MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 9

If, in a Poisson distribution

$P(X= 0)=k$ then the variance is:

0%
$e^{\lambda}$
0%
$\log \dfrac{1}{k}$
0%
$\dfrac{1}{k}$
0%
$\log k$

Explanation

Poisson distribution

$P\left(X\right)=\dfrac{{e}^{-\lambda}{\lambda}^{x}}{x!}$

$P\left(X=0\right)=k\Rightarrow \dfrac{{e}^{-\lambda}1}{1}=k$

$\Rightarrow {e}^{-\lambda}=k$

$\therefore k={e}^{-\lambda}$

$\Rightarrow \log e^{-\lambda}=\log k$

$\Rightarrow$

$-\lambda \log e=\log k$

$\Rightarrow$

$-\lambda=\log k$

$Variance=\lambda=-\log k$

$\Rightarrow$

$\lambda=\log\left(\dfrac{1}{k}\right)$

Let a random variable X have a binomial distribution with mean

$8$ and variance

$4$ . If

$P(x\leq 2)=\dfrac{k}{2^{16}}$ , then k is equal to?

0%
$17$
0%
$1$
0%
$121$
0%
$137$

Explanation

$np=8$

$npq=4$

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

$n=16$

$p(x=r)={^{16}C_r}\left(\dfrac{1}{2}\right)^{16}$

$p(x\leq 2)=\dfrac{^{16}C_0+{^{16}C_1}+{^{16}C_2}}{2^{16}}$

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

A die is thrown

$7$ times. The chance that an odd number turns up exactly 4 times, is given by

0%
$\dfrac{1}{{128}}$
0%
$\dfrac{{35}}{{128}}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{{31}}{{128}}$

Explanation

Die thrown

$7$ times

No. of chances of odd number

$=1,3,5=3$

Probability (odd no.)

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

p(odd)

$=\cfrac { 1 }{ 2 }$

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

$p\left( x=4 \right) =^{ 7 }{ C }_{ 4 }{ p }^{ 4 }{ q }^{ 3 }$

$=\cfrac { 7! }{ 4!3! } \times { \left( \cfrac { 1 }{ 2 } \right) }^{ 7 }=\cfrac { 7\times 6\times 5 }{ 3\times2 } \left( \cfrac { 1 }{ { 2 }^{ 7 } } \right)$

$=\cfrac { 35 }{ 2^7 } =\cfrac { 35 }{ 128 }$

If n different apples are to be distributed among m children, find the chance that the particular child receives p apples.

0%
$\displaystyle\frac{^{ n }{ C_p }(m-1)^{n-p }}{m^n}$
0%
$\displaystyle\frac{^{ n }{ C_p }(m-1)^{n-p }}{n^m}$
0%
$\displaystyle\frac{^{ n }{ P_p }(m-1)^{n-p }}{n^m}$
0%
$\displaystyle\frac{^{ n }{ P_p }(m-1)^{n-p }}{m^n}$

Explanation

Total ways of distributing

$n$ apples to

$m$ children =

$m^n$
Ways of giving

$p$ apples to particular child = Select

$p$ apples out of

$n$ apples

$\times$ Total ways of distributing

$(n-p)$ apples to

$(m-1)$ children
=

${ _{ }^{ n }{ C } }_{ p }{ (m-1) }^{ n-p }$
so

$P = \cfrac { { _{ }^{ n }{ C } }_{ p }{ (m-1) }^{ n-p } }{ { m }^{ n } }$

The minimum number of times one has to toss fair coins so that probability of observing at least one head is at least

$90$ % is:

0%
$5$
0%
$3$
0%
$2$
0%
$4$

Explanation

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.

The mean and variance of a random variable

$X$ having a binomial distribution are

$6$ and

$3$ respectively. The probability of variable

$X$ less than

$2$ is

0%
$\cfrac { 13 }{ 2048 }$
0%
$\cfrac { 13 }{ 4096 }$
0%
$\cfrac { 15 }{ 4096 }$
0%
$\cfrac { 25 }{ 2048 }$

Explanation

mean =6

variance=3

$np=6,npq=3$

p+q=1

$\Rightarrow q=\cfrac { 1 }{ 2 } ,$

$p=\cfrac { 1 }{ 2 } ,$

$n=12\quad$
so required probability

${ _{ }^{ 12 }{ C } }_{ 0 }{ p }^{ 0 }{ q }^{ 12 }+{ _{ }^{ 12 }{ C } }_{ 1 }{ p }^{ 1 }{ q }^{ 11 }$

$=\cfrac { 13 }{ { 2 }^{ 12 } }$

$=\cfrac { 13 }{ 4096 }$

A coin is biased so that the head is $3$ times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tails.

0%
$P(T=0)=\dfrac{11}{16}$
0%
$P(T=1)=\dfrac{6}{16}$
0%
$P(T=2)=\dfrac{13}{16}$
0%
none of these

Explanation

Given a biased coin such that heads is

$3$ times as likely as tails. The coin is tossed twice.

$P{\left( H \right)} = \cfrac{3}{4}$ and

$P{\left( T \right)} = \cfrac{1}{4}$

Let

$T$ be the random variable for the number of tails.

$P{\left( T = 0 \right)} = P{\left( HH \right)} = \cfrac{3}{4} \times \cfrac{3}{4} = \cfrac{9}{16}$

$P{\left( T = 1 \right)} = P{\left( HT, TH \right)} = \cfrac{3}{4} \times \cfrac{1}{4} + \cfrac{3}{4} \times \cfrac{1}{4} = \cfrac{3}{16} + \cfrac{3}{16} = \cfrac{3}{8} = \cfrac{6}{16}$

$P{\left( T = 2 \right)} = P{\left( TT \right)} = \cfrac{1}{4} \times \cfrac{1}{4} = \cfrac{1}{16}$

Hence

$\left( B \right)$ is the correct answer.

In a workshop, there are five machines and the probability of any one of them to be out of service on day is

$\dfrac{1}{4}$ . If the probability that at most two machines will be out of service on the same day is

$\left(\dfrac{3}{4} \right)^3 k$ , then k is equal to :

0%
$4$
0%
$\dfrac{17}{2}$
0%
$\dfrac{17}{8}$
0%
$\dfrac{17}{4}$

Explanation

Given

$n=5,p=\dfrac 14,q=1-p=\dfrac 34$

Required probability

= when no machine has fault + when only one machine has fault + when only two machine has fault

$= \, ^5C_0 \left(\dfrac{3}{4} \right)^5 + \,^5C_1 \left(\dfrac{1}{4} \right) \left(\dfrac{3}{4} \right)^4 + \, ^5C_2 \left(\dfrac{1}{4} \right)^2 \left(\dfrac{3}{4} \right)^3$

$= \dfrac{243}{1024} + \dfrac{405}{1024} + \dfrac{270}{1024}$

$= \dfrac{459}{512}$

$= \left(\dfrac{3}{4} \right)^3 \times k = \left(\dfrac{3}{4} \right)^3 \times \dfrac{17}{8}$

by comparing we get

$k = \dfrac{17}{8}$

The mean and variance of a binomial distribution are

$8$ and

$4$ respectively. What is

$(X = 1)$ ?

0%
$\dfrac{1}{2^8}$
0%
$\dfrac{1}{2^{12}}$
0%
$\dfrac{1}{2^6}$
0%
$\dfrac{1}{2^4}$
0%
$\dfrac{1}{2^5}$

Explanation

Let n and p be the parameters of the binomial distribution.

Mean

$= 8$ and variance

$= 4$

$\Rightarrow np = 8$ and

$npq = 4$

$\Rightarrow q = \dfrac{1}{2} = p$ and

$n = 16$

$\therefore$ Required probability

$= P(X = 1)$

$= \ ^{16} C_1 \left(\dfrac{1}{2} \right)^1 \left(\dfrac{1}{2}\right)^{15} = 16 \times \left(\dfrac{1}{2} \right)^{16}$

$= \dfrac{2^4}{2^{16}} = \dfrac{1}{2^{12}}$

A poison variate X satisfies

$P(X - 1) = P (X = 2)$

$P(X = 6)$ is equal to

0%
$\dfrac{4}{45} e^{-2}$
0%
$\dfrac{1}{45} e^{-1}$
0%
$\dfrac{1}{9} e^{-2}$
0%
$\dfrac{1}{4} e^{-2}$
0%
$\dfrac{1}{45} e^{-2}$

Explanation

Given that,

$P(X=1)=P(X=2)$

$=\dfrac{e^{-\lambda}\lambda^1}{1!}=\dfrac{e^{-\lambda}\lambda^2}{2!}$

$\Rightarrow \lambda=2$

$\therefore P(X=6)=\dfrac{e^{-2}(2)^6}{6!}$

$=\dfrac{e^{-2}\times 4\times 2^4}{6\times 5\times 4\times 3\times 2}$

$=\dfrac{4\times e^{-2}\times 2^4}{45\times 2^4}=\dfrac{4e^{-2}}{45}$

An unbiased coin is tossed

$5$ times. Suppose that a variable

$X$ is assigned the value

$k$ when

$k$ consecutive heads are obtained for

$k=3,4,5,$ otherwise

$X$ takes the value

$-1$ . The the expected value of

$X$ , is :

0%
$\dfrac{1}{8}$
0%
$\dfrac{3}{16}$
0%
$-\dfrac{1}{8}$
0%
$-\dfrac{3}{16}$

Explanation

$\textbf{Step-1: Prepare a table as per information given.}$

$\text{If a coin is tossed n number of times, the total sample space}$

$= 2^n$

$\text{Given, }$

$\text{A coin tossed 5 times, so total outcomes = }$

$2^5 = 32$

$\text{k=no. of times head occur simultaneously.}$

$\textbf{Step-2: Use the above results & find the required unknown.}$

$\displaystyle \sum Xp(k)$

$=(-1)\times \dfrac{1}{32}+(-1)\times \dfrac{12}{32}+(-1)\times \dfrac{(11)}{32}+3\times \dfrac{5}{32}+4\times \dfrac{2}{32}+5\times \dfrac{1}{32}\\$

$=\dfrac{-1}{32}-\dfrac{12}{32}-\dfrac{11}{32}+\dfrac{15}{32}+\dfrac{8}{32}+\dfrac{5}{32}\\$

$=\dfrac{28-24}{32}\\$

$=\dfrac{4}{32}\\$

$=\dfrac{1}{8}$

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

If the mean and variance of a binomial variate

$X$ are

$\dfrac {7}{3}$ and

$\dfrac {14}{9}$ respectively. Then probability that

$X$ takes value

$6$ or

$7$ is equal to

0%
$\dfrac {1}{729}$
0%
$\dfrac {5}{729}$
0%
$\dfrac {7}{729}$
0%
$\dfrac {13}{729}$

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%
$1/5$

A coin is tossed

$5$ times. What is the probability that tail appears an odd number of times?

0%
$\cfrac{3}{5}$
0%
$\cfrac{2}{15}$
0%
$\cfrac{1}{2}$
0%
$\cfrac{1}{3}$

A die is thrown

$5$ times. If getting an odd number is a success, then what is the probability of getting atleast

$4$ successes?

0%
$\cfrac{4}{5}$
0%
$\cfrac{7}{16}$
0%
$\cfrac{3}{16}$
0%
$\cfrac{3}{20}$

A coin is tossed

$5$ times. What is the probability that head appears an even number of times?

0%
$\cfrac{2}{5}$
0%
$\cfrac{3}{5}$
0%
$\cfrac{4}{15}$
0%
$\cfrac{1}{2}$

$X$ follows binomial distribution with parameters

$n=8$ and

$p =1/2$ then

$p\left(|x-4| < 2\right)=$

0%
$121/128$
0%
$119/128$
0%
$117/128$
0%
$115/128$

A fair coin is tossed

$6$ times. What is the probability of getting at least

$3$ heads?

0%
$\cfrac{11}{16}$
0%
$\cfrac{21}{32}$
0%
$\cfrac{1}{18}$
0%
$\cfrac{3}{64}$

The probability of the safe arrival of one ship out of

$5$ is

$\cfrac{1}{5}$ . What is the probability of the safe arrival of at least

$3$ ships?

0%
$\cfrac{1}{31}$
0%
$\cfrac{3}{52}$
0%
$\cfrac{181}{3125}$
0%
$\cfrac{184}{3125}$

$8$ coins are tossed simultaneously. The probability of getting at least

$6$ heads is

0%
$\cfrac{7}{64}$
0%
$\cfrac{57}{64}$
0%
$\cfrac{37}{256}$
0%
$\cfrac{249}{256}$

Suppose random variable X follows the binomial distribution with parameters n and p, where

$0<p<1$ . If

$P(x=r)/P(x=n-r)$ is independent of n and r, then p equals

0%
1/2
0%
1/3
0%
1/5
0%
1/7

Explanation

$\because P(X=r) ^nC_r(p)^r(q)^{n-r}= \dfrac{n!}{(n-r)!r!} (p)^r(1-p)^{n-r} [\because q=1-p] ....(i)$

$P(X=0)=(1-P)^n$
And

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

$\dfrac{n!}{(n-r)!r!}(p)^{n-r} [\because q=1-p] [\because^nC_r=^nC_{n-r}] ...(ii)$

$\dfrac { P(x=r) }{ P(x=n-r) } =\dfrac { \dfrac { n! }{ (n-r)!r! } { p }^{ r }(1-p)^{ n-r } }{ \dfrac { n! }{ (n-r)!r! } { p }^{ n-r }(1-p)^{ +r } }$ [using Eqs. (i) and (ii)]

$\left( \dfrac{1-p}{p}\right)^{n-r}\times \dfrac{1}{\left(\dfrac{1-p}{p}\right)^r}$
Above expression is independent of n and r, if

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

2K coins (K is an integer) each with probability P(O<P<1) of getting head are tossed together. If the probability of getting K heads is equal to the probability of getting K+1 heads, the value of P is

0%
$\dfrac{K}{2K+1}$
0%
$\dfrac{K+1}{2K}$
0%
$\dfrac{K+1}{2K+1}$
0%
$\dfrac{2K}{K+1}$

Explanation

Let

$X$ be the number of heads occuring

$\ P(X=k+1) = (\begin{matrix} 2k \\ k+1 \end{matrix}){ p }^{ k+1 }{ (1-p) }^{ k-1 }$

Given

$P(X=k) = P(X=k+1)\\ (\begin{matrix} 2k \\ k \end{matrix}){ p }^{ k }{ (1-p) }^{ k } =(\begin{matrix} 2k \\ k+1 \end{matrix}){ p }^{ k+1 }{ (1-p) }^{ k-1 }\\ \dfrac { k+1 }{ k } (1-p) =\ p\\ p\ =\ \dfrac { k+1 }{ 2k+1 } \\$

A discrete random variable

$X$ can take all possible integer values from

$1$ to

$K$ , each with a probability

$\dfrac{1}{k}$ . Its variance is

0%
$\dfrac{k^{2}}{4}$
0%
$\dfrac{(k+1)^{2}}{4}$
0%
$\dfrac{k^{2}-1}{12}$
0%
$\dfrac{k^{2}-1}{6}$

Explanation

Variance =

$\sum{(X_{i}- \mu)^2.P_{i}}$ ,
where

$\mu$ is the mean of the given distribution.

$\displaystyle \mu = \sum { { X }_{ i }{ P }_{ i } } =\frac { 1 }{ k } \left( 1+2+3+...k \right) =\frac { 1 }{ k } .\frac { k\left( k+1 \right) }{ 2 } \\ =\dfrac { k+1 }{ 2 }$
So, variance

$V \displaystyle= \frac{1}{k} \sum\left[{X_{i} - \left(\frac{k + 1}{2} \right)}\right]^2$

$\displaystyle = \frac{1}{k} \left[\sum{X_{i}^2} - \sum{(k + 1)X_{i}} + \sum{\left (\frac{k + 1}{2} \right)^2}\right]$

$\displaystyle = \frac{(k + 1)(2k + 1)}{6} - \frac{(k + 1)^2}{2} + \frac{(k + 1)^2}{4}$

$\displaystyle = \frac{(k + 1)(2k + 1)}{6} - \frac{(k + 1)^2}{4}$

$\displaystyle = \frac{k + 1}{12} (4k + 2 - 3k - 3)$

$\displaystyle = \frac{k^2 - 1}{12}$

A multiple choice examination has

$5$ questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get

$4$ or more correct answers just by guessing is

0%
$\dfrac{13}{3^{5}}$
0%
$\dfrac{11}{3^{5}}$
0%
$\dfrac{10}{3^{5}}$
0%
$\dfrac{17}{3^{5}}$

Explanation

Total number of questions

$n=5$
Probability of correct answer

$p=\displaystyle \frac{1}{3}$
Probability of incorrect answer

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

Required probablility

$=P(X\ge 4)$

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

$\displaystyle =^{ 5 }{ C }_{ 4 }{ (\frac { 1 }{ 3 } ) }^{ 4 }{ (\frac { 2 }{ 3 } ) }^{ 1 }+^{ 5 }{ C }_{ 5 }{ (\frac { 1 }{ 3 } ) }^{ 5 }{ (\frac { 2 }{ 3 } ) }^{ 0 }$

$\displaystyle =\frac { 10 }{ { 3 }^{ 5 } } +\frac { 1 }{ { 3 }^{ 5 } }$

$\displaystyle =\frac { 11 }{ { 3 }^{ 5 } }$

Numbers are selected at random, one at a time, from the two digit numbers

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

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

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

$3$ times is

0%
$\dfrac{96}{(25)^{4}}$
0%
$\dfrac{97}{(25)^{4}}$
0%
$\dfrac{95}{(25)^{4}}$
0%
$\dfrac{94}{(28)^{4}}$

Explanation

Let's call the occurrence of

$E$ at least

$3$ times to be

$S$

$E$ will occur when one of

${29,36,63,92}$ is selected.

$P\left( E \right) =\dfrac { 4 }{ 100 } \quad and\quad P\left( \overline { E } \right) =1-\dfrac { 4 }{ 100 } =\dfrac { 96 }{ 100 }$
Note that replacement is allowed
Hence,

$P\left( S \right) ={ P\left( E \right) }^{ 4 }+\left( \begin{matrix} 4 \\ 1 \end{matrix} \right) \times { P\left( E \right) }^{ 3 }P\left( \overline { E } \right) \\ \Rightarrow P\left( S \right) =\dfrac { 256 }{ { 100 }^{ 4 } } +\dfrac { 256\times 96 }{ { 100 }^{ 4 } } =\dfrac { 97 }{ { 25 }^{ 4 } }$

$3$ % of electric bulbs manufactured by a company are defective, the probability that in a sample of

$100$ bulbs exactly five are defective is

0%
$\displaystyle \frac{e^{-0.03}(0.03)^{5}}{ 5!}$
0%
$\displaystyle \frac{e^{-0.03}(0.3)^{5}}{ 5!}$
0%
$\displaystyle \frac{e^{-3}3^{5}}{ 5!}$
0%
$\displaystyle \frac{e^{-0.03}3^{5}}{ 5!}$

Explanation

By using Poisson distribution,

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

$\mu$ : The mean number of successes that occur in a specified region.

$x:$ The actual number of successes that occur in a specified region.

$P(x;$

$\mu$ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is

$\mu$ .

$\mu = 100 \times 0.03 = 3$

$x = 5$

$P(5; 3)=\dfrac { { e }^{ -3 }{ 3 }^{ 5 } }{ 5! }$

N coins, each with probability

$p$ of getting head are tossed together. In the options q = 1-p, The probability of getting odd number of heads is

0%
$\displaystyle \frac{1-(q-p)^{n}}{2}$
0%
$\displaystyle \frac{1+(q-p)^{n}}{2}$
0%
$\displaystyle \frac{(q-p)^{n}}{2}$
0%
$\displaystyle \frac{(q+p)^{n}}{2}$

Explanation

In the expansion of

$(x+a)^n$

if the sum of odd - term be 'P' and sum of even be'Q'

$(x+a)^n=x^n+^nC_1x^{n-1}a+^nC_2x^{n-2}a^2+^nC_3x^{n-3}a^3+...........+^nC_na^n$

$=(n^n+^nC_2x^{n-2}a^2+........)+(^C_1x^{n-1}a+^nC_3x^{n-3}a^3+.........)$

$(x+a)^n=P+Q -------> (1)$ and

$(x-a)^n$ =

$x^n-^nC_1x^{n-1}a+^nC_2x^{n-2}a^2-^nC_3x^{n-3}a^3 +......+(-1)^{n \,n}C_na^n$

$=(x^n+^nC_2x^n-2a^2+......)-(^nC_1x^{n-1}a+^nC_3x^{n-3}a^3+........)$

$(x-a)^n$ =

$P-Q$ ----------> (2)

Hence sum of t add terms =P=

$(x+a)^n -(x-a)^n$

Probability of getting head = p
Probability of getting tail = 1-p =q
Now, P(odd no. of heads in N tosses = P(1 head)+P(3 heads) +P(5 heads)+.....

$= ^NC_1pq^{N-1}+^NC_3p^3q^{N-3}+^NC_5p^5q^{N-5}+. .$

$= \dfrac{(p+q)^N-(p-q)^N}{2}$
Using the property of binomial theorem
as shown in the image below

since

$p+q =1$

$=\dfrac{1-(p-q)^N}{2}$

There are

$500$ boxes each containing

$1000$ ballot papers for election. The chance that a ballot paper is defective is

$0.002$ . Assuming that the number of defective ballot papers follow Poisson distribution, the number of boxes containing at least one defective ballot paper given that

$e^{-2}=0.1353$ is

0%
$216$
0%
$432$
0%
$648$
0%
$234$

Explanation

Here

$\lambda = .002\times 1000=2$
Thus probability that a box contain at least one defective ballot

$=1-P(X=0)=1-e^{-2}=1-0.1353=.8647$
Hence number of box out of

$500$ which contain at least one defective ballot is

$=.8647\times 500 =432$

A telephone switch board receiving number of phone calls follows Poisson distribution with parameter

$3$ per hour. The probability that it receives

$5$ calls in

$2$ hours duration is

0%
$\displaystyle \frac{e^{-3}3^{5}}{ 5!}$
0%
$\displaystyle1-\frac{e^{-3}3^{5}}{ 5!}$
0%
$\displaystyle \frac{e^{-6}3^{5}}{ 5!}$
0%
$\displaystyle 1-\frac{e^{-6}3^{5}}{5!}$

Explanation

By using Poisson distribution,

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

$\mu$ : The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x;

$\mu$ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is

$\mu$
Poisson distribution with parameter 3 per hour. The probability that it receives 5 calls in 2 hours duration so,

$\mu = 3\times 2 = 6$ and

$x = 5$

$P(x;\mu)=\dfrac { { e }^{ -6} 3^{ 5} }{5!}$

The incidence of an occupational disease to the workers of a factory is found to be

$\displaystyle \frac{1}{5000}$ . If there are

$10,000$ workers in a factory then the probability that none of them will get the disease is

0%
$e^{-1}$
0%
$e^{-2}$
0%
$e^{3}$
0%
$e^{4}$

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, then the value of P is

0%
$\dfrac{50}{100}$
0%
$\dfrac{51}{101}$
0%
$\dfrac{52}{101}$
0%
$\dfrac{53}{101}$

Explanation

Assume total

$2k$ coins are tossed

Let

$X =$ Number of heads occuring

$P(X=k)\ =\ (\begin{matrix} 2k \\ k \end{matrix}){ p }^{ k }{ (1-p) }^{ k }\\ P(X=k+1)\ =\ (\begin{matrix} 2k \\ k+1 \end{matrix}){ p }^{ k+1 }{ (1-p) }^{ k-1 }\\$

Given,

$\\ P(X=k)\ =\ P(X=k+1)\\ (\begin{matrix} 2k \\ k \end{matrix}){ p }^{ k }{ (1-p) }^{ k }\ =\ (\begin{matrix} 2k \\ k+1 \end{matrix}){ p }^{ k+1 }{ (1-p) }^{ k-1 }\\ \dfrac { k+1 }{ k } (1-p)\ \ =\ p\\ p\ =\ \dfrac { k+1 }{ 2k+1 }$

Here

$\ k=50$

Therefore

$\ P\ =\ \dfrac { 50+1 }{ 2(50)+1 } =\dfrac { 51 }{ 101 } \\$

The probability that atmost

$5$ defective fuses will be found in a box of

$200$ fuses, if experience shows that

$20 \%$ of such fuses are defective, is

0%
$\displaystyle \frac{e^{-40}40^{5}}{ 5!}$
0%
$\displaystyle \sum_{x=0}^{5}\frac{e^{-40}40^{x}}{ x!}$
0%
$\displaystyle \sum_{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$
0%
$1-\displaystyle \sum_{x=6}^{\infty}\frac{e^{-40}40^{x}}{ x!}$

Explanation

${\textbf{Step - 1: Finding mean}}$

${\text{It is given that there are total 200 bulbs,}}$

$\Rightarrow {\text{ n = 200}}$

${\text{Also, 20% of given bulbs are effective,}}$

$\Rightarrow {\text{ Probability of defective bulbs, p = }}\dfrac{{{\text{20}}}}{{{\text{100}}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{5}}}$

${\text{We know that, mean = np}}$

$\therefore {\text{ Mean, m = 200 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{5}}}$

$\Rightarrow {\text{ m = 40}}$

${\textbf{Step - 2: Finding probability that atmost 5 bulbs will be effective}}$

${\text{According to Poisson theorem, P(X = x) = }}\dfrac{{{{\text{e}}^{{\text{ - m}}}}{{\text{m}}^{\text{x}}}}}{{{\text{x!}}}}{\text{, where m is the mean}}$

${\text{We have to find P(X}} \leqslant {\text{5),}}$

$\therefore {\text{ P(X}} \leqslant {\text{5) = P(X = 0) + P(X = 1) + P(X = 2) + p(X = 3) + P(X = 4) + p(X = 5)}}$

$\Rightarrow {\text{ P(X}} \leqslant {\text{5) = }}\sum\limits_{{\text{x = 0}}}^{\text{5}} {\dfrac{{{{\text{e}}^{{\text{ - 40}}}}{{{\text{(40)}}}^{\text{x}}}}}{{{\text{x!}}}}}$

$\textbf{Hence option B is correct}$

Six unbiased coins are tossed

$6400$ times. Using Poisson distribution, the approximate probability of getting six heads

$2$ times is

0%
$\displaystyle \frac{e^{-64}(64)^{2}}{ 2!}$
0%
$\displaystyle \frac{e^{-100}(100)^{2}}{ 2!}$
0%
$1-\displaystyle \frac{e^{-100}(100)^{x}}{ x!}$
0%
$\displaystyle \frac{e^{-100}(100)^{x}}{x!}$

Explanation

Here

$\lambda = \left(\cfrac{1}{2}\right)^6\times 64100=100$
Hence probability of getting six heads two times

$=P(X=2)=\cfrac{e^{-100}(100)^2}{2!}$

In a big city,

$5$ accidents take place over a period of

$100$ days. If the numebr of accidents follows P.D., the probability that there will be

$2$ accidents in a day is

0%
$\displaystyle \frac{e^{-5}5^{2}}{ 2!}$
0%
$\displaystyle \frac{e^{-05}5^{2}}{ 2!}$
0%
$\displaystyle \frac{e^{-005}(0.05)^{2}}{ 2!}$
0%
$\displaystyle \frac{e^{5}5^{2}}{ 2!}$

Explanation

By using Poisson distribution,

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

$\mu\ :$ The mean number of successes that occur in a specified region.

$x\ :$ The actual number of successes that occur in a specified region.

$P(x;\mu )\ :$ The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is

$\mu = \dfrac {5}{100} = 0.05$

$x = 2$

$P(2; 005)=\dfrac { { e }^{ -0.05 }{ 0.05 }^{ 2 } }{ 2! }$

$2$ % of pipes manufactured by a company are defective, the probability that in a sample of

$1000$ pipes, exactly

$6$ pipes are defective is (Assume Poisson distribution for the result)

0%
$\displaystyle \frac{e^{-2}2^{6}}{ 6!}$
0%
$\displaystyle \frac{e^{-20}(20)^{6}}{ 6!}$
0%
$e^{-20}$
0%
$e^{-2}$

Explanation

By using Poisson distribution,

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

$\mu$ : The mean number of successes that occur in a specified region.

$x$ : The actual number of successes that occur in a specified region.

$P(x ; \mu )$ : The Poisson probability that exactly

$x$ successes occur in a Poisson experiment, when the mean number of successes is

$\mu$ .

$\mu = 1000 \times 0.02 = 20$

$x = 6$

$P(6; 20)=\dfrac { { e }^{ -20 }{ 20 }^{ 6 } }{ 6! }$

A bag contains a very large number of white and black marbles in the ratio 1 :Two samples of marbles, 5 each are picked up. The probability that the first sample contains exactly one black and the second exactly three marbles is

0%
$\dfrac{10}{3^{9}}$
0%
$\dfrac{1600}{3^{10}}$
0%
$\dfrac{800}{3^{10}}$
0%
$\dfrac{10}{3^{5}}$

Explanation

Since there are large no. of marbles present, therefore we will not consider any change in bag even after drawing 10 marbles

Probability of drawing black marble = 2/3

Probability of drawing white marble = 1/3

Required Probability =

$^{5}C_{1}$ *

$({ \dfrac { 2 }{ 3 } ) }^{ 1 }\times {( \dfrac { 1 }{ 3 } ) }^{ 4 }$ *

$^{5}C_{2}$ *

${ (\dfrac { 2 }{ 3 } ) }^{ 2 }\times { (\dfrac { 1 }{ 3 }) }^{ 3 }$

In a precision bombing attack, there is a

$50$ % chance that any one bomb will strike the target. Two direct hits are required to destroy the target completely. The number of bombs which should be dropped to give a 99% chance or better of completely destroying the target can be

0%
$12$
0%
$11$
0%
$10$
0%
$13$

Explanation

The probability of success in one strike is

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

$\Rightarrow$ Probability of failure

$= q = \dfrac{1}{2}$
Now, probability of

$r$ successes

$P(x = r) = ^nC_r. \left(\dfrac{1}{2} \right)^{r}. \left(\dfrac{1}{2} \right)^{n - r} = ^nC_r \left (\dfrac{1}{2} \right)^n$
According to the given condition,

$P(x \geq 2) \geq 0.99$

$\Rightarrow 1 - P(x < 2) \geq 0.99$

$\Rightarrow 1 - P(x = 0) - P(x = 1) \geq 0.99$

$\Rightarrow 1 - 0.99 \geq \dfrac{1 + n}{2^n}$

$\Rightarrow 2^n \geq 100 + 100n$
For

$n =10$ ,

$2^n < 100+100n$
For

$n =11,12,13$ ,

$2^n > 100+100n$ .

A company knows on the basis of past experience that

$2$ % of the blades are defective. The probability of having 3 defective blades in a sample of

$100$ blades is

0%
$e^{-2}2^{2}$
0%
$\displaystyle \frac{e^{-2}2^{3}}{3!}$
0%
$\displaystyle \frac{e^{-2}2^{3}}{ 2!}$
0%
$\displaystyle \frac{e^{-4}2^{-1}}{ 2!}$

Explanation

By using Poisson distribution,

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

$\mu$ : The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x;

$\mu$ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is

$\mu$
Here,

$\mu = 100 \times 0.02 = 2$

$x = 3$ (defective blades)

$P(3;2)=\dfrac { { e }^{ -2 }{ 2 }^{3 } }{ 3! }$

A car hire firm has

$2$ cars which it hires out day by day. If the number of demands for a car on each day follows poisson distribution with parameter

$1.5$ , then the probability that neither car is used is

0%
$e^{-1.5}$
0%
$1.5\times e^{-1.5}$
0%
$1-2.5\times e^{-1.5}$
0%
$1-1.5\times e^{-1.5}$

Explanation

By using Poisson distribution,

$P(x;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ x } }{ x! }$

$\mu$ : The mean number of successes that occur in a specified region(parameter)
x: The actual number of successes that occur in a specified region.
P(x;

$\mu$ ): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is

$\mu$
Here,

$\mu = 1.5$ and

$x = 0$ (neither car is used)

$P(0;1.5)=\dfrac { { e }^{ -1.5 }{ 1.5 }^{0 } }{ 0! } = {e}^{-1.5}$

$X$ and

$Y$ are independent binomial variates

$B(5,1/2)$ and

$B(7,1/2)$ , then find the value of

$P(X+Y=3)$ .

0%
$P(X+Y=3)=\cfrac{45}{1024}$
0%
$P(X+Y=3)=\cfrac{65}{1024}$
0%
$P(X+Y=3)=\cfrac{55}{1024}$
0%
$P(X+Y=3)=\cfrac{75}{1024}$

Explanation

For

$X=0$ and

$Y=3$ ,
we get

$P=\:^{5}C_{0}\dfrac{1}{2}^{5}.\:^{7}C_{3}\dfrac{1}{2}^{7}$

$=\dfrac{35}{4096}$ ...(i)
For

$X=1$ and

$Y=2$ ,
we get

$P=\:^{5}C_{1}\dfrac{1}{2}^{5}.\:^{7}C_{2}\dfrac{1}{2}^{7}$

$=\dfrac{105}{4096}$ ...(ii)
For

$X=2$ and

$Y=1$ ,
we get

$P=\:^{5}C_{2}\dfrac{1}{2}^{5}.\:^{7}C_{1}\dfrac{1}{2}^{7}$

$=\dfrac{70}{4096}$ ...(iii)
For

$X=3$ and

$Y=0$ ,
we get

$P=\:^{5}C_{3}\dfrac{1}{2}^{5}.\:^{7}C_{0}\dfrac{1}{2}^{7}$

$=\dfrac{10}{4096}$ ...(iv)
Adding i ii, iii and iv.
we get

$\dfrac{35+105+70+10}{4096}$

$=\dfrac{220}{4096}$

$=\dfrac{55}{1024}$

A die is thrown

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

$4$ times?

0%
$\cfrac{1}{2}$
0%
$\cfrac{35}{128}$
0%
$\cfrac{37}{128}$
0%
$\cfrac{63}{128}$

Explanation

P(odd)

$=$ P(even)

$=$

$\dfrac{1}{2}$
Now considering that getting an odd number is a success.
Hence In 7 trials, getting exactly four odd numbers is

$=\:^{7}C_{4}\left (\dfrac{1}{2} \right )^{4}.\left (\dfrac{1}{2} \right)^{3}$

$=\dfrac{\:^{7}C_{4}}{2^{7}}$

$=\dfrac{35}{128}$ .

A can take a step forward with probability

$0.4$ and backward with probability

$0.6$ . find the probability that at the end of eleven steps he is one step away from the starting point.

0%
probability $=0.57$
0%
probability $=0.63$
0%
probability $=0.43$
0%
probability $=0.37$

Explanation

Probability of going forward

$=P(F)=0.4$
Probability of going backward

$=P(B)=0.6$

After

$11$ steps if he has to be one step ahead of starting point,then he must take in total

$6$ steps ahead and

$5$ backwards. This can be done in

${ ^{ 11 }{ C }_{ 6 } }$ ways.
Hence,

$P(Forward)={ ^{ 11 }{ C }_{ 6 } }{ P(F) }^{ 6 }{ P(B) }^{ 5 }$

After

$11$ steps if he has to be one step behind the starting point,then he must take in total

$5$ steps ahead and

$6$ backwards. This can be done in

${ ^{ 11 }{ C }_{ 5 } }$ ways.
Hence,

$P(Backward)={ ^{ 11 }{ C }_{ 5 } }{ P(F) }^{ 5 }{ P(B) }^{ 6 }$

Total probability

$= P(Forward)+P(Backward)=0.37$

In a series of n independent trials for an event of constant probability p, the most probable number r of successes is given by

$\left ( n+1 \right )p-1< r< \left ( n+1 \right )p$ . Hence, the most probable number of successes is the integral part of

$\left ( n+1 \right )p$ . But if

$\left ( n+1 \right )p$ is an integer, the chance of r successes is equal to that of

$r+1$ successes and both

$r,r+1$ are most probable numbers of successes.
A bag contains 2 white balls and 1 black ball. A ball is drawn at random and returned to the bag.

The experiment is done 10 times. The probability that a white ball is drawn exactly 5 times is

0%
$\dfrac{10!}{5!5!}\left ( \dfrac{2}{3} \right )^5$
0%
$\dfrac{10!}{5!5!}\left ( \dfrac{1}{3} \right )^5$
0%
$\dfrac{10!}{\left(5! \right)^2}\left(\dfrac{2}{9} \right )^5$
0%
$\dfrac{10!}{5!}\left(\dfrac{2}{9} \right )^5$

Explanation

We need to select which 5 of the 10 are white. Also, probability of a white is

$\dfrac { 2 }{ 3 }$ and that of a black is

$\dfrac { 1 }{ 3 }$
Probability of choosing a white ball 5 times out of 10 is

$_{ 5 }^{ 10 }{ C }*({ \dfrac { 2 }{ 3 } ) }^{ 5 }*{ (\dfrac { 1 }{ 3 } ) }^{ 5 }=\dfrac { 10! }{ { (5!) }^{ 2 } } *{ (\dfrac { 2 }{ 9 } ) }^{ 5 }$
Hence, (c) is correct.

$n$ different apples are to be distributed among

$m$ children, find the chance that a particular child recieves

$p$ apples.

0%
$\cfrac { { _{ }^{ n }{ C } }_{ p }{ m }^{ n-p } }{ { m }^{ n } }$
0%
$\cfrac { { _{ }^{ n }{ C } }_{ p }{ (m+1) }^{ n-p } }{ { m }^{ n } }$
0%
$\cfrac { { _{ }^{ n }{ C } }_{ p }{ (m-1) }^{ n-p } }{ { m }^{ n } }$
0%
$\cfrac { { _{ }^{ n }{ C } }_{ p }{ (m-2) }^{ n-p } }{ { m }^{ n } }$

Explanation

Total ways of distributing

$n$ apples to

$m$ children =

$m^n$
Ways of giving

$p$ apples to particular child = Select

$p$ apples out of

$n$ apples

$\times$ Total ways of distributing

$(n-p)$ apples to

$(m-1)$ children
=

${ _{ }^{ n }{ C } }_{ p }{ (m-1) }^{ n-p }$
so,

$P = \cfrac { { _{ }^{ n }{ C } }_{ p }{ (m-1) }^{ n-p } }{ { m }^{ n } }$

From a box containing

$20$ tickets of value

$1$ to

$20$ , four tickets are drawn one by one. After each draw, the ticket is replaced. The probability that the largest value of tickets drawn is

$15$ is

0%
$\displaystyle { \left( \frac { 3 }{ 4 } \right) }^{ 4 }$
0%
$\displaystyle \frac { 27 }{ 320 }$
0%
$\displaystyle \frac { 27 }{ 1280 }$
0%
none of these

Explanation

The probability of drawing a number less than or equal to

$15$ in a draw

$\displaystyle =\frac { 15 }{ 20 } =\frac { 3 }{ 4 }$
The probability of drawing the ticket of value

$15$ in a draw

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

$\therefore$ The required probability

$\displaystyle =^{ 4 }{ { C }_{ 1 } }\times \frac { 1 }{ 20 } \times { \left( \frac { 3 }{ 4 } \right) }^{ 3 }=\frac { 27 }{ 320 }$

Let

$A$ be a set containing

$n$ elements. A subset

$P$ of the set

$A$ is chosen at random.
The set

$A$ is reconstructed by replacing the element of

$P$ , and another subsets

$Q$ of

$A$ is chosen at random. The probability that

$\left( P\cap Q \right)$ contains exactly

$m(m<n)$ elements is

0%
$\displaystyle \frac { { 3 }^{ n-m } }{ { 4 }^{ n } }$
0%
$\displaystyle \frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ m } }{ { 4 }^{ n } }$
0%
$\displaystyle \frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ n-m } }{ { 4 }^{ n } }$
0%
none of these

Explanation

We know that the number of subsets of a set containing

$n$ elements is

${ 2 }^{ n }$ .
Therefore, the number of ways of choosing

$P$ and

$Q$ is

$^{ { 2 }^{ n } }{ { C }_{ 1 } }\times _{ }^{ { 2 }^{ n } }{ { C }_{ 1 } }={ 2 }^{ n }\times { 2 }^{ n }={ 4 }^{ n }$ .
Out of

$n$ elements,

$m$ elements can be chosen in

$^{ n }{ { C }_{ m } }$ ways.
If

$\left( P\cap Q \right)$ contains exactly in elements,

then from the remaining

$n-m$ elements either an element belongs to

$P$ or

$Q$ but not both

$P$ and

$Q$ .
Suppose

$P$ contains

$r$ elements from the remaining

$n-m$ elements.
Then

$Q$ may contain any number of elements from the remaining

$(n-m)-r$ elements.
Therefore,

$P$ and

$Q$ can be chosen in

$^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r }$
But

$r$ can vary from

$0$ to

$(n-m)$ . So,

$P$ and

$Q$ can be chosen in general in

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

$\displaystyle =\dfrac { ^{ n }{ { C }_{ m } }\times { 3 }^{ n-m } }{ { 4 }^{ n } }$

An unbiased 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

0%
$\displaystyle \frac { 16 }{ 81 }$
0%
$\displaystyle \frac { 1 }{ 81 }$
0%
$\displaystyle \frac { 80 }{ 81 }$
0%
$\displaystyle \frac { 65 }{ 81 }$

Explanation

Let

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

$2$ and not more than

$5$ in a single throw of die

$\displaystyle =\frac { 4 }{ 6 } =\frac { 2 }{ 3 }$

$n=$ number of times dice is rolled

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

$2$ and not more than

$5$ .
Then,

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

$\therefore$ Required probability

$\displaystyle =P\left( X=4 \right) =_{ }^{ 4 }{ { C }_{ 4 } }.{ p }^{ 4 }={ \left( \frac { 2 }{ 3 } \right) }^{ 4 }=\frac { 16 }{ 81 }$

A die is thrown

$(2n+1)$ times,

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

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

Explanation

Probability of showing even number in a throw

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

$\therefore$ Required probability

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

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

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

Suppose the probability for A to win a game against B is 0.If A has an option of playing either "best of 3 games" or a "best of 5 games" match against B, which option should A choose so that the probability of his winning the match is higher ? (No game ends in draw)

0%
best of 3 games
0%
best of 5 games
0%
same probability in both
0%
cannot win

Explanation

The probability

${ p }_{ 1 }$ or winning the best of three games is = the probability of winning two games + the probability of inning three games
=

$_{ }^{ 3 }{ { C }_{ 2 } }\left( 0.6 \right) { \left( 0.4 \right) }^{ 2 }+^{ 3 }{ { C }_{ 3 } }{ \left( 0.4 \right) }^{ 3 }$
Similarly the probability of winning the best five games

${ p }_{ 2 }$ = the probability of winning three games + the probability of winning 5 games

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

${ p }_{ 1 }=0.288+0.064=0.352$
and

${ p }_{ 2 }=0.2304+0.0768+0.01024=0.31744$
As

${ p }_{ 1 }>{ p }_{ 2 }$
Therefore A must choose the best of three games

Suppose the probability for

$A$ to win a game against

$B$ is

$0.4$ . If

$A$ has an option playing either a "best of

$3$ games" or a "best of 5 games" match against

$B$ , which option should

$A$ choose so that the probability of his winning the match is higher? (No game ends in a draw).

0%
Best of $5$ games
0%
Best of $3$ games
0%
Either best of $3$ games or best of $5$ games
0%
None of the above

Explanation

Case 1:

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,

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

$\therefore { P }_{ 1 }=$ Probability of wining the best of

$3$ games.

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

$=\ ^{ 3 }{ { C }_{ 2 } }{ \left( 0.4 \right)}^{ 2 }{ \left( 0.6 \right)}^{ 1 }{ + }\ ^{ 3 }{ { C }_{ 3 } }{ \left( 0.4 \right) }^{ 3 }{ \left( 0.6 \right) }^{ 0 }\\ =0.288+0.064=0.352$

Case 2:

When

$A$ plays

$5$ games against

$B$ .

In this case, we have

$n=5,p=0.4$ and

$q=0.6$
Let

$X$ denotes the number of wins in

$5$ games.
Then,

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

$\therefore { P }_{ 2 }=$ Probability of winning the best of

$5$ games

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

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

$=0.2304+0.0768+0.1024=0.31744$

Clearly

${ P }_{ 1 }>{ P }_{ 2 }$ .

Therefore, first option i.e. 'best

$3$ games' has higher probability of winning the match.

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 9 - 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 9 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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