$ takes a value greater than

$1$ is equal to.....

0%
$11/16$
0%
$5/16$
0%
$9/16$
0%
$7/16$

Explanation

${\textbf{Step - 1: Finding n, p and q}}$

${\text{We know that, mean = np and variance = np(1 - p)}}$

$\therefore {\text{ np = 2 and np(1 - p) = 1}}$

${\text{Dividing, }}\dfrac{{{\text{np(1 - p)}}}}{{{\text{np}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{2}}}$

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

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

$\because {\text{ q = 1 - p}}$

$\Rightarrow {\text{ q = 1 - }}\dfrac{{\text{1}}}{{\text{2}}}$

$\Rightarrow {\text{ q = }}\dfrac{{\text{1}}}{{\text{2}}}$

${\text{Substituting value of p in np = 2}}$

$\Rightarrow {\text{ n }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ = 2}}$

$\Rightarrow {\text{ n = 4}}$

${\textbf{Step - 2: Calculating probability}}$

${\text{Probability of value greater than 1, P(X > 1) = 1 - P(X = 0) - P(X = 1)}}$

$\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{1}}}{{{{\text{2}}^{\text{4}}}}}{\text{ - }}{{\text{ }}^{\text{4}}}{{\text{C}}_{\text{1}}}\dfrac{{\text{1}}}{{{{\text{2}}^{\text{4}}}}}$

$\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{1}}}{{{\text{16}}}}{\text{ - }}\dfrac{{\text{4}}}{{{\text{16}}}}$

$\Rightarrow {\text{ P(X > 1) = 1 - }}\dfrac{{\text{5}}}{{{\text{16}}}}$

$\Rightarrow {\text{ P(X > 1) = }}\dfrac{{{\text{11}}}}{{{\text{16}}}}{\text{ }}$

$\textbf{Hence option A is correct}$

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

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

$51$ 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

$\textbf{Step-1: Assume the required to be equal to some variable & then}$

$\textbf{apply all information given.}\\$

$\text{Let X be the number of coins showing heads}$

$\text{Then}$

$X\sim B\left( 100,p \right)$

$\text{We have}$

$P\left( X=51 \right) =P\left( X=50 \right)$

$\text{Since, } P\left( X=r \right)= { n }_{ { C }_{ r } }{ p }^{ r }{ q }^{ n-r }$

$\text{we get}$

$\Rightarrow { 100 }_{ { C }_{ 51 } }{ p }^{ 51 }{ q }^{ 49 }={ 100 }_{ { C }_{ 50 } }{ p }^{ 50 }{ q }^{ 50 }$

$\displaystyle \Rightarrow \frac { p }{ q } =\frac { 100! }{ 50!.50! } \times \frac { 51!49! }{ 100! }$

$\displaystyle \Rightarrow \frac { p }{ 1-p } =\frac { 51 }{ 50 }$

$\text{where}$

$q=1-p$

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

$\displaystyle \Rightarrow P=\dfrac { 51 }{ 101 }$

$\textbf{Hence, the value of p is }$

$\boldsymbol{\dfrac { 51 }{ 101 } }$

A biased coin with probability p, 0 < p < 1, of head is tossed until a head appears for the first time. If the probability that the number of tossess required is even is 2/5, then p = .......

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{3}$

Explanation

Let

$x$ denote the number of tosses required.
Then

$P(X=r)={ \left( 1-p \right) }^{ r-1 }p$ for

$r=1,2,3,...$
Let

$E$ denote the event that the number of tosses required are even. Then,

$P\left( E \right) =P\left[ \left( X=2 \right) \cup \left( X=4 \right) \cup \left( X=6 \right) \cup ... \right]$

$=P\left( X=2 \right) +P\left( X=4 \right) +P\left( X=6 \right) +...$

$=\left( 1-p \right) p+{ \left( 1-p \right) }^{ 3 }p+{ \left( 1-p \right) }^{ 5 }p+...$

$\displaystyle=\frac { p\left( 1-p \right) }{ 1-{ \left( 1-p \right) }^{ 2 } } =\frac { p\left( 1-p \right) }{ 2p-{ p }^{ 2 } } =\frac { 1-p }{ 2-p }$
As we are given that

$\displaystyle P(E)=\frac { 2 }{ 5 },$ we get

$5\left( 1-p \right) =2\left( 2-p \right)\Rightarrow 1-3p=0$ or

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

$6$ ordinary dice are rolled. The probability that at least half of them will show at least

$3$ is

0%
$\displaystyle 41 \times \frac {2^4}{3^6}$
0%
$\displaystyle \frac {2^4}{3^6}$
0%
$\displaystyle 20 \times \frac {2^4}{3^6}$
0%
none of these

Explanation

If one dice is rolled then probability that it will show at least 3 is,

$p =\cfrac{4}{6}=\cfrac{2}{3}$
Here number of dices rolled is,

$n = 6$
Also number appeared on dices are independent to each other.
Thus probability of getting at least three on half of them is,

$=P(X=3)+P(X=4)+P(X=5)+P(X=6)$

$=^6C_3\left(\cfrac{2}{3}\right)^3\left(\cfrac{1}{3}\right)^3 + ^6C_4\left(\cfrac{2}{3}\right)^4\left(\cfrac{1}{3}\right)^2 + ^6C_5\left(\cfrac{2}{3}\right)^5\left(\cfrac{1}{3}\right)+^6C_6\left(\cfrac{2}{3}\right)^6=41.\cfrac{2^4}{3^6}$

A coin is tossed $n$ times, the probability that head will turn up an even number of times is

0%
$\displaystyle \frac{n+1}{2n}$
0%
$\displaystyle \frac{n}{n+1}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle 2^{n-1}$

Explanation

Let x denote the number of heads in n trials then

$P(x=r)=^nC_rp^rq^{n−r}=^nC_r(\dfrac{1}{2})^r(\dfrac{1}{2})^{n−r} =^nC_r(\dfrac{1}{2})^n$

now required probability is

$p(x=0,2,4,6,⋯) ∴P(X=0,2,4,6,⋯)=P(0)+P(2)+P(4)+⋯ =(\dfrac{1}{2})^n[c_0+c_2+c_4+⋯] =(\dfrac{1}{2})^{2n−1}=\dfrac{1}{2}$

Let

$X$ denote the number of time tail appear in

$n$ tosses of a fair coin. If

$P\left ( X\:=\:1 \right ), \: P\left ( X\:= \:2 \right )$ and

$P\left ( X\:=\:3 \right )$ are in A.P., then value of

$n$ is

0%
$9$
0%
$2$
0%
$7$
0%
None of these

Explanation

$\displaystyle P\left ( T \right )= P\left ( \bar{T} \right )=\frac{1}{2}$

$\therefore$ Using

$\displaystyle P\left ( X=r \right )=^{n}C_{r}\left ( \frac{1}{2} \right )^{2}$
Given,

$\displaystyle 2P\left ( x=2 \right )=P\left ( X=1 \right )+P\left ( X=3 \right )$

$\displaystyle 2.^{n}C_{2}=^{0}C_{1}+^{n}C_{3}$

$\displaystyle \Rightarrow n^{2}-9n+14=0$

$\displaystyle \Rightarrow n=7,2$

$\displaystyle \therefore n=7$ if

$n=2$ there exist only three terms

$\displaystyle \therefore n\neq 2.$

The sum mean and variance of a B.D. for

$5$ trials is

$1.8$ , then the Binomial distribution (B.D.) is given by

0%
$\displaystyle\left (0.2+0.8 \right )^{5}$
0%
$\displaystyle\left (0.3+0.7 \right )^{5}$
0%
$\displaystyle\left (0.4+0.6 \right )^{5}$
0%
None of these

Explanation

mean

$(m)$ for

$B.D.=np$
Variance

$\displaystyle \left ( o^{2} \right )$ for

$B.D.=npq$

$\displaystyle \therefore np\left ( 1+q \right )=\frac{18}{10,}p\left ( 1+q \right )=\frac{18}{50}$

$\displaystyle \left ( 1-q \right )\left ( 1+q \right )=\frac{18}{50}$

$\displaystyle \therefore q=0.8(q=-0.8$ can't possible as

$0\leq q\leq 1)$

$\displaystyle \therefore q=0.8, p=0.2$

$\displaystyle \therefore B.D.=\left ( 0.2+.8 \right )^{5}$

For a Poission distribution which pair has same value.

0%
(Mean, Std. Deviation)
0%
(Variance, Standard Deviation)
0%
(Mean, Variance)
0%
None of these

Explanation

For Poission distribution we have

$\displaystyle P\left ( r \right )=\dfrac{e^{-\pi }\lambda ^{r}}{r^{i}};\left ( r=0, 1, 2,...\infty \right)$

$\displaystyle$ mean

$\displaystyle =\dfrac{\sum f_{i}x_{i}}{\sum f_{i}}=\sum_{r=0}^{\infty }rP\left ( r \right ),\sum f_{i}P\left ( r \right )=1$

$\displaystyle =0+\lambda e^{-\lambda }+2\dfrac{\lambda ^{2}e^{-\lambda }}{2!}+3\dfrac{\lambda ^{3}e^{-\lambda }}{3!}+.....\infty$

$\displaystyle =\lambda e^{-\lambda }\left ( 1+\dfrac{\lambda }{11}+\dfrac{\lambda ^{2}}{2!}+\dfrac{\lambda ^{3}}{3!}+.....\infty \right )$

$\displaystyle =\lambda e^{-\lambda }.e^{\lambda }=\lambda$

Similarly,

$\displaystyle o^{2}=$ Variance

$=\sum_{r=0}^{\infty }r^{2}P\left ( r \right )-\left ( \sum_{r=0}^{\infty } rP\left ( r \right ) \right )^{2}$

$=\displaystyle \lambda e^{\lambda }\left ( e^{-\lambda }+\lambda e^{-\lambda } \right )-\lambda ^{2}=\lambda =$ mean

$\therefore$ mean

$=$ variance

Numbers are selected at random one at a time, from the number

$00,01,02,...,99$ with replacement. An event

$E$ occurs if and only if the product of the two digits of a selected number is

$18.$ If four numbers are selected, then the probability that

$E$ occurs at least

$3$ times, is

0%
$\displaystyle \frac{97}{390625}$
0%
$\displaystyle \frac{68}{390625}$
0%
$\displaystyle \frac{72}{390625}$
0%
None of these

Explanation

Out of the numbers

$00,01,02,..., 99,$ those numbers the product of whose digits is

$18$ are

$29,36,63,92$ i.e., only

$4.$

$\displaystyle p=P\left( E \right) =\dfrac { 4 }{ 100 } =\dfrac { 1 }{ 25 } ,q=P\left( \overline { E } \right) =1-\dfrac { 1 }{ 25 } =\dfrac { 24 }{ 25 }$

Let

$X$ be the random variable, showing the number of times

$E$ occurs in

$4$ selectios.

Then

$P(E$ occurs at least

$3$ times

$)$

$=P(X=3$ or

$X=4)$

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

$=^{ 4 }{ { C }_{ 3 } }{ p }^{ 3 }{ q }^{ 1 }+_{ }^{ 4 }{ { C }_{ 4 } }{ p }^{ 4 }{ q }^{ 0 }=4{ p }^{ 3 }q+{ p }^{ 4 }$

$\displaystyle =4\times { \left( \dfrac { 1 }{ 25 } \right) }^{ 3 }\times \dfrac { 24 }{ 25 } +{ \left( \dfrac { 1 }{ 25 } \right) }^{ 4 }=\dfrac { 97 }{ 390625 } .$

If X follows a B.D. with parameter

$n= 6$ . If

$\:4P\left ( X= 4 \right )= P\left ( X= 2 \right )$ then P equals

0%
$\displaystyle \frac{1}{6}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{8}$
0%
$\displaystyle \frac{1}{3}$

Explanation

Given,

$\displaystyle 4.P\left ( X=4 \right )=P\left ( X=2 \right )$

$\displaystyle \Rightarrow 4 . ^{6}C_{4}p^{4}q^{2}=^{6}C_{2}p^{2}q^{4}$

$\displaystyle \Rightarrow 4p^{2}=q^{2}=(1-p)^2$

$\displaystyle \Rightarrow 3p^{2}+2p-1=0$

$\displaystyle \Rightarrow p=1/3$

$X$ and

$Y$ are the independent random variables

$\displaystyle B\left( 5,\frac { 1 }{ 2 } \right)$ 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%
$\displaystyle \frac { 4032 }{ 4096 }$
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

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

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

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

$[\therefore X$ and

$Y$ are independent

$]$

$\displaystyle =1-^{ 5 }{ { C }_{ 0 } }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }._{ }^{ 7 }{ { C }_{ 0 } }{ \left( \frac { 1 }{ 2 } \right) }^{ 7 }=1-{ \left( \frac { 1 }{ 2 } \right) }^{ 12 }=\frac { 4095 }{ 4096 } .$

For a Binomial distribution whose mean is 9 and S.D. is

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

0%
$12$
0%
$36$
0%
$9$
0%
none of these

Explanation

Fact: for Binomial Distribution
mean

$=np=9$ and S.D.

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

$np=9$ and

$\displaystyle npq=\frac{9}{4}$

$\therefore$

$\displaystyle q=\frac{1}{4}$ ,

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

$\therefore$

$np=9$

$\displaystyle n=\frac{9\times 4}{3}=12$

A die is thrown

$2n$ times. The probability that the number greater than 4 appears at least once in

$2n$ throws is

0%
$\displaystyle \left ( \frac{1}{3} \right )^{2n}$
0%
$1-\displaystyle \left ( \frac{1}{3} \right )^{2n}$
0%
$\displaystyle \frac{3^{2n}-\:2^{2n}}{3^{2n}}$
0%
None of these

Explanation

Probability of success,

$p = \displaystyle P ( X> 4 )=P ( 5$ or

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

$\displaystyle \\ \\ \therefore q=1-\frac{1}{3} =\frac{2}{3}$
Now using binomial distribution,

$\displaystyle P\left ( X=r \right )^{2n}C_{r}\left ( \frac{1}{3} \right )^{r}\left ( \frac{2}{3} \right )^{2n-r}$

$P(1$ or

$2$ or

$3...$ or

$2n)=P(1)+P(2)+...+p(2n)=1-P(0)$

$\displaystyle =1-^{2n}C_{0}\left ( \frac{1}{3} \right )^{0}\left ( \frac{2}{3} \right )^{2n}$

$\displaystyle =\left ( 1-\frac{2^{2n}}{3^{2n}} \right )=\frac{3^{2n}-2^{2n}}{3^{2n}}$

For a poission distribution variable

$X$ is such that

$P(X = 2) = 9 P(X= 4) + 90 P(X= 6)$ the mean is

0%
$2$
0%
$3$
0%
$1$
0%
None of these

Explanation

For

$P.D. P(X=r)=\displaystyle \dfrac{e^{-\lambda}\lambda ^{r}}{r!},r=0,1,2,\cdots$

$\therefore \displaystyle \dfrac{e^{-\lambda}\lambda ^{2}}{2!}=\dfrac{9e^{-\lambda}\lambda ^{4}}{4!}+90 \dfrac{e^{-\lambda}\lambda ^{6}}{6!}$ (given)

$\Rightarrow \displaystyle\dfrac{\lambda ^{2}}{2}=\dfrac{9}{24}\lambda ^{4}+\dfrac{90}{720}\lambda ^{6}$

$\Rightarrow \lambda ^{4}+3\lambda ^{2}-4=0$

$\Rightarrow \left ( \lambda^{2}+4 \right )\left ( \lambda^{2}-1 \right )=0=>\lambda= \pm 1$

$\Rightarrow \lambda=1$ as

$\lambda>0$ and

$(\lambda^{2}+4=0$ impossible

$)$

$\therefore$ mean

$=\lambda=1$

For a B,D, variance is given by the formula

0%
$\displaystyle \sigma = npq$
0%
$\displaystyle \sigma = \frac{\sqrt{npq}}{2}$
0%
$\displaystyle \sigma ^{2} = npq$
0%
$\displaystyle \sigma = n^{2}p^{2}q^{2}$

Explanation

It is fundamental concept that variance in Binomial distribution is calculated by ,

$\sigma^2 = npq$ .

For a Poission distribution, which of the following is true

0%
$Mean = Mode$
0%
$Median = S.D.$
0%
$Mean = Variance$
0%
$Median = Variance$

Explanation

For Poission distribution we have

$\displaystyle P\left ( r \right )=\dfrac{e^{-\pi }\lambda ^{r}}{r^{i}}\left ( r=0, 1, 2,...\infty \right)$

$\displaystyle$ mean

$\displaystyle =\dfrac{\sum f_{i}x_{i}}{\sum f_{i}}=\sum_{r=0}^{\infty }rP\left ( r \right ),\sum f_{i}P\left ( r\right )=1$

$\displaystyle =0+\lambda e^{-\lambda }+2\dfrac{\lambda ^{2}e^{-\lambda }}{2!}+3\dfrac{\lambda ^{3}e^{-\lambda }}{3!}+.....\infty$

$\displaystyle =\lambda e^{-\lambda }\left ( 1+\dfrac{\lambda }{11}+\dfrac{\lambda ^{2}}{2!}+\dfrac{\lambda ^{3}}{3!}+.....\infty \right )$

$\displaystyle =\lambda e^{\lambda }.e^{\lambda }=\lambda$

similarly,

$\displaystyle o^{2}=$ Variance

$=\sum_{r=0}^{\infty }r^{2}P\left ( r \right )-\left ( \sum_{r=0}^{\infty } rP\left ( r \right ) \right )^{2}$

$=\displaystyle \lambda e^{\lambda }\left ( e^{-\lambda }+\lambda e^{-\lambda } \right )-\lambda ^{2}=\lambda =$ mean

$\therefore \text{mean}=\text{variance}$

A pair of dice is thrown 4 times, then the probability of getting doublets at least twice is

0%
$\displaystyle \frac{19}{44}$
0%
$\displaystyle \frac{21}{44}$
0%
$\displaystyle \frac{31}{44}$
0%
$\displaystyle \frac{39}{44}$

Explanation

$\displaystyle n(S)=6^{2}=36$

$\displaystyle p=$ probability of getting a doublet on the pair of dice

$\displaystyle \therefore P=\frac{6}{36}=\frac{1}{6},q=1-p=1-\frac{1}{6}=\frac{5}{6}, n=4$

$\displaystyle \therefore P(X=r)=^{4}C_{r}\left(\frac{1}{6}\right)^{r}\left(\frac{5}{6}\right)^{4-r}$ ,

$\displaystyle r=0,1,2,3,4$

$\displaystyle \therefore P(X=r\geq 2)=1-P(X=r<2)$

$\displaystyle =1-\left\{P(X=r= 0)+P(X=r=1)\right\}$

$\displaystyle =1-\left\{^{4}C_{0}\left(\frac{1}{6}\right)^{0}\left(\frac{5}{6}\right)^{4}+^{4}C_{1}\left(\frac{1}{6}\right)^{1}\left(\frac{5}{6}\right)^{3}\right\}$

$\displaystyle =1-\left(\frac{5}{6}+\frac{2}{3}\right)\left(\frac{5}{6}\right)^{3}=1-\frac{375}{432}=\frac{19}{44}$
Hence choice (a) is correct answer.

A die is thrown 5 times. Getting an odd number considered as success, then variance of distribution is

0%
$\displaystyle \frac{8}{3}$
0%
$\displaystyle \frac{3}{8}$
0%
$\displaystyle \frac{4}{5}$
0%
$\displaystyle \frac{5}{4}$

Explanation

$\displaystyle n=5$ ,

$\displaystyle p=q=1/2$ since probability of getting even and odd are same.

$\therefore$ Variance is

$\sigma^2 = npq = \cfrac{5}{4}$

The mean and variance of Binomial Distribution are

$4$ and

$2$ respectively, then the probability of success is

0%
$\displaystyle \frac{128}{256}$
0%
$\displaystyle \frac{219}{256}$
0%
$\displaystyle \frac{37}{256}$
0%
$\displaystyle \frac{28}{256}$

Explanation

Given

$\displaystyle np=4$ and

$\displaystyle npq=2,$

$\displaystyle q=\frac{npq}{np}=\frac{2}{4}=\frac{1}{2}$ so

$\displaystyle p=1-\frac 12=\frac{1}{2}$

Now

$\displaystyle npq=2$

$\therefore n=8$

$\displaystyle \therefore$ BD is given by

$\displaystyle P(X=r)=^{8}C_{r}p^{r} q^{n-r}$

$\displaystyle \therefore P(X=r=2)=^{8}C_{2}\left(\frac{1}{2}\right)^{8}$

$\displaystyle = \frac{28}{256}$

The mean and the variance of a random variable X having Binomial Distribution are 4 and 2 respectively then

0%
$\displaystyle \frac{1}{16}$
0%
$\displaystyle \frac{1}{8}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{32}$

Explanation

Given mean

$=4$

Variance

$=2$

$\therefore np=4$

and

$npq=2$

$\therefore4q=2,q=\cfrac { 2 }{ 4 } =\cfrac { 1 }{ 2 }$

$p=1-q=\cfrac { 1 }{ 2 }$

$\therefore n\left( { 1 }/{ 2 } \right) =4;n=8$

Thus

$p(X=1)=^{ 8 }{ C }_{ 1 }\left( \cfrac { 1 }{ 2 } \right) { \left( \cfrac { 1 }{ 2 } \right) }^{ 7 }$

$=8{ \left( \cfrac { 1 }{ 2 } \right) }^{ 8 }=\cfrac { 8 }{ 256 } =\cfrac { 1 }{ 32 }$

$n$ 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 the above information answer the following question.

A product is supposed to contain 5% defective items. What is the probability that a sample of 8 items will contain less than 2 defeclive items?

0%
$\displaystyle \frac{25\times \left ( 19 \right )^{7}}{\left ( 20 \right )^{8}}$
0%
$\displaystyle \frac{27\times \left ( 19 \right )^{7}}{\left ( 20 \right )^{8}}$
0%
$\displaystyle \frac{ \left ( 19 \right )^{7}}{\left ( 20 \right )^{8}}$
0%
None of these

Explanation

Here

$\displaystyle n=8,p=\frac{5}{100}=\frac{1}{20}.q=\frac{19}{20}$

Now using

$\displaystyle P(X=r)=C(8,r)p^{r}q^{8-r}$

$\displaystyle \therefore P(X=r<2)=P(0)+P(1)$

$\displaystyle =\left(\frac{19}{20}\right)^{8}+\frac{8\times (19)^{7}}{(20)^{8}}=\frac{(19)^{7}}{(20)^{8}}[19+8]=\frac{27\times (19)^{7}}{(20)^{8}}$

Hence choice (b) is the correct answer.

The mean variance of a binomial variable

$X$ are

$2$ and

$1$ respectively. The probability that

$X$ takes values greater than

$1$ , is

0%
$\displaystyle \dfrac { 5 }{ 16 }$
0%
$\displaystyle \dfrac { 9 }{ 16 }$
0%
$\displaystyle \dfrac { 11 }{ 16 }$
0%
None of these

Explanation

Given: mean

$np=2$ ...(1)

and variance

$npq=144$ ...(2)

Dividing (2) by (1), then

$\displaystyle q=\dfrac { 1 }{ 2 }$

$\displaystyle \therefore p=1-q=\dfrac { 1 }{ 2 }$

From (1),

$\displaystyle n\times \dfrac { 1 }{ 2 } =2\Rightarrow n=4$

The binomial distribution is

$\displaystyle { \left( \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 2 } \right) }^{ 4 }$

Now,

$P\left( X>1 \right) =P\left( X=2 \right) +P\left( X=3 \right) +P\left( X=4 \right)$

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

$\displaystyle =\dfrac { 6+4+1 }{ 16 } =\dfrac { 11 }{ 16 }$

A random variable

$X$ has Poisson distribution with mean

$2$ The

$\displaystyle P(X>1.5)$ equals

0%
$0$
0%
$\displaystyle 2/e^{2}$
0%
$\displaystyle 3/e^{2}$
0%
$\displaystyle 1-\frac{3}{e^{2}}$

Explanation

We know

$\displaystyle P(X=r)=\frac{e^{-\lambda}\lambda^{r}}{r!}, (\lambda =$ mean

$)$

$\therefore \displaystyle P(X=r>1.5)=P(2)+P(3)+....\infty=1-[P(0)+P(1)]$

$\displaystyle =1-\left[ e^{-2}+\frac{e^{-2}\times 2^{2}}{2}\right]=1-\frac{3}{e^{2}}$

At a telephone enquiry system the number of phone calls regarding relevant enquiry follow Poisson distribution with a average of 5 phone calls during IO-minute time intervals. The probability that there is at the most one phone call during a 10-minute time period is

0%
$\displaystyle \frac{6}{5^{e}}$
0%
$\displaystyle \frac{5}{6}$
0%
$\displaystyle \frac{6}{55}$
0%
$\displaystyle \frac{6}{e^{5}}$

Explanation

According to Poisson distribution

$\displaystyle P(X=r)=\frac{e^{-m}m^{r}}{r!}$

$\displaystyle \therefore P(X\leq 1)=P(X=0)+P(X=1)$

$\displaystyle =e^{-m}+\frac{e^{-m} m}{1!}$
Given m

$\displaystyle =$ mean

$= 5$

$\displaystyle \therefore P(x \leq 1)=e^{-5}+5\times e^{-5}=e^{-5}(1+5)=\frac{6}{e^{5}}$

The probability that a student is not a swimmer is

$1/5$ . The probability that out of

$5$ students exactly

$4$ are swimmer is

0%
$\left (\dfrac {4}{5}\right )^3$
0%
$\left (\dfrac {4}{5}\right )^4$
0%
$5\left (\dfrac {4}{5}\right )^4$
0%
$3\left (\dfrac {4}{5}\right )^3$

Explanation

Let

$p=$ probability that a student selected at random is a swimmer

$=1-1/5=4/5$

$n=$ number of students selected

$=5$
and

$X=$ number of swimmers
Then

$X~B(n, p)$ where

$n=5$ and

$p=4/5$
Probability that exactly 4 students are swimmer is

$\displaystyle = ^5C_4 p^4 q=5\left (\frac {4}{5}\right )^4\left (\frac {1}{5}\right )= \left (\frac {4}{5}\right )^4$ .

If the mean and variance of a binomial variate

$X$ and

$2$ and

$1$ respectively, then the probability that

$X$ takes a value at most

$3$ is

0%
$2/3$
0%
$4/5$
0%
$7/8$
0%
$15/16$

Explanation

We have

$E(X)=np=2$ and

$var (X)=npq=1$ .
Therefore,

$q=1=1/2$ or

$p=1/2$ , and

$n=4$ .

Thus,

$\displaystyle P(X\leq 3)=1-P(X-4)=1-^4C_4\left (\frac {1}{2}\right )^4=1-\frac {1}{16}=\frac {15}{16}$

An unbaised die with faces marked

$1, 2, 3, 4, 5$ and

$6$ is rolled four times. Out of four face values obtained, the probability that the minimum face value is not less than

$2$ and the maximum face value is not greater than

$5$ is then

0%
$16/81$
0%
$1/81$
0%
$80/81$
0%
$65/81$

Explanation

Let

$p=$ probability of getting face value not less than

$2$ and not more than

$5$ in a single throw of die

$=4/6=2/3$

$n=$ number of times die is rolled

$X=$ number of times we get a number not less than

$2$ and not more than

$5$ .
Then by bernoulli distribution

$X~B(n, p)$
Required probability

$=P(X=4)=^4C_4p^4=(2/3)^4=16/81$ .

Suppose

$X$ follows a binomial distribution with parameters

$n$ and

$p$ , where

$0 < p < 1$ . If

$P(X-r)/P(X=n-r)$ is independent of

$n$ for every value of

$r$ , then

0%
$p=1/2$
0%
$p=1/3$
0%
$p=1/4$
0%
$p=1/5$

Explanation

We have

$\displaystyle \frac {P(X=r)}{P(X=n-r)}=\frac {^nC_rp^r(1-p)^{n-r}}{^nC_{n-r}p^{n-r}(1-p)^r}=\frac {(1-p)^{n-2r}}{p^{n-2r}}=\left (\frac {1}{p}-1\right )^{n-2r}$
Note that

$(1/p) - 1 > 0$ . Therefore the ratio will be independent of

$n$ for each

$r$ if

$(1/p)-1=1$ , or

$p=1/2$ .

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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