MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 4

For a binomial distribution if

$P=\dfrac{2}{3}$ and

$n=10$ the probability of mode is

0%
$^{10}C_{4}\left ( \dfrac{1}{3} \right )^{7}\left ( \dfrac{2}{3} \right )^{3}$
0%
$^{10}C_{7}\left ( \dfrac{2}{3} \right )^{7}\left ( \dfrac{1}{3} \right )^{3}$
0%
$\left ( \dfrac{2}{3} \right )^{7}\left ( \dfrac{1}{3} \right )^{3}$
0%
$\left ( \dfrac{1}{3} \right )^{7}\left ( \dfrac{2}{3} \right )^{3}$

Explanation

$(n+1)(p)$

$=(10+1)(\dfrac{2}{3})$

$=\dfrac{22}{3}$
Hence

$(n+1)(p)$ is non-integral.
Therefore
mode

$=[\dfrac{22}{3}]$ ; where

$[x]$ denotes greatest integer function.
mode

$=[7.33]$

$=7$
Hence the require value is

$\:^{10}C_{7}(\dfrac{2}{3})^{7}(\dfrac{1}{3})^{3}$

$8$ unbiased coins are tossed at a time

$512$ times.The expected frequency of getting one head is

0%
$4$
0%
$8$
0%
$16$
0%
$20$

Explanation

The probability of one head, when 8 unbiased coins are tossed.

$\:^{8}C_{1}(\dfrac{1}{2})^{1}(\dfrac{1}{2})^{7}$

$=\dfrac{8}{256}$
Hence the expected frequency is

$=512(\dfrac{8}{256})$

$=2(8)$

$=16$

Out of

$2000$ families with

$4$ children, the number of families you expect to have at least one boy is

0%
$1875$
0%
$750$
0%
$1250$
0%
$625$

Explanation

The probability of

$4$ families having at-least one boy will be

$1-\:^{4}C_{0}(\dfrac{1}{2})^{4}$

$=1-\dfrac{1}{16}$

$=\dfrac{15}{16}$
Hence the expected frequency for

$2000$ families is

$=2000(\dfrac{15}{16})$

$=125(15)$

$=1875$

Out of

$2000$ families with

$4$ children, the number of families you expect to have two male children

0%
$1875$
0%
$750$
0%
$1250$
0%
$625$

Explanation

The probability of

$4$ families having two boys will be

$\:^{4}C_{2}(\dfrac{1}{2})^{4}$

$=\dfrac{6}{16}$
Hence the expected frequency for

$2000$ families is

$=2000(\dfrac{6}{16})$

$=125(6)$

$=750$

If for a

$B.D.$ ,

$P=\dfrac{2}{3}$ ,

$n = 8$ , then the probability of mode is

0%
$^{8}C_{6}\left ( \dfrac{2}{3} \right )^{6}\left ( \dfrac{1}{3} \right )^{2}$
0%
$^{8}C_{4}\left ( \dfrac{2}{3} \right )^{4}\left ( \dfrac{1}{3} \right )^{4}$
0%
$^{10}C_{6}\left ( \dfrac{2}{3} \right )^{6}\left ( \dfrac{1}{3} \right )^{4}$
0%
$^{10}C_{4}\left ( \dfrac{2}{3} \right )^{4}\left ( \dfrac{1}{3} \right )^{4}$

$201$ coins each with probability

$P(0<P<1)$ of showing head are tossed together. If the probability of getting

$100$ heads is equal to the probability of getting

$101$ heads, then the value of

$P$ is

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

Explanation

Probability of

$100$ heads and

$101$ tails

$=$ probability of

$101$ heads and

$100$ tails (out of

$201$ coins)
Thus,

$^{ 201 } C _{101}*{ P}^{ 101 }*{ (1-P) }^{ 100 }=^{ 201 }C_{100}*{ P }^{ 100 }*{ (1-P) }^{ 101 }\quad =>\quad P=1-P\quad =>\quad P=\dfrac { 1 }{ 2 }$

$x$ is a binomial variable with

$P=\dfrac{1}{4}$ , then the smallest value of

$n$ so that

$P(x\geq 1)> 0.70$ is

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

For

$P(x\geq1)$
That is at-least

$1$ .

$p=1-(\dfrac{3}{4})^{n}$

$1-(\dfrac{3}{4})^{n}=0.7$

$0.3=(\dfrac{3}{4})^{n}$

$n=\dfrac{ln(3)}{ln(4)-ln(3)}$

$n=4.18$
For a probability

$>0.7$ , the least value of

$n$ has to be

$5$ .

Twenty coins each with probability

$P(0<P<1)$ of showing head are tossed together. If the probability of getting

$10$ heads is equal to the probability of getting

$11$ heads, the value of

$P$ is

0%
$\dfrac{9}{20}$
0%
$\dfrac{11}{20}$
0%
$\dfrac{11}{21}$
0%
$\dfrac{9}{21}$

Explanation

$P(X=n)=(\begin{matrix} 20 \\ n \end{matrix})*{ p }^{ n }{ (1-p) }^{ 20-n }\\ (\begin{matrix} 20 \\ 10 \end{matrix})*{ p }^{ 10 }{ (1-p) }^{ 10 }=(\begin{matrix} 20 \\ 11 \end{matrix})*{ p }^{ 11 }{ (1-p) }^{ 9 }\\ \dfrac { 11(1-p) }{ 10 } =p\\ p=\dfrac { 11 }{ 21 }$

In a market region half of the households is known to use a particular brand of soap. In a household survey, a sample of

$10$ house holds are alloted to each investigator and

$2048$ investigators are appointed for the survey. The number of investigators likely to report that there are at least

$4$ users is

0%
$240$
0%
$352$
0%
$1696$
0%
$120$

Explanation

$E\{ X=n\} =(\begin{matrix} 10 \\ n \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ n }*{ \dfrac { 1 }{ 2 } }^{ 10-n }*(no.\quad of\quad investigator)\\ E\{ X=0\} +E\{ X=1\} +E\{ X=2\} +E\{ X=3\} = \\(\begin{matrix} 10 \\ 0 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 0 }*{ \dfrac { 1 }{ 2 } }^{ 10 }*2048+(\begin{matrix} 10 \\ 1 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 1 }*{ \dfrac { 1 }{ 2 } }^{ 9 }*2048(\begin{matrix} 10 \\ 2 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 2 }*{ \dfrac { 1 }{ 2 } }^{ 8 }*2048+(\begin{matrix} 10 \\ 3 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 3 }*{ \dfrac { 1 }{ 2 } }^{ 7 }*2048\\ E\{ X=0\} +E\{ X=1\} +E\{ X=2\} +E\{ X=3\} =352\\ So\quad for\quad atleast\quad 4\quad users\quad =\quad 1-\quad atmost\quad 3\\ =2048-352\\ =1696$

If we take

$1280$ sets each of

$10$ tosses of a fair coin, the number of sets we expect to get

$7$ heads and

$3$ tails is

0%
$450$
0%
$300$
0%
$150$
0%
$75$

Explanation

$P$ (head with a fair coin) is

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

$P$ (tail with a fair coin) is

$\displaystyle q=\frac { 1 }{ 2 }$
Probability of

$7$ heads and

$3$ tails in

$10$ tosses of a fair coin is equal to

$\displaystyle_{ }^{ 10 }{ { C }_{ 7 }{ \left( p \right) }^{ 7 }{ \left( q \right) }^{ 3 }=\frac { 10! }{ \left( 10-7 \right) !7! } { \left( \frac { 1 }{ 2 } \right) }^{ 7 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 } }=\frac { 120 }{ 1024 }$
Thus, out of

$1280$ sets, each of

$10$ tosses of a fair coin we expect

$7$ heads and

$3$ tails in

$\displaystyle1280\times \left( \frac { 120 }{ 1024 } \right) =150$ sets

In a market region half of the households is known to use a particular brand of soap. In a house hold survey, a sample of

$10$ house holds are alloted to each investigator and

$2048$ investigators are appointed for the survey. The number of investigators likely to report that there are three households is

0%
$240$
0%
$352$
0%
$1696$
0%
$120$

Explanation

$E\{ X=n\} =(\begin{matrix} 10 \\ n \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ n }*{ \dfrac { 1 }{ 2 } }^{ 10-n }*(no.\quad of\quad investigator)\\ E\{ X=3\} =(\begin{matrix} 10 \\ 3 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 3 }*{ \dfrac { 1 }{ 2 } }^{ 7 }*2048\\ E\{ X=3\} =240$

An irregular six faced die is thrown and the expectation that in

$10$ throws it will give

$5$ even numbers is twice the expectation that it will give

$4$ even numbers. The number of times in

$10000$ sets of

$10$ throws you expect to give no even numbers is

0%
$6$
0%
$1$
0%
$2$
0%
$9$

Explanation

Let the random variable

$X$ denote the number of even numbers.

$P$ (getting x even numbers)

$=P(X=x)$

$^{ n }{ { C }_{ x } }{ p }^{ x }{ q }^{ 10-x },x=0,1,2,...,n$
Given

$n=10$

$\therefore P(X=x)=^{ 10 }{ { C }_{ x } }{ p }^{ x }{ q }^{ 10-x },x=0,1,2,...10$

$P$ (getting

$5$ even numbers)

$=2P$ (getting

$4$ even numbers)

$P(X=5)=2P(X=4)=^{ 10 }{ { C }_{ 5 } }{ p }^{ 5 }{ q }^{ 5 }$

$=2\left( ^{ 10 }{ { C }_{ 4 } }{ p }^{ 4 }{ q }^{ 6 } \right)$

$\displaystyle \frac { p }{ 4 } =\frac { q }{ 3 } \Rightarrow 3p=5q=5\left( 1-p \right)$

$\Rightarrow 8p=5\Rightarrow p=\dfrac { 5 }{ 8 }$

$\displaystyle\therefore q=1-p=1-\dfrac { 5 }{ 8 } =\dfrac { 3 }{ 8 }$

$\therefore P$ (getting x even numbers)

$\displaystyle^{ 10 }{ { C }_{ x } }{ \left( \frac { 5 }{ 8 } \right) }^{ x }{ \left( \frac { 3 }{ 8 } \right) }^{ 10-x },x=0,1,2,...n$

$\therefore$ the required number of times that in

$10000$ sets of

$10$ throws each we get no even number.

$\displaystyle=1000\times P(X=0)=1000\times^{ 10 }{ { C }_{ 0 } }{ \left( \frac { 5 }{ 8 } \right) }^{ 0 }{ \left( \frac { 3 }{ 8 } \right) }^{ 10 }=1$ (approx)

The largest possible variance of a binomial variate is

0%
$n$
0%
$\dfrac{n}{2}$
0%
$\dfrac{n}{4}$
0%
$\dfrac{n}{6}$

Explanation

Variance =

${ \sigma }^{ 2 } = np(1-p)$
For the largest possible variance =>

$\dfrac { d{ \sigma }^{ 2 } }{ dp } = 0$

$n [ (1-p) + p (-1) ] = 0$

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

${ \sigma }^{ 2 } = np(1-p) = n \times \dfrac {1}{2}\times ( 1-\dfrac {1}{2}) = \dfrac {n}{4}$

Suppose

$A$ and

$B$ are two equally strong table tennis players. Which of the following two events is more probable

$a)$

$A$ beats

$B$ in exactly

$3$ games out of

$4$

$b)$

$A$ beats

$B$ in exactly

$5$ games out of

$8$

0%
$a$
0%
$b$
0%
$a$ & $b$
0%
neither $a$ nor $b$

Explanation

Since

$A$ and

$B$ are equally strong players,

the probability of

$A's$ winning a game of table tennis is

$\displaystyle \frac { 1 }{ 2 }$ .

$\displaystyle \therefore p=\frac { 1 }{ 2 }$ and

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

$(\because p+q=1)$
This is a case of binomial distribution
a) Here,

$\displaystyle n=4,r=3,p=\frac { 1 }{ 2 }$ and

$\displaystyle P\left( X=r \right) =_{ }^{ 4 }{ C_{ r } }{ \left( \frac { 1 }{ 2 } \right) }^{ r }{ \left( \frac { 1 }{ 2 } \right) }^{ 4-r }$

$\displaystyle \therefore P\left( X=3 \right) =^{ 4 }{ C_{ 3 } }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }{ \left( \frac { 1 }{ 2 } \right) }^{ 1 }=4\times \frac { 1 }{ 16 } =\frac { 1 }{ 4 }$

$\therefore P(A$ wins

$3$ games out of

$4)=\displaystyle \frac { 1 }{ 4 }$
b) Now,

$\displaystyle n=8,p=\frac { 1 }{ 2 } ,q=\frac { 1 }{ 2 } ,r=5$

$\displaystyle \therefore P\left( X=5 \right) =^{ 8 }{ C_{ 5 } }{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }=\frac { 7 }{ 32 }$

$\therefore P(A$ wins

$5$ games out of

$8)=\displaystyle \frac { 7 }{ 32 }$
Since

$\displaystyle \frac { 1 }{ 4 } >\frac { 7 }{ 32 }$ therefore event (a) is more probable.

Assuming that half of the population are consumers of rice so that the chance of an individual being a consumer is

$1/2$ and assuming that

$1024$ investigators each take

$10$ individuals to see whether they are consumers, the number of investigators you expect to report that three or less are consumers is

0%
$360$
0%
$240$
0%
$176$
0%
$60$

Explanation

$E\{ X=n\} =(\begin{matrix} 10 \\ n \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ n }*{ \dfrac { 1 }{ 2 } }^{ 10-n }*(no.\quad of\quad investigator)\\ E\{ X=0\} +E\{ X=1\} +E\{ X=2\} +E\{ X=3\} = \\(\begin{matrix} 10 \\ 0 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 0 }*{ \dfrac { 1 }{ 2 } }^{ 10 }*1024+(\begin{matrix} 10 \\ 1 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 1 }*{ \dfrac { 1 }{ 2 } }^{ 9 }*1024+(\begin{matrix} 10 \\ 2 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 2 }*{ \dfrac { 1 }{ 2 } }^{ 8 }*1024+(\begin{matrix} 10 \\ 3 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 3 }*{ \dfrac { 1 }{ 2 } }^{ 7 }*1024\\ E\{ X=0\} +E\{ X=1\} +E\{ X=2\} +E\{ X=3\} =176$

The least number of times a fair coin is to be tossed in order that the probability of getting atleast one head is at least

$0.99$ is

0%
$5$
0%
$6$
0%
$7$
0%
$8$

Explanation

${\textbf{Step -1: Calculate probability of getting head and tail.}}$

${\text{When a coin is tossed, probability of getting head = }}\dfrac{1}{2}.$

${\text{Probability of getting tail = }}\dfrac{1}{2}.$

${\textbf{Step -2: Calculate probability of getting atleast one head when a coin is tossed n times.}}$

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

${\text{Probability of getting all tails in n tosses = }}{\left( {\dfrac{1}{2}} \right)^n}$

${\therefore\text{Probability of getting atleast one head in n tosses = 1}} - {\left( {\dfrac{1}{2}} \right)^n}$

${\textbf{Step -3: Solve according to question.}}$

${\text{Putting,}}$

$\Rightarrow1 - {\left( {\dfrac{1}{2}} \right)^n} \geqslant 0.99$

$\Rightarrow1 - {\left( {\dfrac{1}{2}} \right)^n} \geqslant \dfrac{{99}}{{100}}$

$\Rightarrow \dfrac{1}{{{2^n}}} \leqslant \dfrac{1}{{100}}$

$\Rightarrow {2^n} \geqslant 100$

$\Rightarrow {2^n} \geqslant 128$ $\mathbf{[\because 2^6<100<2^7,}\textbf{ and n is an integer]}$

$\Rightarrow {2^n} \geqslant 2^7$

$\Rightarrow n \geqslant 7$

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

If the difference between the mean and variance of binomial distribution for

$5$ trials is

$5/9$ , the distribution is of the form

0%
$\left( 5,\ \cfrac { 1 }{ 2 } ,\ \cfrac { 1 }{ 2 } \right)$
0%
$\left( 5,\ \cfrac { 1 }{ 3 } ,\ \cfrac { 2 }{ 3 } \right)$
0%
$\left( 5,\ \cfrac { 3 }{ 4 } ,\ \cfrac { 1 }{ 4 } \right)$
0%
$\left( 5,\ \cfrac { 3 }{ 5 } ,\ \cfrac { 2 }{ 5 } \right)$

Explanation

Difference between mean and variance

$=$

$np(1-(1-p))$

$n{ p }^{ 2 }$

$=$

$\dfrac{5}{9}$

${ p }^{ 2 }$

$=$

$1/9$ (give n=5)
Hence,

$p =$

$\dfrac { 1 }{ 3 }$

$1-p =$

$\dfrac { 2 }{ 3 }$

$(n,p,q)=(5,$

$\dfrac { 1 }{ 3 }$ ,

$\dfrac { 2 }{ 3 }$ )

A coin is tossed

$n$ times. If the probability of getting head

$6$ times is equal to the probability of getting head 8 times then

$n \: =$

0%
$6$
0%
$8$
0%
$14$
0%
$10$

Explanation

Probability of getting heads exactly 6 times in n tosses of a coin

$=\:{n}C_{6}.\dfrac{1}{2^{n}}$ .
Probability of getting heads exactly 8 times in n tosses of a coin

$=\:{n}C_{8}.\dfrac{1}{2^{n}}$ .
It is given that both are equal.
Hence

$\:{n}C_{6}.\dfrac{1}{2^{n}}=\:{n}C_{8}.\dfrac{1}{2^{n}}$
Or

$\:^{n}C_{6}=\:^{n}C_{8}$
Or

$\:^{n}C_{n-6}=\:^{n}C_{8}$

$n-6=8$

$n=14$ .
Hence no of trials are 14.

In a market region half of the households is known to use a particular brand of soap. In a household survey, a sample of

$10$ house holds are alloted to each investigator and

$2048$ investigators are appointed for the survey. The number of investigators likely to report that there are not more than

$3$ households is

0%
$240$
0%
$352$
0%
$1696$
0%
$120$

Explanation

$E\{ X=n\} =(\begin{matrix} 10 \\ n \end{matrix})*{ \frac { 1 }{ 2 } }^{ n }*{ \dfrac { 1 }{ 2 } }^{ 10-n }*(no.\quad of\quad investigator)\\ E\{ X=0\} +E\{ X=1\} +E\{ X=2\} +E\{ X=3\} = \\(\begin{matrix} 10 \\ 0 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 0 }*{ \dfrac { 1 }{ 2 } }^{ 10 }*2048+(\begin{matrix} 10 \\ 1 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 1 }*{ \dfrac { 1 }{ 2 } }^{ 9 }*2048(\begin{matrix} 10 \\ 2 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 2 }*{ \dfrac { 1 }{ 2 } }^{ 8 }*2048+(\begin{matrix} 10 \\ 3 \end{matrix})*{ \dfrac { 1 }{ 2 } }^{ 3 }*{ \dfrac { 1 }{ 2 } }^{ 7 }*2048\\ E\{ X=0\} +E\{ X=1\} +E\{ X=2\} +E\{ X=3\} =352$

A machine manufacturing screws is known to produce

$5\%$ defectives. In a random sample of

$15$ screws the probability that there are exactly

$3$ defectives is

0%
$\dfrac{16}{20}$
0%
$\dfrac{17}{20}$
0%
$^{15}C_{3}\left ( \dfrac{1}{20} \right )^{3}\left ( \dfrac{19}{20} \right )^{12}$
0%
$\dfrac{18}{20}$

Explanation

Total screws

$= 15$

Machine produce

$5\%$ defective screws, so probability of defective screws

$= p =$

$\dfrac {5}{100} = \dfrac{1}{20}$
Probability for

$k$ success is

$= ^{n}C_{k}\left ( p \right )^{k}\left ( 1-p \right )^{n-k}$
Probability that there are exactly

$3$ defectives is

$= ^{n}C_{3}\left ( p \right )^{3}\left ( 1-p \right )^{n-3}$

$=>$

$^{15}C_{3}\left ( \dfrac{1}{20} \right )^{3}\left ( 1-\dfrac{1}{20} \right )^{15-3}$

$=>$

$^{15}C_{3}\left ( \dfrac{1}{20} \right )^{3}\left ( \dfrac{19}{20} \right )^{12}$

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

$24$ and

$128$ respectively.The binomial distribution is

0%
$\left ( \dfrac{1}{2} +\dfrac{1}{2}\right )^{32}$
0%
$\left ( \dfrac{3}{10} +\dfrac{7}{10}\right )^{32}$
0%
$\left ( \dfrac{1}{50} +\dfrac{49}{50}\right )^{32}$
0%
$\left ( \dfrac{1}{3} +\dfrac{2}{3}\right )^{32}$

Explanation

Let the binomial distribution be

$(p+q)^{n}$
We have,
Mean

$=np$
Variance

$=npq$
Given

$np+npq=24$

$np(1+q)=24$

$n^{2}p^{2}(1+q)^{2}=576$ .....(1)
and

$np.npq=128$ ......(2)
Dividing (1) by (2), we get

$\dfrac{(1+q)^{2}}{q}=\displaystyle \frac{9}{2}$

$\Rightarrow 2q^2+4q+2=9q$

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

$\Rightarrow q=2$ (not possible) ,

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

$\Rightarrow p=\displaystyle \frac{1}{2}$
So, by (1), we get

$n=32$
So, the distribution is

$\displaystyle (\frac{1}{2}+\frac{1}{2})^{n}$

A coin tossed

$n$ times. If the probability that

$4, 5, 6$ heads occur are in

$A.P.$ , then

$n =$

0%
$14$
0%
$8$
0%
$15$
0%
$11$

Explanation

$P(X=r)=(\begin{matrix} n \\ r \end{matrix}){ \dfrac { 1 }{ 2 } }^{ r }{ \dfrac { 1 }{ 2 } }^{ n-r }=(\begin{matrix} n \\ r \end{matrix}){ \dfrac { 1 }{ 2 } }^{ n }\\ P(X=4)=(\begin{matrix} n \\ 4 \end{matrix}){ \dfrac { 1 }{ 2 } }^{ 4 }{ \dfrac { 1 }{ 2 } }^{ n-4 }\\ P(X=5)=(\begin{matrix} n \\ 5 \end{matrix}){ \dfrac { 1 }{ 2 } }^{ 5 }{ \dfrac { 1 }{ 2 } }^{ n-5 }\\ P(X=6)=(\begin{matrix} n \\ 6 \end{matrix}){ \dfrac { 1 }{ 2 } }^{ 6 }{ \dfrac { 1 }{ 2 } }^{ n-6 }\\ \\ P(X=5)=\dfrac { P(X=4)+P(X=6) }{ 2 } \\ \\ (\dfrac { n }{ 5 } )=\quad [(\dfrac { n }{ 4 } )+(\dfrac { n }{ 6 } )]/2\\ 12*(n-5)=(n-5)*(n-4)+30\\ n=7\quad or\quad 14\\ \\ \\$

A symmetric die is thrown

$(2n+1)$ times. The probability of getting a prime score on the upturned face at most

$n$ times is

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

Explanation

Probability of getting prime no. in single throw

$= 1/2$
Probability of not getting prime no. in single throw

$= 1/2$
Probability of getting

$n$ throws

$=^{ 2n+1 }C_{ 0 }*({ 1/2 }^{ 0 })*{ (1/2 }^{ 2n+1 })+^{ 2n+1 }C_{ 1 }*({ 1/2 }^{ 1 })*{ (1/2 }^{ 2n })+........^{ 2n+1 }C_{ n-1 }*({ 1/2 }^{ n-1 })*{ (1/2 }^{ n+2 })+^{ 2n+1 }C_{ n }*({ 1/2 }^{ n })*{ (1/2 }^{ n+1 })\\ ={ 1/2 }^{ 2n+1 }(^{ 2n+1 }C_{ 0 }+^{ 2n+1 }C_{ 1 }+........^{ 2n+1 }C_{ n-1 }+^{ 2n+1 }C_{ n })\\ ={ 1/2 }^{ 2n+1 }*{ 2 }^{ n+1 }/2\quad =\quad 1/2$
Sum of

$^{ 2n+1 }C_{ 0 }+^{ 2n+1 }C_{ 1 }+........^{ 2n+1 }C_{ n-1 }+^{ 2n+1 }C_{ n }$ can be found by putting

$x =1$ in binomial expansion of

$(1+x)^(2n+1)$ and is equal to half of that sum.
Hence, option 'A' is correct.

If for a binomial distribution with

$n = 5, 4P(X=1)=P(X=2),$ the probability of success is

0%
$\displaystyle\frac{1}{3}$
0%
$\displaystyle\frac{2}{3}$
0%
$\displaystyle\frac{1}{4}$
0%
$\displaystyle\frac{1}{8}$

Explanation

$n = 5,$

$4P(X=1)=P(X=2)$

$P(X=k)$

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

$=> 4P(X=1)=P(X=2)$

$=> 4$

$^{5}C_{1} { p }^{1}{ (1-p) }^{5-1} = ^{5}C_{2} { p }^{2}{ (1-p) }^{5-2}$

$=> 20$

$p { (1-p) }^{4} = 10 { p }^{2} { (1-p) }^{3}$

$=>$

$20 (1-p) = 10 p$

$=> p =$

$\dfrac{2}{3}$

If the difference between the mean and variance of a binomial distribution for

$5$ trials is

$\dfrac{5}{9}$ , then the distribution is

0%
$\left ( \dfrac{1}{9}+\dfrac{2}{9} \right )^{5}$
0%
$\left ( \dfrac{1}{4}+\dfrac{3}{4} \right )^{5}$
0%
$\left ( \dfrac{2}{3}+\dfrac{1}{3} \right )^{5}$
0%
$\left ( \dfrac{3}{4}+\dfrac{1}{4} \right )^{5}$

Explanation

$np-np(1-p)=\dfrac{5}{9}$

$np(p)=\dfrac{5}{9}$

$n(p^2)=5(\dfrac{1}{3})^{2}$
By comparing we get

$p=\dfrac{1}{3}$ and

$n=5$
Where, p is the probability of success and n is the no. of trials.

For a binomial distribution :

$n = 6$ and

$9$ .

$P(X = 4) = P (X = 2)$ . Then we have

0%
$P(X=1)\leq P(X=5)< P(X=3)$
0%
$P(X=5)> P(X=3)> P(X=1)$
0%
$P(X=5)< P(X=3)< P(X=1)$
0%
$P(X=5)+ P(X=3)> P(X=1)$

Explanation

Given,

$n=6$ ,

$9P(X=4)=P(X=2)$ .

which implies

$9\times { { ^{ 6 }{ C } } }_{ 4 }\times { p }^{ 4 }\times { q }^{ 2 }={ { ^{ 6 }{ C } } }_{ 2 }\times { p }^{ 2 }\times { q }^{ 4 }$ , where

$p$ is success and

$q$ is failure. such that

$p+q=1$ .

therefore

$9\times { p }^{ 2 }={ q }^{ 2 },\quad 3\times p=q$

so we get

$p=1/4,q=3/4$

Now

$P(X=5)={ ^{ 6 }{ C } }_{ 5 }\times { (1/4) }^{ 5 }\times (3/4)=18/{ 4 }^{ 6 }$

$P(X=3)={ ^{ 6 }{ C } }_{ 3 }\times { (1/4) }^{ 3 }\times { (3/4) }^{ 3 }=540/{ 4 }^{ 6 }$

$P(X=1)={ ^{ 6 }{ C } }_{ 1 }\times { (1/4) }^{ 1 }\times { (3/4) }^{ 5 }=1458/{ 4 }^{ 6 }$

Therefore

$P(X=5)<P(X=3)<P(X=1)$

$X$ is a binomial variate with

$n=6$ and

$9P(X=4)=P(X=2)$ , the parameter

$p$ is

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

Explanation

$9P(X=4)=P(X=2)$

$9\:^{6}C_{4}p^{4}(1-p)^{2}=\:^{6}C_{2}p^{2}(1-p)^{4}$

$9p^2=(1-p)^{2}$

$3p=(1-p)$

$4p=1$

$p=\dfrac{1}{4}$

For a binomial distribution mean is

$6$ and S.D. is

$2$ . The distribution is

0%
$\left( 18,\ \cfrac { 1 }{ 3 } ,\ \cfrac { 2 }{ 3 } \right)$
0%
$\left( 9,\ \cfrac { 2 }{ 5 } ,\ \cfrac { 3 }{ 5 } \right)$
0%
$\left( 9,\ \cfrac { 1 }{ 4 } ,\ \cfrac { 3 }{ 4 } \right)$
0%
$\left( 9,\ \cfrac { 5 }{ 6 } ,\ \cfrac { 1 }{ 6 } \right)$

Explanation

$SD$ of Binomial Distribution =

$\sqrt{np(1-p)}$

Mean of Binomial Distribution =

$np$

Therefore

$S.D=\sqrt{Mean (1-p)}$

Hence

$2=\sqrt{6(1-p)}$

$4=6(1-p)$

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

$n=18$

If for a B.D. with

$n=12$ , the ratio of variance to mean is

$\displaystyle \frac{1}{3}$ , then the probability of

$10$ successes is

0%
${ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 10 }$
0%
$_{ 10 }^{ 12 }{ C{ \left( \cfrac { 1 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 2 }{ 3 } \right) }^{ 10 } }$
0%
$_{ 10 }^{ 12 }{ C{ \left( \cfrac { 2 }{ 3 } \right) }^{ 2 }{ \left( \cfrac { 1 }{ 3 } \right) }^{ 10 } }$
0%
$_{ 10 }^{ 12 }{ C{ \left( \cfrac { 1 }{ 4 } \right) }^{ 2 }{ \left( \cfrac { 3 }{ 4 } \right) }^{ 10 } }$

Explanation

If for a B.D. with n=12,
The ratio of variance to mean is

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

$\dfrac { { \sigma }^{ 2 } }{ \mu } = \dfrac {1}{3}$

${ \sigma }^{ 2 } = np(1-p)$ and

$\mu = np$ =>

$\dfrac {np(1-p)}{np} = \dfrac {1}{3}$
=>

$(1-p) = \dfrac {1}{3}$ => p =

$\dfrac {2}{3}$
Probability for k success is =

$^{n}C_{k}\left ( p \right )^{k}\left ( 1-p \right )^{n-k}$
Then the probability of 10 successes =

$^{n}C_{k}\left ( p \right )^{k}\left ( 1-p \right )^{n-k}$
=

$^{12}C_{10}\left ( \dfrac{2}{3} \right )^{10}\left ( 1-\dfrac{2}{3} \right )^{12-10}$
=

$^{12}C_{10}\left ( \dfrac{2}{3} \right )^{10}\left ( \dfrac{1}{3} \right )^{2}$

The mean or average number of heads when we toss

$10$ unbiased coins is

0%
$20$
0%
$10$
0%
$5$
0%
$15$

Explanation

Probability of getting head

$= \dfrac {1}{2}$

$\Rightarrow p = \dfrac {1}{2}$
For toss of

$10$ unbiased coin

$n = 10$
Mean

$= \mu = np$

$\Rightarrow \mu = 10 \times \dfrac {1}{2} = 5$

If the mean and variance of a binomial variate

$X$ are respectively

$\displaystyle \frac{35}{6}$ and

$\displaystyle \frac{35}{36}$ , then the probability of

$X>6$ is

0%
$1-{ \left( \cfrac { 5 }{ 6 } \right) }^{ 7 }$
0%
${ \left( \cfrac { 5 }{ 6 } \right) }^{ 7 }$
0%
$1-{ \left( \cfrac { 1 }{ 6 } \right) }^{ 7 }$
0%
${ \left( \cfrac { 1 }{ 6 } \right) }^{ 7 }$

Explanation

Here,

$\displaystyle np = \frac{35}{6}, npq = \frac{35}{36}$

$\therefore\displaystyle q = \frac{1}{6}, p = \frac{5}{6}, n = 7$
Therefore

$P(X> 6)=P(X=7) = ^7C_7 p^{7}q^{7-7}=\left(\cfrac{5}{6}\right)^7$

Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. The variance of the number of aces is

0%
$24/169$
0%
$\sqrt{24/169}$
0%
$\sqrt{24/173}$
0%
$24/173$

Explanation

Let

$X$ denote the number of aces

$X$ can tare values of

$0,1,2$
Let

$S$ be the sample space

$\Rightarrow n\left( S \right) =52$
Let

$E$ be the event of getting an ace probability of getting an ace

$(p)$

$\displaystyle =\frac { n\left( E \right) }{ n\left( S \right) } =\frac { 4 }{ 52 } =\frac { 1 }{ 13 }$
Probability of not getting an ace

$(q)=1-p$

$\displaystyle =1-\frac { 1 }{ 13 } =\frac { 12 }{ 13 }$
we know that

$n=2$
Mean

$\displaystyle =np=2\left( \frac { 1 }{ 13 } \right) =\frac { 2 }{ 13 }$
variance

$\displaystyle =npq=2\left( \frac { 1 }{ 13 } \right) \left( \frac { 12 }{ 13 } \right) =\frac { 24 }{ 169 }$

The probability that a student is not a swimmer is

$\displaystyle\frac{1}{5}$ . Out of

$5$ students the probability that exactly four are swimmers is

0%
$\left ( \displaystyle\frac{4}{5} \right )^{4}$
0%
$\left ( \displaystyle\frac{1}{5} \right )^{4}$
0%
$^{5}C_{4}\left ( \displaystyle\frac{4}{5} \right )^{4}\left ( \displaystyle\frac{1}{5} \right )^{1}$
0%
$^{5}C_{4}\left ( \displaystyle\frac{1}{5} \right )^{4}\left ( \displaystyle\frac{4}{5} \right )^{1}$

Explanation

The probability that a student is not a swimmer is

$\dfrac {1}{5}$ .

The probability that a student is a swimmer

$=$

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

Probability for

$k$ successes is

$= { }^{n}C_{k}\left ( p \right )^{k}\left ( 1-p \right )^{n-k}$
Total swimmers

$=$

$n= 5$ ;

$p = \dfrac {4}{5}$
The probability that exactly four are swimmers out of

$5$ is

$=$

$^{5}C_{4}\left ( \dfrac {4}{5} \right )^{4}\left ( 1-\dfrac {4}{5} \right )^{5-4}$

$=$

$^{5}C_{4}\left ( \dfrac {4}{5} \right )^{4}\left ( \dfrac {1}{5} \right )^{1} = \left(\dfrac45\right)^4$

Hence, both options

$A$ and

$C$ are correct.

$\mathrm{A}$ and

$\mathrm{B}$ play a game in which

$\mathrm{A}^{'}\mathrm{s}$ chance of winning is

$\displaystyle \frac{1}{5}$ in a series of

$6$ games, the probability that

$\mathrm{A}$ will win all the

$6$ games is

0%
$_{ 2 }^{ 6 }{ C }\left (\displaystyle\frac { 1 }{ 5 } \right )^{ 6 }$
0%
$_{ 6 }^{ 6 }{ C }\left (\displaystyle\frac { 1 }{ 5 } \right )^{ 6 }\left (\displaystyle\frac { 4 }{ 5 }\right )^{ 0 }$
0%
${ \left( \displaystyle\frac { 4 }{ 5 } \right) }^{ 6 }$
0%
$_{ 5 }^{ 6 }{ C }{ \left( \displaystyle\cfrac { 1 }{ 5 } \right) }^{ 5 }\left( \displaystyle\cfrac { 4 }{ 5 } \right)$

Explanation

As chance of winning is

$\dfrac {1}{5}$ .
For a series of

$6$ games, the probability that $$A$$ will win all the

$6$ games,

$n= 6$ and

$k = 6$ probability of win

$= p =$

$\dfrac {1}{5}$

Probability for

$k$ success is =

$^{n}C_{k}\left ( p \right )^{k}\left ( 1-p \right )^{n-k}$

Probability

$=$

$^{6}C_{6}\left ( \dfrac {1}{5} \right )^{6}\left ( 1- \dfrac {1}{5}\right )^{6-6}$

$=$

$^{6}C_{6}\left ( \dfrac {1}{5} \right )^{6}\left ( \dfrac {4}{5}\right )^{0}$

The binomial distribution whose mean is

$9$ and the variance is

$2.25$ is

0%
$\left( 12,\ \cfrac { 1 }{ 2 } ,\ \cfrac { 1 }{ 2 } \right)$
0%
$\left( 12,\ \cfrac { 2 }{ 3 } ,\ \cfrac { 1 }{ 3 } \right)$
0%
$\left( 12,\ \cfrac { 5 }{ 6 } ,\ \cfrac { 1 }{ 6 } \right)$
0%
$\left( 12,\ \cfrac { 3 }{ 4 } ,\ \cfrac { 1 }{ 4 } \right)$

Explanation

Mean

$=$

$np$
Variance

$=$

$np(1-p)$
Therefore according to question

$1-p=q =$

$\dfrac{\text{Variance}}{\text{Mean}}=$

$\cfrac{2.25}9$

$=$

$0.25$

$p =1-0.25=$

$0.75$

$n =$

$\cfrac{\text{Mean}}p=$

$\cfrac9{0.75} = 12$
Therefore Binomial Distribution

$(n,p,1-p)= (12,.75,.25) = \left(12,\dfrac34,\dfrac14\right)$

The probability of answering

$6$ out of

$10$ questions correctly in a true or false examination is

0%
$\displaystyle ^{10}C_{4}\left ( \frac{1}{2} \right )^{4}$
0%
$\displaystyle ^{10}C_{6}\left ( \frac{1}{2} \right )^{6}$
0%
$\displaystyle ^{10}C_{6}\left ( \frac{1}{2} \right )^{10}$
0%
$\displaystyle ^{10}C_{6}\left ( \frac{1}{2} \right )^{8}$

Explanation

Probability of answering a question correctly in a true or false questionnaire is

$\dfrac{1}{2}$ .
Hence, for

$10$ questions, the probability of answering

$6$ of them correctly is

$\:^{10}C_{6}(\dfrac{1}{2})^{6}(1-\dfrac{1}{2})^{4}$

$=\:^{10}C_{6}(\dfrac{1}{2})^{10}$

$X$ follows a binomial distribution with parameters

$n$ and

$p$ , and

$Y$ follows a binomial distribution with parameters

$m$ and

$p$ . Then, if

$X$ and

$Y$ are independent

$\displaystyle P\left( \frac { X=r }{ X+Y=r+s } \right) =$

0%
$\displaystyle \frac { \left( _{ }^{ m }{ { C }_{ r } } \right) \left( _{ }^{ n }{ { C }_{ s } } \right) }{ _{ }^{ m+n }{ { C }_{ r+s } } }$
0%
$\displaystyle \frac { \left( _{ }^{ m }{ { C }_{ s } } \right) \left( _{ }^{ n }{ { C }_{ r } } \right) }{ _{ }^{ m+n }{ { C }_{ r+s } } }$
0%
$\displaystyle \frac { \left( _{ }^{ m }{ { C }_{ s } } \right) }{ _{ }^{ m+n }{ { C }_{ r } } }$
0%
none of these

Explanation

We have

$\displaystyle P\left( \frac { X=r }{ X+Y=r+s } \right) =\frac { P\left[ \left( X=r \right) \cap \left( X+Y=r+s \right) \right] }{ P\left( X+Y=r+s \right) }$

$\displaystyle =\frac { P\left[ \left( X=r \right) \cap \left( Y=s \right) \right] }{ P\left( X+Y=r+s \right) } =\frac { P\left( X=r \right) P\left( Y=s \right) }{ P\left( X+Y=r+s \right) }$

$\displaystyle P\left( X+Y=r+s \right) =\sum _{ k=0 }^{ r+s }{ P\left[ \left( X=k \right) \cap \left( Y=r+s-k \right) \right] }$

$\displaystyle =\sum _{ k=0 }^{ r+s }{ \left( _{ }^{ n }{ { C }_{ k } }.{ p }^{ k }.{ q }^{ n-k } \right) \left( _{ }^{ m }{ { C }_{ r+s-k } }.{ p }^{ r+s-k }.{ q }^{ m-r-s+k } \right) }$

$\displaystyle ={ p }^{ r+s }.{ q }^{ m+n-r-s }.\sum _{ k=0 }^{ r+s }{ \left( _{ }^{ n }{ { C }_{ k } } \right) \left( _{ }^{ m }{ { C }_{ r+s-k } } \right) }$
Now the last sum is the expression for the number of ways of choosing

$r+s$ persons out of

$n$ men and

$m$ women, which is

$^{ m+n }{ { C }_{ r+s } }$ .
Therefore

$P\left( X+Y=r+s \right) =_{ }^{ m+n }{ { C }_{ r+s } }.{ p }^{ r+s }.{ q }^{ m+n-r-s }$ so that

$\displaystyle P\left( \frac { X=r }{ X+Y=r+s } \right) =\frac { \left( _{ }^{ m }{ { C }_{ r } }.{ p }^{ r }.{ q }^{ n-r } \right) \left( _{ }^{ n }{ { C }_{ s } }.{ p }^{ s }.{ q }^{ m-s } \right) }{ _{ }^{ m+n }{ { C }_{ r+s } }.{ p }^{ r+s }.{ q }^{ m+n-r-s } } =\frac { \left( _{ }^{ m }{ { C }_{ r } } \right) \left( _{ }^{ n }{ { C }_{ s } } \right) }{ _{ }^{ m+n }{ { C }_{ r+s } } }$

$\mathrm{m}$ is the variance of a Poisson Distribution, then the sum of the terms in odd places is:

0%
$e^{-m}$
0%
$e^{-m}\cosh m$
0%
$e^{-m}\sinh m$
0%
$e^{-m}\coth m$

Explanation

$m$ is the variance of P. D, then

$P(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!}$
Sum of the terms in odd places

$=$

$[P(0; \mu) + P(2; \mu) + P(4; \mu) + ...]$

$= [$

$\dfrac { { e }^{-\mu}{\mu}^{0}}{0!} + \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} + \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} + ... ]$

$=$

${ e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + ....]$ ----- (1)

Since

${ e }^{ x }=1+\dfrac { x }{ 1! } +\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{4} }{ 4! } +....$

${ e }^{ -x }=1+\dfrac {( -x )}{ 1! } +\dfrac { { (-x) }^{ 2 } }{ 2! } +\dfrac { {( -x) }^{ 3 } }{ 3! } +\dfrac { { (-x) }^{4} }{ 4! } +....$

on adding , we get

${e}^{x} + { e }^{-x} = 2[ 1 + \dfrac {{x}^{2}}{2!} + \dfrac {{x}^{4}}{4!} + .....]$

$\dfrac {{e}^{x} + { e }^{-x}}{2} = [ 1 +\dfrac { { x }^{ 2} }{ 2! } + \dfrac {{x}^{4}}{4!} + .....]$
So,

$\dfrac {{e}^{\mu} + { e }^{-\mu}}{2} = [ 1+\dfrac { { \mu }^{ 2 } }{ 2! } + \dfrac {{\mu}^{4}}{4!} + .....]$
put this value in equation (1),
Sum of the terms in odd places

$=$

${ e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$

$=$

${ e }^{-\mu} (\dfrac {{e}^{\mu} + { e }^{-\mu}}{2})$

$=$

${ e }^{-\mu} \cosh { \mu }$

$=$

${ e }^{-m} \cosh {m}$ [Variance

$(m)$ is equal to mean

$(\mu)$ in Poisson distribution]

The variance of P.D. with parameter

$\lambda$ is

0%
$\lambda$
0%
$\sqrt{\lambda }$
0%
$\dfrac{1}{\lambda}$
0%
$\dfrac{1}{\sqrt {\lambda}}$

Explanation

$E[{ X }^{ 2 }]= \sum _{ k=0 }^{ \infty }{ k^{ 2 } } \sum _{ }^{ }{ } \dfrac { 1 }{ k! } \lambda ^{ k }e^{ -\lambda }\\ \\ = ^{ k }e^{ -\lambda }\sum _{ 1 }^{ \infty }{ k\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } \\= ^{ k }e^{ -\lambda }(\sum _{ 1 }^{ \infty }{ (k-1)\dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } ) +\sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\\ =^{ k }e^{ -\lambda }(\sum _{ 2 }^{ \infty }{ (\lambda \dfrac { 1 }{ (k2)! } \lambda ^{ k-2 } } )+ \sum _{ 1 }^{ \infty }{ \dfrac { 1 }{ (k1)! } \lambda ^{ k-1 } } )\\ = ^{ k }e^{ -\lambda }(\sum _{ i=0 }^{ \infty }{ (\lambda \dfrac { 1 }{ i! } \lambda ^{ i } } )+ \sum _{ j=0 }^{ \infty }{ \dfrac { 1 }{ j! } \lambda ^{ j } } )\\ = ^{ k }e^{ -\lambda }(\lambda e^{ \lambda }+e^{ \lambda })\\ \\ = { \lambda }^{ 2 }+ \lambda \\ \\ Variance= E[{ X }^{ 2 }]- { E[X] }^{ 2 }\\ = { \lambda }^{ 2 }+ \lambda - { \lambda }^{ 2 }\\ = \lambda \\ \\ \\$

A : the sum of the times in odd places in a P.D is

$e^{-\lambda }$ cosh

$\lambda$
R : cosh

$\lambda =\frac{\lambda ^{1}}{1!}+\frac{\lambda ^{3}}{3!}+\frac{\lambda ^{5}}{5!}+......$

0%
Both A and R are true and R is the correct

explanation of A
0%
Both A and R are true but R is not correct

explanation of A
0%
A is true but R is false
0%
A is false but R is true

Explanation

Considering all odd places in Poisson's Distribution, we get

$e^{-\lambda}+\dfrac{e^{-\lambda}.\lambda^{2}}{2!}+\dfrac{e^{-\lambda}.\lambda^{4}}{4!}....$

$=e^{-\lambda}[1+\dfrac{\lambda^{2}}{2!}+\dfrac{\lambda^{4}}{4!}+....]$

$=e^{-\lambda}[\dfrac{e^{-\lambda}+e^{\lambda}}{2}]$

$=e^{-\lambda}(\cosh\lambda)$ .
Hence reason is false.

The probability of r successes in case of poissons distrbution is

0%
$\dfrac{e^{\gamma }m}{\angle \gamma }$
0%
$\dfrac{\gamma ^{m}e^{m}}{\angle \gamma }$
0%
$\dfrac{e^{m}\gamma }{\angle \gamma }$
0%
$\dfrac{e^{-m}m^{r}}{\angle \gamma }$

Explanation

in Poisson distribution, probability

$P(x; m )= \dfrac { { e }^{ - m }{ m }^{ x } }{ \angle x }$

$m$ = mean ;

$x$ = number of success

$\angle x =$ factorial

$x$
For

$r$ success

$x= r$ ;

$P(r; m )= \dfrac { { e }^{ -m }{ m }^{ r } }{ \angle r }$

A box contains 6 tickets. Two of the tickets carry a prize of Rs. 5/- each, the other four a prize of Rs.1/-. If one ticket is drawn. The mean value of prize is

0%
Rs. $2.5$
0%
Rs. $7/3$
0%
Rs. $5/3$
0%
Rs. $4/3$

Explanation

The probability of drawing a ticket out of six is

$\dfrac{1}{6}$
Hence the mean prize if one ticket is drawn is

$(5(\dfrac{1}{6}))+(5(\dfrac{1}{6}))+(\dfrac{1}{6})+((\dfrac{1}{6}))+(\dfrac{1}{6}))+((\dfrac{1}{6}))$

$=2\times[5\times\dfrac{1}{6}]+\dfrac{4}{6}$

$=\dfrac{10}{6}+\dfrac{4}{6}$

$=\dfrac{14}{6}$

$=\dfrac{7}{3}$ rupees.

A random variable

$X$ has the following probability distribution

$X=x:\quad \quad \quad1 \quad \quad 2 \quad\quad 3 \quad\quad 4$

$p(X=x): \ \ \ \ \ \ 0.4 \quad 0.3 \quad \ 0.2 \quad \ \ 0.1$ , then its mean is

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

Explanation

Here,

$n = 4$ ,

$P(X = 1) = 0.4,P(X = 2) = 0.3, P(X = 3) = 0.2, P(X = 4) =0 .1$

We know,

$P(X=r) = ^nC_r p^rq^{n-r}$

$\Rightarrow P(X = 1) =^4C_1 pq^3 = .4\Rightarrow pq^3 =0 .1$ ..(i)

$\Rightarrow P(X = 4) =^4C_4 pq^3 = .1\Rightarrow p^3q = 0.1$ ..(ii)

Divide (i) and (ii)

$\dfrac{p^3q}{pq^3}=1\Rightarrow p^2=q^2\Rightarrow p=q$

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

$p+q=1$

Hence mean

$=np = 4\cdot \cfrac{1}{2} = 2$

If m is the variance of P.D., then the ratio of sum of the terms in odd places to the sum of the terms in even places is

0%
$e^{-m}\cosh m$
0%
$e^{-m}\sinh m$
0%
$\coth m$
0%
$\tanh m$

Explanation

If m is the variance of P. D, then P

$(x; \mu) = \frac { { e }^{-\mu}{\mu}^{x}}{x!}$
Sum of the terms in odd places = [ P

$(0; \mu) + P(2; \mu) + P(4; \mu) + .........$ ]

= [

$\dfrac { { e }^{-\mu}{\mu}^{0}}{0!} + \dfrac { { e }^{-\mu}{\mu}^{2}}{2!} + \dfrac { { e }^{-\mu}{\mu}^{4}}{4!} + .......$ ]

${ e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$ --------------------- (1)

Since

${ e }^{ x }=1+\dfrac { x }{ 1! } +\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{4} }{ 4! } +....$

${ e }^{ -x }=1+\dfrac {( -x )}{ 1! } +\dfrac { { (-x) }^{ 2 } }{ 2! } +\dfrac { {( -x) }^{ 3 } }{ 3! } +\dfrac { { (-x) }^{4} }{ 4! } +....$

on adding , we get

${e}^{x} + { e }^{-x} = 2[ 1 + \dfrac {{x}^{2}}{2!} + \dfrac {{x}^{4}}{4!} + .....]$

$\dfrac {{e}^{x} + { e }^{-x}}{2} = [ 1 +\dfrac { { x }^{ 2} }{ 2! } + \dfrac {{x}^{4}}{4!} + .....]$
So,

$\dfrac {{e}^{\mu} + { e }^{-\mu}}{2} = [ 1+\dfrac { { \mu }^{ 2 } }{ 2! } + \dfrac {{\mu}^{4}}{4!} + .....]$

put this value in equation (1),
Sum of the terms in odd places =

${ e }^{-\mu} [ 1 + \dfrac {{\mu}^{2}}{2!} + \dfrac {{\mu}^{4}}{4!} + .......]$
=

${ e }^{-\mu} (\dfrac {{e}^{\mu} + { e }^{-\mu}}{2})$
=

${ e }^{-\mu} \cosh { \mu }$
=

${ e }^{-m} \cosh {m}$ [ Variance (m) is equal to mean (

$\mu$ ) in Poisson distribution ]
Similarly Sum of the terms in even places =

${ e }^{-m} \sinh {m}$
The ratio of sum of the terms in odd places to the sum of the terms in even places is =

$\dfrac { { e }^{ -m }\cosh { m } }{ { e }^{ -m }\sinh { m } } =\coth { m }$

${m}$ is the variance of Poisson distribution, then sum of the terms in even places is

0%
$e^{-m}$
0%
$e^{-m}\cosh m$
0%
$e^{-m}\sinh m$
0%
$e^{-m}\coth m$

Explanation

If m is the variance of P. D, then P

$(x; \mu) = \dfrac { { e }^{-\mu}{\mu}^{x}}{x!}$
Sum of the terms in even places = [ P

$(1; \mu) + P(3; \mu) + P(5; \mu) + .........$ ]

= [

$\dfrac { { e }^{-\mu}{\mu}^{1}}{1!} + \dfrac { { e }^{-\mu}{\mu}^{3}}{3!} + \dfrac { { e }^{-\mu}{\mu}^{5}}{5!} + .......$ ]

${ e }^{-\mu} [ \mu + \dfrac {{\mu}^{3}}{3!} + \dfrac {{\mu}^{5}}{5!} + .......]$ --------------------- (1)

Since

${ e }^{ x }=1+\dfrac { x }{ 1! } +\dfrac { { x }^{ 2 } }{ 2! } +\dfrac { { x }^{ 3 } }{ 3! } +\dfrac { { x }^{4} }{ 4! } +....$

${ e }^{ -x }=1+\dfrac {( -x )}{ 1! } +\dfrac { { (-x) }^{ 2 } }{ 2! } +\dfrac { {( -x) }^{ 3 } }{ 3! } +\dfrac { { (-x) }^{4} }{ 4! } +....$

on subtracting , we get

${e}^{x} - { e }^{-x} = 2[ x + \dfrac {{x}^{3}}{3!} + \dfrac {{x}^{5}}{5!} + .....]$

$\dfrac {{e}^{x} - { e }^{-x}}{2} = [ x +\dfrac { { x }^{ 3 } }{ 3! } + \dfrac {{x}^{5}}{5!} + .....]$
So,

$\dfrac {{e}^{\mu} - { e }^{-\mu}}{2} = [ x +\dfrac { { \mu }^{ 3 } }{ 3! } + \dfrac {{\mu}^{5}}{5!} + .....]$
put this value in equation (1),
Sum of the terms in even places =

${ e }^{-\mu} [ \mu + \dfrac {{\mu}^{3}}{3!} + \dfrac {{\mu}^{5}}{5!} + .......]$
=

${ e }^{-\mu} (\dfrac {{e}^{\mu} - { e }^{-\mu}}{2})$
=

${ e }^{-\mu} \sinh { \mu }$
=

${ e }^{-m} \sinh {m}$ [ Variance (m) is equal to mean (

$\mu$ ) in Poisson distribution ]

If m is the variance of P.D., then the ratio of sum of the terms in even places to the sum of the terms in odd places is

0%
$e^{-m}\cosh m$
0%
$e^{-m}\sinh m$
0%
$\coth m$
0%
$\tanh m$

Explanation

Sum of the terms in Even places will be

$=e^{-m}[m+\dfrac{m^{3}}{3!}+\dfrac{m^{5}}{5!}+...]$

$=e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]$ ...(i)
Sum of the terms in Odd places will be

$=e^{-m}[1+\dfrac{m^{2}}{2!}+\dfrac{m^{4}}{4!}+...]$

$=e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]$ ...(ii)
Taking the ratio of i and ii, we get

$\dfrac{i}{ii}$

$=\dfrac{e^{-m}\dfrac{1}{2}[e^{m}-e^{-m}]}{e^{-m}\dfrac{1}{2}[e^{m}+e^{-m}]}$

$=\dfrac{e^{m}-e^{-m}}{e^{m}+e^{-m}}$

$=\dfrac{2sinhm}{2coshm}$

$=tanh(m)$ .

A random variable

$X$ follows poisson distribution such that

$P(X=k)=P(X=k+1)$ then the parameter of the distribution

$\lambda =$

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

Explanation

$P(X=k)=P(X=k+1)$

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

$P(k;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ k } }{ k! } =P(k+1;\mu)=\dfrac { { e }^{ -\mu }{ \mu }^{ k+1 } }{ (k+1)! }$

$1 = \dfrac {\mu}{k+1}$

$\mu = k+1$

If X is a poisson variate such that

$P(X=2)=9p(X=4)+90p(X=6)$ , then the mean of x is

0%
$3$
0%
$2$
0%
$1$
0%
$0$

Explanation

For Poisson's distribution,

$P(X)$ is given by

$\displaystyle P(X)=\frac { { e }^{ -\lambda }.{ \lambda }^{ x } }{ x! }$
It is given that

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

$\Rightarrow \displaystyle \frac { { e }^{ -\lambda }.{ \lambda }^{ 2 } }{ 2! } =9.\frac { { e }^{ -\lambda }.{ \lambda }^{ 4 } }{ 4! } +90.\frac { { e }^{ -\lambda }.{ \lambda }^{ 6 } }{ 6! }$
Cancelling

${ e }^{ -\lambda }$ from both sides, we get

$\displaystyle \frac { { \lambda }^{ 2 } }{ 2! } =9.\frac { { \lambda }^{ 4 } }{ 4! } +90.\frac { { \lambda }^{ 6 } }{ 6! }$
Since

$\lambda$ can not be zero, cancel out

$\lambda^2$ from both sides to obtain

${ \lambda }^{ 4 }+3{ \lambda }^{ 2 }-4=0$ .
Solving , we get

$\lambda = 1$ as a valid solution.

If a random variable

$X$ has a poisson distributionsuch that

$P(X=1)=P(X=2)$ , its mean and varianceare

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

Explanation

In Poisson distribution such that

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

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

$P(1; \mu) = P(2; \mu)$
=>

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

$1 =$

$\dfrac {\mu}{2}$
=>

$\mu = 2$
In Poisson distribution Variance

$(m)$ is equal to mean.
Mean = Variance =

$2$

If in a poisson frequency distribution, the frequency of

$3$ successes is

$\displaystyle \frac{2}{3}$ times the frequency of

$4$ successes, the mean of the distribution is

0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{1}{3}$
0%
$6$
0%
$\sqrt{6}$

Explanation

According to question

$P(X=3; \mu )= \dfrac {2}{3}P(X=4; \mu)$

$\dfrac { { e }^{-\mu}{\mu}^{3}}{3!} = \dfrac {2}{3} \dfrac { { e }^{-\mu}{\mu}^{4}}{4!}$
=>

$\dfrac { { e }^{-\mu}{\mu}^{3}}{6} = \dfrac { { e }^{-\mu}{\mu}^{4}}{36}$
=>

${ e }^{ -\mu }{ \mu }^{ 3 }\left[ \dfrac { 1 }{ 6 } -\dfrac { \mu }{ 36 } \right]$
=>

$\mu = 0 \ or\ \mu = \dfrac{36}{6} = 6$

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 4 - 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 4 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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