MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 2

If X following the Binomial distribution with parameters n=6 and p and 9P (X = 4) = P (X = 2), then p is:

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

Explanation

$9p(x = 4) = p(x = 2)$

$9{ \times ^6}{C_1}{p^4}{(1 - p)^2}{ = ^6}{C_2}{p^2}{(1 - p)^4}$

${{9 \times 6 \times 5} \over 2}{p^2} = {{6 \times 5} \over 2}{\left( {1 - p} \right)^2}$

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

${p \over {1 - p}} = {1 \over 3}$

$3p = 1 - p$

$4p = 1$

$p = {1 \over 4}$ .

The probability of A's a Tenis game against

$B$ is

$\dfrac{1}{3}$ . Find the probability that A wins at least

$1$ of the total

$3$ set of games.

0%
$\dfrac{1}{19}$
0%
$\dfrac{19}{27}$
0%
$\dfrac{1q}{19}$
0%
$\dfrac{7}{19}$

Explanation

Given: Probability of

$A$ winning the tennis game against

$B$ is

$\dfrac 13$ and a total of

$3$ games is played.

To find: the probability that

$A$ wins at least

$1$ of the total

$3$ set of games

Probability of

$A$ winning at least

$1$ game

$=$ P (A winning exactly

$1$ game)

$+$ P(A winning exactly

$2$ games)

$+$ P(A winning exactly

$3$ games)

This is Bernoulli's' theorem, hence

P(Success in each trial)

$={}^nC_k p^k(1-p)^{n-k}$ , in this case

$n = 3,$

$p=\dfrac 13$ and

$k = 1, 2, 3$

$= {}^3C_1\left( \dfrac 13\right)^1\left(1-\dfrac 13\right)^{3-1}+{}^3C_2\left( \dfrac 13\right)^2\left(1-\dfrac 13\right)^{3-2}+{}^3C_3\left( \dfrac 13\right)^3\left(1-\dfrac 13\right)^{3-3}$

$= 3\times \dfrac 13\times \dfrac {2^2}{3^2}+3\times \dfrac {2}{3^2}\times \dfrac 1{3}+1\times \dfrac 1{3^3}\times 1$

$= \dfrac{12}{27}+\dfrac{6}{27}+\dfrac{1}{27}=\dfrac {12+6+1}{27}=\dfrac{19}{27}$

If the mean and variance of a binomial distribution are

$4$ and

$2$ respectively, then the probability of

$2$ successes of that binomial variate

$X$ , is

0%
$\dfrac {1}{2}$
0%
$\dfrac {219}{256}$
0%
$\dfrac {37}{256}$
0%
$\dfrac {7}{64}$

Explanation

$mean = 4\,\,$

$np = 4$

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

$np = 4 \Rightarrow n = \frac{4}{p} = 4 \times 2 = 8$

$p\left( {2succes} \right){ = ^2}{c_2} = 8{c_2}{\left( {\dfrac{1}{2}} \right)^{8 - 2}}{\left( {\dfrac{1}{2}} \right)^2}$

$= \dfrac{{8 \times 7}}{2} \times {\left( {\dfrac{1}{2}} \right)^8} = \dfrac{{4 \times 7}}{{4 \times 2}} = \dfrac{7}{{64}}$

$\sum_{r=1}^{11} r.5^{r} =\dfrac{(43\times 5^{a}+5)}{b}$ , then

$(a+b)$ is

0%
$18$
0%
$28$
0%
$15$
0%
$38$

$X$ follows a binomial distribution with parameters

$n=6$ and

$p$ . If

$4P(X=4)=P(X=2)$ , then

$p$ is:

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

Explanation

For binomial distribution,

$P(X=r)={^nC_rp^r(1-p)^{n-r}}$ . Here, n =6.

According to question,

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

$\Rightarrow4\times{^6C_4p^4(1-p)^2}={^6C_2p^2(1-p)^4}$

$\Rightarrow4\times{15p^4(1-p)^2}={15p^2(1-p)^4}$

$\Rightarrow4{p^2}={(1-p)^2}\Rightarrow 3p^2+2p-1=0$

$\Rightarrow3p^2+3p-p-1=0\Rightarrow3p(p+1)-1(p+1)=0$

$\Rightarrow (p+1)(3p-1)=0$

$\therefore$ Either

$p=-1$ or

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

$\because\quad 0\le p \le1$

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

Find the Binomial probability distribution whose mean is

$3$ and variance is

$2$

0%
${\left( \dfrac { 2 } { 3 } + \dfrac { 1 } { 3 } \right) }^ { 9 }$
0%
${\left( \dfrac { 5 } { 3 } + \dfrac { 2 } { 3 } \right)} ^ { 9 }$
0%
${\left( \dfrac { 3 } { 3 } + \dfrac { 1 } { 2 } \right) }^ { 9 }$
0%
None of these

Explanation

Mean =

$np = 3$

variance =

$npq = 2$

$np(q) = 2$

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

$p = 1-q = 1-\dfrac{2}{3}= \dfrac{1}{3}$

$\therefore n = 3\times 3= 9$

Hence the distribution is

$\left(\dfrac{2}{3}+\dfrac{1}{3}\right)^{9}$ .

The probability that a certain kind of component will survive a given shock test is

$3/4$ . Find the probability that among

$5$ components tested exactly

$2$ will survive.

0%
$\cfrac{90}{124}$
0%
$\cfrac{9}{14}$
0%
$\cfrac{90}{1024}$
0%
None of these

Explanation

Component will survive the test =

$\cfrac{3}{4}$

Component will not survive the test =

$\cfrac{1}{4}$

2 survive the test =

$(\cfrac{3}{4})^2 \times (\cfrac{1}{4})^3 \times \cfrac{5!}{3!\times 2!}= \cfrac{90}{1024}$

In a poisson distribution, the variance is

$m$ . The sum of terms in odd places in the distribution is

0%
$e^{-m}$
0%
$e^{-m} \cos \, h \, (m)$
0%
$e^{-m} \sin \, h \, (m)$
0%
$e^{-m} \cot \, h \, (m)$

Explanation

$e^{-m}\cos h(m)$

The probability that a certain kind of component will survive a given shock test is

$3/4$ . Find the probability that among

$5$ components tested at most

$3$ will survive.

0%
$\cfrac{75}{1024}$
0%
$\cfrac{375}{1024}$
0%
$\cfrac{275}{1024}$
0%
None of these

Explanation

Component will survive the test =

$\cfrac{3}{4}$

Component will not survive the test =

$\cfrac{1}{4}$

1 survive the test =

$(\cfrac{3}{4})^1 \times (\cfrac{1}{4})^4 \times \cfrac{5!}{4!\times 1!}$

2 survive the test =

$(\cfrac{3}{4})^2 \times (\cfrac{1}{4})^3 \times \cfrac{5!}{3!\times 2!}$

3 survive the test =

$(\cfrac{3}{4})^3 \times (\cfrac{1}{4})^2 \times \cfrac{5!}{2!\times 3!}$

Required Probability =

$\cfrac{15+90+270}{1024} = \cfrac{375}{1024}$

Eight coins are tossed at a time, the probability of getting at least 6 heads up, is

0%
$\frac{7}{{64}}$
0%
$\frac{{57}}{{64}}$
0%
$\frac{{37}}{{256}}$
0%
$\frac{{229}}{{256}}$

Explanation

$x\le6\\P(x=6)+P(x=7)+P(x=8)\\={ { ^{ 8 }{ C } } }_{ 6 }(\cfrac{1}{2})^8+{ { ^{ 8 }{ C } } }_{ 7 }(\cfrac{1}{2})^8+{ { ^{ 8 }{ C } } }_{ 8 }(\cfrac{1}{2})^8 \\=(\cfrac{1}{2})^8[28+8+1]\\=(37/256)$

If the sum of the mean and variance of a binomial distribution for 5 trials is 1.8, then p=......

0%
0.18
0%
0.2
0%
0.8
0%
0.4

Explanation

Given number of trials

$,n=5$

Sum of mean and variance of a binomial distribution is

$1.8$

$\implies n{p}+n{p}{q}=1.8$

$\implies p(1+1-p)=\dfrac{1.8}{5}=0.36$

$\implies 2{p}-p^2=0.36\implies p^2-2{p}+0.36=0$

$(p-1)^2=0.64\implies p=1-\sqrt{0.64}=1-0.8=0.2$

In a binomial distribution, the probability of getting a success is

$\dfrac{1}{4}$ and the standard deviation is

$3$ . Then its mean is. (Eamcet 2002)

0%
$12$
0%
$10$
0%
$8$
0%
$6$

Explanation

Hence,

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

Variance

$=S.D^{2}$

$np(1-p)=9$

$n(\dfrac{1}{4})(\dfrac{3}{4})=9$

$n=3(16)$

$n=48$
Hence, Mean

$=np$

$=\dfrac{48}{4}$

$=12$

For a binomial variate with n = 6, if

$P(X=2)=9P(X=4),$ then the variance is

0%
$\dfrac{8}{9}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{9}{8}$
0%
$4$

Explanation

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

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

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

$3p=(1-p)$

$4p=1$

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

Hence Variance

$=np(1-p)$

$=6(\dfrac{3}{16})$

$=\dfrac{9}{8}$

$X$ follows a binomial distribution with parameters

$n=8$ and

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

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

0%
$\dfrac{119}{128}$
0%
$\dfrac{119}{228}$
0%
$\dfrac{19}{128}$
0%
$\dfrac{18}{28}$

Explanation

We have

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

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

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

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

The probability that a bomb dropped from a plane strikes the target is

$\displaystyle \frac{1}{5}$ . The probability that out of

$6$ bombs dropped atleast

$2$ bombs strike the target

0%
$0.345$
0%
$0.246$
0%
$0.543$
0%
$0.426$

Explanation

The required probability is

$\sum\:^{6}_{k=2}\:^{6}C_{k}(p)^{k}(1-p)^{6-k}$ where

$p=\frac{1}{5}$
Upon expanding we get

$\displaystyle \frac{1}{5^{6}}[\:^{6}C_{2}4^{4}+\:^{6}C_{3}4^{3}+\:^{6}C_{4}4^{2}+\:^{6}C_{5}4^{1}+\:^{6}C_{6}4^{0}]$

$=\displaystyle \frac{1}{5^{6}}[15(256)+20(64)+15(16)+6(4)+1]$

$=\displaystyle \frac{5385}{5^{6}}$

$=0.345$

A binomial distribution has a mean of

$5$ and variance

$4$ . The number of trials is

0%
$20$
0%
$16$
0%
$25$
0%
$10$

Explanation

Mean

$= \mu = np$ = 5;

$n$ = number of trials ;

$p$ = probability of success ;

$q = (1-p) =$ probability of failure
Variance

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

$4 = np(1-p) = 5(1-p)$ ;

$p = \dfrac {1 }{ 5 }$

$\mu = np$ = 5

$n = \dfrac { 5}{ p } = 5 \times \dfrac { 5 }{ 1 } = 25$

$X$ is binomial variate with

$E(X) = 5$ and variance

$4$ . The parameters of the distribution are

0%
$\dfrac{1}{4},20$
0%
$\dfrac{1}{5},20$
0%
$25,\dfrac{1}{5}$
0%
$\dfrac{1}{25},\dfrac{1}{5}$

Explanation

Mean

$= E(X) = \mu = np$

$= 5$ ;

$n$ = number of trials ;

$p =$ probability of success ;

$q = (1-p) =$ probability of failure
Variance

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

$4 = np(1-p) = 5(1-p)$ ;

$p = \dfrac {1 }{ 5 }$

$\mu = np$ = 5

$n = \dfrac { 5}{ p } = 5 \times \dfrac { 5 }{ 1 } = 25$
Parameters are

$n =25$ and

$p = \dfrac {1 }{ 5 }$

If the mean of the binomial distribution is

$25$ . Then standard deviation lies in the interval

0%
$[0, 5)$
0%
$(0, 5]$
0%
$[0, 25)$
0%
$(0, 25]$

Explanation

Given, mean of binomial distribution

$\mu=np=25$

We have ,Variance

$= {\sigma }^ {2} = npq$

Now,

$0 \le p<1$
and

$0 \le q \le 1$

$\Rightarrow 0 \le npq \le np$

$\Rightarrow 0 \le \sigma^2 \le 25$

$\Rightarrow -5 \le \sigma \le 5$

But

$\sigma \ge 0$

$\therefore 0 \le \sigma <5$

$\sigma \in [0,5)$

The mean of binomial distribution is 6 and its S.D. is

$\sqrt{2}$ , then the number of trials

$n$ is

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

Mean

$= \mu = np$ = 6; n = number of trials ; p = probability of success ; q = (1-p) = probability of failure
Standard deviation

$= \sigma = \sqrt { npq } = \sqrt { 2 }$

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

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

$\mu = np$ = 6

$n = \dfrac { 6}{ p } = 6 \times \dfrac { 3 }{ 2 } = 9$

A man take a step forward with probability

$0.4$ and backward with probability

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

0%
$0.37$
0%
$0.57$
0%
$0.3$
0%
None of these

Explanation

Since the man is one step away from starting point mean that either.

(i) man has taken

$6$ steps forward and

$5$ steps backward.

(ii) man has taken

$5$ steps forward and

$6$ steps backward.

Taking movement

$1$ step forward as success and

$1$ step backward as failure

$\therefore p=$ Probability of success

$=0.4$

and

$q=$ Probability of failure

$=0.6$

$\therefore$ Required probability

$=P\left\{ X=6orX=5 \right\} =P\left( X=6 \right) +P\left( X=5 \right)$

$=\,_{ }^{ 11 }{ { C }_{ 6 }{ p }^{ 6 }{ q }^{ 5 } }+_{ }^{ 11 }{ { C }_{ 5 }{ p }^{ 5 }{ q }^{ 6 }=_{ }^{ 11 }{ { C }_{ 5 }\left( { p }^{ 6 }{ q }^{ 5 }+{ p }^{ 5 }{ q }^{ 6 } \right) } }$

$\displaystyle =\frac { 11.10.9.8.7 }{ 1.2.3.4.5 } \left\{ { \left( 0.4 \right) }^{ 6 }{ \left( 0.6 \right) }^{ 5 }+{ \left( 0.4 \right) }^{ 5 }{ \left( 0.6 \right) }^{ 6 } \right\}$

$\displaystyle =\frac { 11.10.9.8.7 }{ 1.2.3.4.5 } { \left( 0.24 \right) }^{ 5 }=0.37$

Hence the required probability

$=0.37$

A symmetrical die is rolled

$720$ times. Getting a face with four points is considered to be a success. The mean and variance of the number of successes is

0%
$20, 120$
0%
$120, 100$
0%
$100, 100$
0%
$50, 50$

Explanation

Here,

$n = 720, p = \cfrac{1}{6}, q = \cfrac{5}{6}$

$\therefore$ mean

$= np = 120$ and variance

$=npq = 100$

The mean and the variance of a random variable

$X$ having a binomial distribution are

$4$ and

$2$ respectively, then

$P(X = 1)$ is

0%
$1/32$
0%
$1/16$
0%
$1/8$
0%
$1/4$

Explanation

Mean =

$np$

$=4$
Variance

$=np(1-p)$

$=2$
Hence

$4=2(1-p)$

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

$n=4(2)$

$=8$
Hence

$P(1)$

$=\:^{8}C_{1}\dfrac{1}{2^{8}}$

$=\dfrac{8}{256}$

$=\dfrac{1}{32}$

The probability of obtaining

$2$ heads when an unbiased coin is tossed

$5$ times is

0%
$\displaystyle \frac{5}{8}$
0%
$\displaystyle \frac{4}{9}$
0%
$\displaystyle \frac{5}{16}$
0%
$\displaystyle \frac{4}{16}$

Explanation

The probability of getting

$2$ heads in

$5$ coin tosses

$= 5C_2 \times (\displaystyle \frac{1}{2}) ^5 = \displaystyle \frac {5}{16}$

The mean and the variance of a binomial distribution are 4 and 2 respectively. Then the probability of 2 successes is

0%
$\dfrac{128}{256}$
0%
$\dfrac{219}{256}$
0%
$\dfrac{37}{256}$
0%
$\dfrac{28}{256}$

Explanation

Mean =

$np$

$=4$
Variance

$=np(1-p)$

$=2$
Hence

$4=2(1-p)$

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

$n=4(2)$

$=8$
Hence

$P(2)$

$=\:^{8}C_{2}\dfrac{1}{2^{8}}$

$=\dfrac{28}{256}$

For a binomial distribution

$n = 10$ ,

$q = 0.4$ , then its mean is

0%
$1$
0%
$4$
0%
$6$
0%
$10$

Explanation

$q=0.4$

$=1-p$
Hence,

$p=1-q=0.6$
Now, Mean

$=np$

$=(0.6)(10)$

$=6$

If for a binomial distribution mean is

$\dfrac{10}{3}$ and sum of mean and variance is

$\dfrac{40}{9}$ . Then the parameters are

0%
$\dfrac{2}{3},10$
0%
$\dfrac{2}{3},20$
0%
$5,\dfrac{2}{3}$
0%
$4,\dfrac{2}{3}$

Explanation

Mean

$=np=\dfrac{10}{3}$
Variance

$=npq=\dfrac{40}{9}$

The sum of Mean and variance,

$\Rightarrow np+np(1-p)=\dfrac{40}{9}$

$\Rightarrow np(2-p)=\dfrac{40}{9}$

$\Rightarrow \dfrac{10}{3}(2-p)=\dfrac{40}{9}$

$\Rightarrow 2-p=\dfrac{4}{3}$

$\therefore p=\dfrac{2}{3}$
Hence,

$n=5$

If for a binomial distribution with n = 16, the ratio of mean to variance is

$\dfrac{5}{4}$ , then the probability of success is

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

Explanation

The ratio of mean to variance will be

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

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

$=\dfrac{5}{4}$

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

$p=\dfrac{1}{5}$ .
where

$p$ being the probability of success.

If the mean of binomial distribution with

$9$ trials is

$6$ , then its variance is

0%
$2$
0%
$3$
0%
$4$
0%
$\sqrt{2}$

Explanation

$\textbf{Step1:Find the variance using formula.}$

$\text{Given}$

$\text {Mean=np}$

$\text{np=6 }$

$9 \times \text{p}=6$

$\text{p}=2 / 2$

$\text{p}= 1$

$\text{Hence, variance}$

$\text{=n p(1-p)}$

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

$=6(\dfrac{1}{3} )$

$=2$

$\textbf{Hence the variance value is 2}$

The mean and variance of a random variable

$X$ having binomial distribution are

$4$ and

$2$ respectively. Then

$P(X>6)$ =

0%
$\dfrac{9}{128}$
0%
$\dfrac{81}{128}$
0%
$\dfrac{9}{256}$
0%
$\dfrac{3}{11}$

Explanation

$Mean=np$

$=4$

$Variance=np(1-p)$

$=2$
Hence

$4(1-p)=2$

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

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

No.of trials will be

$n$

$=2(4)$

$=8$ .
Therefore for

$P(X>6)$ we get

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

$=\dfrac{1}{2^{8}}[8+1]$

$=\dfrac{9}{2^{8}}$

$=\dfrac{9}{256}$

The probability of happening of an event in an experiment is 0.The probability of happening of the event atleast once, if the experiment is repeated 3 times under similar conditions is

0%
$0.216$
0%
$0.784$
0%
$0.64$
0%
$0.32$

Explanation

$p=0.4$

$q=1-0.4=0.6$
Hence the required probability is

$\:^{3}C_{1}(0.4)^{1}(0.6)^{2}+\:^{3}C_{2}(0.4)^{2}(0.6)^{1}+\:^{3}C_{3}(0.4)^{3}(0.6)^{0}$

$=3(0.4)(0.36)+3(0.16(0.6))+(0.064)$

$=0.432+0.288+0.064$

$=0.784$

If in a binomial distribution the mean is

$20$ , standard deviation is

$\sqrt{15}$ , then

$p=$

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

Explanation

Mean

$=np$

$=20$

$S.D^{2}=$ Variance

$=np(1-p)$
Hence,

$15=np(1-p)$

$15=20(1-p)$

$\dfrac{3}{4}=1-p$

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

If the mean and variance of a binomial distribution are

$\dfrac{15}{4}$ and

$\dfrac{15}{16}$ . The number of trials is

0%
$5$
0%
$2$
0%
$4$
0%
$6$

Explanation

$Mean=np$

$=\dfrac{15}{4}$

$Variance$

$=np(1-p)$

$=\dfrac{15}{16}$

Hence

$\dfrac{15}{16}=\dfrac{15}{4}(1-p)$

$p=\dfrac{3}{4}$
Therefore

$n=5$

A symmetrical die is rolled

$6$ times. If getting an odd number is a success, the probability of atmost

$5$ successes is

0%
$\dfrac{60}{64}$
0%
$\dfrac{63}{64}$
0%
$\dfrac{120}{164}$
0%
$\dfrac{10}{14}$

Explanation

Probability of getting atmost

$5$ success in

$6$ throw

$= 1-$ (probability of

$6$ success in

$6$ throw)

$= 1- [\dfrac{1}{2}]^6$

$= 1 - [\dfrac{1}{64}]$

$= \dfrac{63}{64}$

A symmetrical die is thrown four times and getting
a multiple of 2 is considered to be a success.
The mean and variance of success are

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

Explanation

The probability of getting a multiple of 2 for one throw is

$\dfrac{3}{6}$

$=\dfrac{1}{2}$
Hence the mean will be

$np$

$=4(\dfrac{1}{2})$

$=2$

Now the variance will be

$np(1-p)$

$=2(1-\dfrac{1}{2})$

$=1$

$A$ and

$B$ are two equally strong table tennis players, the probability that

$A$ beats

$B$ in exactly three games out of

$4$ games is

0%
$\displaystyle\frac{1}{6}$
0%
$\displaystyle\frac{1}{5}$
0%
$\displaystyle\frac{1}{4}$
0%
$\displaystyle\frac{1}{2}$

Explanation

$A$ and

$B$ are two equally strong table tennis players, probability of success that

$A$ beats

$B$

$= \displaystyle \frac {1}{2}$
The probability that

$A$ beats

$B$ in exactly three games out of

$4$ games

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

$= ^{4}C_{3}{ (\displaystyle \frac{1}{2}) }^{ 3 }{ (1- \dfrac{1}{2}) }^{4-3}$

$= 4 \times \displaystyle\frac {1}{16}$

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

In a binomial distribution mean is

$5$ , and variance

$4$ , then the number of trials is

0%
$100$
0%
$50$
0%
$25$
0%
$24$

Explanation

$\textbf{Step-1: Apply the relevant formula of mean & variance.}$

$\text{We have,}$

$\text{Mean is 5 and variance is 4}$

$\text{Mean = np =5}$

$\text{Variance=np(1-p)=4}$

$[\because$

$\textbf{Variance = npq & p+q =1]}$

$\textbf{Step-2: Use the above results & simplify.}$

$\Rightarrow \text{np(1-p) = 4}$

$\Rightarrow \text{5(1-p) = 4}$

$\Rightarrow \text{1-p}=\dfrac{4}{5}$

$\therefore \text{p}=\dfrac{1}{5}$

$\text{So, no. of trials (n) will be}$

$n=\dfrac{5}{p}$

$n=5(5)$

$n=25$

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

In six throws of a die, getting

$4$ or

$5$ is considered a success. The mean number of successes is

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

Explanation

Mean number of success =

$\mu = np$ where

$n$

$=$ total number of trials

$p$

$=$ probability of success

$n = 6$ For getting

$4$ or

$5$ is considered a success, probability of success

$=$

$\dfrac {2}{6}$
Mean number of success

$=$

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

If for a binomial distribution

$\bar{x}=\dfrac{6}{5}$ and the difference between mean and variance is

$\dfrac{6}{25}$ . The number of trials is

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

Explanation

$Mean=np=\dfrac{6}{5}$ [given]

Mean- variance

$=\dfrac{6}{25}$

$\Rightarrow np-np(1-p)=\dfrac{6}{25}$

$\Rightarrow np(p)=\dfrac{6}{25}$

$\Rightarrow n(p^2)=6(\dfrac{1}{5^2})$

$\therefore\,n=6$ and

$p=\dfrac{1}{5}$ .
Therefore no.of trials will be

$6$ .

If the standard deviation of the binomial distribution

$(q+p)^{16}$ is 2, then mean is

0%
$2$
0%
$4$
0%
$8$
0%
$7$

Explanation

${\textbf{Step -1: Stating the formulas of standard deviation and mean}}{\text{.}}$

${\text{We are given }}{(x + y)^{16}}{\text{, therefore, n = 16}}{\text{.}}$

${\text{We know that standard variation = }}\sqrt {npq}$

$\sqrt {npq} = 2 \Rightarrow npq = 4$

${\text{Mean for the distribution = }}np.$

${\textbf{Step -2: Finding the value of p}}{\text{.}}$

${\text{We know that }}p + q = 1,{\text{so we can write }}q = p - 1.$

$\Rightarrow np(1 - p) = 4$

$\Rightarrow {\text{16}}p(1 - p) = 4$

$\Rightarrow p(1 - p) = \dfrac{1}{4}$

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

${\text{Therefore, mean = np = 16}} \times \dfrac{1}{2} = 8.$

${\textbf{Hence,}}{\textbf{ The mean of the binomial distribution is 8}}{\textbf{. Thus, option C is correct.}}$

In a binomial distribution mean is

$4.8$ and variance is

$2.88$ , then the parameter

$n$ is

0%
$8$
0%
$12$
0%
$16$
0%
$20$

Explanation

Mean

$=np$

$=4.8$
Variance

$=np(1-p)$

$=2.88$
Therefore

$4.8(1-p)=2.88$

$1-p=0.6$

$p=0.4$
Hence

$n=\dfrac{4.8}{0.4}$

$=\dfrac{48}{4}$

$=12$
'n' being the number of trials.

Six unbiased coins are tossed once. The probability of obtaining atleast two heads is :

0%
$63/64$
0%
$57/64$
0%
$1/64$
0%
$1/32$

Explanation

Six unbiased coins are tossed once.
Probability of head in a coin =

$\frac {1}{2}$
The probability of obtaining at least two heads is given by

$1-$ [ The probability of obtaining no heads + The probability of obtaining one heads ]
The probability of obtaining at least two heads

$= 1- [ ^{6}C_{0} \left (\displaystyle \frac{1}{2}\right )^{0} \left (1-\displaystyle \frac{1}{2}\right )^{6-0} + ^{6}C_{1} \left (\displaystyle \frac{1}{2}\right )^{1} \left (1 - \displaystyle \frac{1}{2}\right )^{6-1}$ ]

$= \displaystyle 1 - [ \frac {1}{64} + \frac {3}{32}$ ]

$=\displaystyle 1 - [ \frac {7}{64}$ ]

$=\displaystyle \frac {57}{64}$

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%
$\left (\displaystyle \frac{1}{2}\right )^3$
0%
$\left (\displaystyle \frac{19}{20}\right )^3$
0%
$^{15}C_{3} \left (\displaystyle \frac{19}{20}\right )^{12} \left (\displaystyle \frac{1}{20}\right )^3$
0%
$\left (\displaystyle \frac{1}{20}\right )^3$

Explanation

A machine manufactures 5% defective screws. Probability of defective screws =

$\dfrac {5}{100} = \dfrac {1}{20}$
In sample of 15 screws;

$n = 15$
Probability that there are exactly 3 defectives =

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

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

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

The mean and variance of Binomial Distribution are 6,Then the parameters of the distribution are

0%
$\displaystyle 12, \frac{1}{2}$
0%
$\displaystyle 9, \frac{2}{3}$
0%
$\displaystyle 10, \frac{3}{5}$
0%
$\displaystyle 18, \frac{1}{3}$

Explanation

Mean

$=np$

$=6$
Variance

$=np(1-p)$

$=4$
Therefore

$4=6(1-p)$

$1-p=\dfrac{2}{3}$

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

$n=6(3)$

$=18$

$6$ symmetrical dice are thrown

$729$ times. The number of times you expect exactly three dice showing a four or five is

0%
$160$
0%
$240$
0%
$12$
0%
$6$

Explanation

$P(4 \ or \ 5) = \dfrac{1}{3} \ \ \, P(not \ 4 \ or \ 5 ) \ = \dfrac{2}{3}$

$\\ P(X=3) \ =\ 6C_3(\dfrac { 1 }{ 3 } *\dfrac { 1 }{ 3 } *\dfrac { 1 }{ 3 } *\dfrac { 2 }{ 3 } *\dfrac { 2 }{ 3 } *\dfrac { 2 }{ 3 } )\ \ =\ \dfrac { 160 }{ 729 }$

So, if Dices are rolled

$729$ times

$\mu \ = 729*(P(X=3))\ \ =\ \dfrac { 160 }{ 729 } *729\ =\ 160$

A fair coin is tossed 6 times. The variance of the number of heads is

0%
$3$
0%
$3/2$
0%
$3/4$
0%
$2$

Explanation

${\textbf{Step - 1: Writing given info}}$

${\text{Given that the coin is tossed 6 times}}$

$\therefore {\text{ Total sample space, n = 6}}$

${\text{Probablity of head appearing in one toss, p = }}\dfrac{{\text{1}}}{{\text{2}}}$

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

${\text{Variance = np(1 - p)}}$

$\Rightarrow {\text{ Variance = 6 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ }} \times {\text{ }}\left( {{\text{1 - }}\dfrac{{\text{1}}}{{\text{2}}}} \right)$

$\Rightarrow {\text{Variance = 6 }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}{\text{ }} \times {\text{ }}\dfrac{{\text{1}}}{{\text{2}}}$

$\Rightarrow {\text{ Variance = }}\dfrac{{\text{3}}}{{\text{2}}}$

$\textbf{Hence option B is correct}$

In a binomial distribution mean

$=\dfrac{11}{4}$ and variance

$=\dfrac{15}{16}$ , then the probability of success is

0%
$\dfrac{1}{2}$
0%
$\dfrac{29}{44}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{3}{4}$

Explanation

$Mean=np$

$=\dfrac{11}{4}$

$Variance=np(1-p)$

$=\dfrac{15}{16}$
Therefore

$\dfrac{11}{4}(1-p)=\dfrac{15}{16}$

$1-p=\dfrac{15}{44}$

$p=\dfrac{29}{44}$

$P$ being the probability of success.

In a binomial distribution with

$n=10, P=\displaystyle \frac{2}{5}$ , the mode of the B.D. is

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

Explanation

Usually the mode of a binomial

$B(N,P)$ distribution is equal to

$[(n+1)p]$

$(n+1)p = 22/5$
which is not an integer
Therefore Mode =

$[22/5] = 4$

$A$ student is given

$6$ questions in a true or false examination. If he gets

$4$ or more correct answers he passes the examination. The probability that he passes the examination is

0%
$5/32$
0%
$7/32$
0%
$11/32$
0%
$3/32$

Explanation

Probability for correct answer:

$\dfrac{1}{2}$
Probability for incorrect answer:

$\dfrac{1}{2}$
Selection of getting 4 correct answers:

$^{6}C_{4}$
Probability if he gets 4 or more correct answers

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

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

$={ \left( \dfrac { 1 }{ 2 } \right) }^{ 6 }\left [_{ 4 }^{ 6 }{ C } + _{ 5 }^{ 6 }{ C } + 1\right]$

$=\dfrac{11}{32}$

Team

$A$ has probability

$\displaystyle \frac{2}{3}$ of winning whenever it plays. If

$A$ plays

$4$ games the probability that

$A$ loses all the games is

0%
$\left (\displaystyle \frac{2}{3}\right )^4$
0%
$\left (\displaystyle \frac{1}{3}\right )^4$
0%
$\left (\displaystyle \frac{3}{4}\right )^4$
0%
$\left (\displaystyle \frac{1}{4}\right )^4$

Explanation

Probability of the team winning is

$\dfrac{2}{3}$ .
Hence the probability of the team losing is

$\dfrac{1}{3}$
Now the probability of A playing 4 games and loosing all will be

$\:^{4}C_{4}(\dfrac{1}{3})^{4}(\dfrac{2}{3})^{0}$

$=\dfrac{1}{3^4}$

$=\dfrac{1}{81}$

The mean of B.D. is 6 and its S.D. is Then the probability of 'x' successes is

0%
$\displaystyle ^{ 18 }C_{ x }{ \left( \frac { 1 }{ 3 } \right) }^{ 18-x }{ \left( \frac { 2 }{ 3 } \right) }^{ x }$
0%
$\displaystyle ^{ 12 }C_{ x }{ \left( \frac { 2 }{ 3 } \right) }^{ 12-x }{ \left( \frac { 1 }{ 3 } \right) }^{ x }$
0%
$\displaystyle ^{ 18 }C_{ x }{ \left( \frac { 2 }{ 3 } \right) }^{ 18-x }{ \left( \frac { 1 }{ 3 } \right) }^{ x }$
0%
$\displaystyle ^{ 18 }C_{ x }{ \left( \frac { 1 }{ 4 } \right) }^{ 18-x }{ \left( \frac { 3 }{ 4 } \right) }^{ x }$

Explanation

$S.D^{2}=$ Variance

$=np(1-p)$

$=2^2$

$=4$
Mean

$=np=6$
Therefore

$6(1-p)=4$

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

$n=3(6)$

$n=18$
Hence the probability of x success

$\:^{18}C_{x}\dfrac{1}{3}^{x}(\dfrac{2}{3})^{18-x}$

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 2 - 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 2 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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