MCQ Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 1

Consider 5 independent bernoulli's trials each with probability p of success. If the probability of one failure is greater than or equal to

${{31} \over {32}}$ , then p lies in the interval

0%
$\left[ {{{11} \over {12}},1} \right]$
0%
$\left[ {{1 \over 2},{3 \over 4}} \right]$
0%
$\left[ {{3 \over 4},{{11} \over {12}}} \right]$
0%
$\left[ {0,{1 \over 2}} \right]$

If the mean of Poisson distribution is

$\displaystyle \frac{1}{2}$ , then the ratio of

$P(X=3)$ to

$P(X=2)$ is

0%
1:2
0%
1:4
0%
1:6
0%
1:8

Explanation

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

$=\dfrac{\lambda^{3}(2!)}{3!(\lambda^{2})}$

$=\dfrac{\lambda}{3}$

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

$=\dfrac{1}{6}$
The ratio is

$1:6$

Mean and variance of binomial variable X are 2 and 1 respectively, then

$\displaystyle P\left ( X\geq 1 \right )$ is

0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{7}{8}$
0%
$\displaystyle \frac{4}{5}$
0%
$\displaystyle \frac{15}{16}$

Explanation

Here,

$np = 2, npq =1$

$\therefore p=q=\cfrac{1}{2}, n =4$

$\therefore P(X\geq 1) =1-P(X =0)=1-^4C_0 \left(\cfrac{1}{2}\right)^4 =1-\cfrac{1}{16}=\cfrac{15}{16}$

If for a

$BD$ the mean is

$6$ and standard deviation is

$\dfrac{1}{\sqrt{2}}$ , then the probability of success is

0%
$\dfrac{11}{12}$
0%
$\dfrac{10}{12}$
0%
$\dfrac{9}{12}$
0%
$\dfrac{8}{12}$

Explanation

$Mean=np$ where

$p$ is the probability of success.

$Variance=np(1-p)$

$Variance=(S.D)^{2}$

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

$\dfrac{1}{2}=6-6p$

$6p=\dfrac{11}{2}$

$p=\dfrac{11}{12}$

In a

$B.D.$

$n = 400, P=\displaystyle \frac{1}{5}$ . Its standard deviation is

0%
$10\times\sqrt{2}$
0%
$\displaystyle \frac{1}{800}$
0%
$4$
0%
$8$

Explanation

$B.D$ , Standard deviation

$=$

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

$n=400$ and

$p =$

$\dfrac { 1 }{ 5 }$

$\sigma = \sqrt {400 \times \dfrac {1}{5} \times \dfrac {4}{5}}$

$\sigma = 8$

$15$ throws of a die

$4$ or

$5$ is considered to be a success. The mean number of success is

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

Explanation

Probability of getting

$4$ or

$5$ in one throw of dice

$= p$

$= \dfrac {2}{6} = \dfrac {1}{3}$
Total no of throwing of dice =

$n = 15$
Mean number of success

$= np = 15 \times \dfrac {1}{3} = 5$

Suppose

$X$ follows binomial distribution with parameters

$n = 100$ and

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

$P(x=r)$ is maximum when

$r =$

0%
$49$
0%
$50$
0%
$33$
0%
$34$

Explanation

Maxima occurs at mean which is np for a binomial random variable.

$np=\dfrac { 100 }{ 3 } =33.33$
Rounding to nearest integer

If the mean of binomial distribution is

$\mu$ , then the variance lies in the interval

0%
$[\mu , 0 ]$
0%
$[0,\displaystyle \frac { \mu }{ 2 } ]$
0%
$\left [ 0,\mu \right ]$
0%
$\left [ 0,\sqrt{\mu } \right ]$

Explanation

Given,

$np=\mu$
we know, variance

$=npq=\mu q$
and

$0\leq q \leq 1$
Hence, range of variance is

$[0,\mu]$ .

If the first two terms of a Poisson distribution are equal to

$k$ , find

$k$ .

0%
$e$
0%
$\displaystyle \frac{1}{e}$
0%
$1$
0%
$2$

Explanation

Poisson's distribution implies

$P(X=n)=\dfrac{\lambda^{n}e^{-\lambda}}{n!}$
Where mean=variance=

$\lambda$
Hence

$P(x=0)=P(x=1)$

$\dfrac{\lambda^{0}e^{-\lambda}}{0!}=\dfrac{\lambda^{1}e^{-\lambda}}{1!}$

$\lambda=1$
Hence

$P(X=0)=P(X=1)$

$=\dfrac{1}{e}$

If the probability of selecting a bolt from

$400$ bolts is

$0.1$ , then the mean for the distribution is ________

0%
$0.09$
0%
$40$
0%
$36$
0%
$360$

Explanation

$\mu_x=np$

where

$\mu_x$ is mean of x

$n=$ sample size

$p=$ probability of success on an individual trial

Here

$n=400$

$p=0.1$

$\mu_x=np=400\times 0.1=40$

The value of

$B\left ( 3,4 ,\displaystyle \frac{1}{4}\right )$ is

0%
$\dfrac{5}{64}$
0%
$\dfrac{1}{16}$
0%
$\dfrac{3}{64}$
0%
$\dfrac{1}{64}$

Explanation

$B(3,4,\dfrac{1}{4})$ implies

$P(X=3)$ for

$4$ trials where the probability of success is

$p=\dfrac{1}{4}$
Hence the required probability is

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

$=\dfrac{1}{256}[4(3)]$

$=\dfrac{3}{64}$
Hence, the required probability is

$\dfrac{3}{64}$ .

Given that

$X\sim B (n = 10, p)$ . If

$E(X) = 8$ , then the value of

$p$ is

0%
$0.6$
0%
$0.7$
0%
$0.8$
0%
$0.4$

Explanation

Given,

$X\sim B (n = 10, p)$ ,

$E(X) = 8$

We know

$E(X) = np$

$\Rightarrow np = 8$

$\Rightarrow 10p = 8$

$\Rightarrow p = 0.8$

Hence, the correct answer from the given alternative is option

$C$ .

The mean and the variance of a binomial distribution are

$4$ and

$2$ respectively. Then the probability of

$2$ successes is:

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

Explanation

$\textbf{Step1: Find the probability of $$2$$ success}$

$\text{Given}$

$\text{Mean}$

$=4$

$\text{Variance}$

$=2$

$\because \text{Mean}$

$=np$

$\text{where }n= \text{number of trials}$

$\text{and }p= \text{probability of success}$

$\therefore \text{Mean}$

$=4$

$\Rightarrow np=4 \ldots(i)$

$\because \text{Variance}$

$=npq$

$\text{where }n= \text{number of trials}$

$\text{and }p= \text{probability of success}$

$\text{and }q= \text{probability of failure}$

$\therefore \text{Variance}$

$=2$

$\Rightarrow npq=2$

$\Rightarrow 4q=2$

$\text{[Using equation}(i)]$

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

$\text{As we know that }$

$p=1-q$

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

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

$\text{Putting the value of p in equation }(i)$

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

$\Rightarrow$

$n=8$

$\text{Probability of $$2$$ success }$

$P(x=2)={}^8C_{2}p^{2}q^{6}$

$=\dfrac{8!}{2!6!}\times$

$\dfrac{1}{4}\times$

$\dfrac{1}{2^{6}}$

$=$

$\dfrac{28}{256}$

$\textbf{Hence, the probability of $$2$$ is}$

$\dfrac{28}{256}$

Consider 5 independent Bernoulli trials each with probability of success

$\mathrm{p}$ . If the probability of at least one failure is greater than or equal to

$\displaystyle \frac{31}{32}$ , then

$\mathrm{p}$ lies in the interval.

0%
$(\displaystyle \frac{1}{2}\frac{3}{4}]$
0%
$(\displaystyle \frac{3}{4}\frac{11}{12}]$
0%
$[0,\displaystyle \frac{1}{2}]$
0%
$(\displaystyle \frac{11}{12},1]$

Explanation

$\mathrm{p}$ (probability of a success)

Probability of at least one failure = 1 - (Probability of no failure)
For 5 events, probability of no failure

$= p^5$
Hence,

$1-\mathrm{p}^5 \displaystyle \geq\frac{31}{32}$

$\Rightarrow \displaystyle \mathrm{p}^{5}\leq(\frac{1}{2})^{5}$

$\Rightarrow \displaystyle \mathrm{p}\leq\frac{1}{2}$
But

$\mathrm{p}\geq 0$
So,

$\mathrm{P}$ lies in the interval

$[0, \displaystyle \frac{1}{2}]$ .

An unbiased coin is tossed eight times. The probability of obtaining at least one head and at least one tail is.

0%
$\displaystyle\frac{63}{64}$
0%
$\displaystyle\frac{1}{2}$
0%
$\displaystyle\frac{127}{128}$
0%
$\displaystyle\frac{255}{256}$

Explanation

$p=P(H)=\dfrac {1}{2}$ ,

$q=P(T)=\dfrac{1}{2}$

$P$ (at least one head & one tail) =

$P(1 \leq H \leq 7$

$=1-[\displaystyle p(H)=0$ and

$p(H)=8]$

$=1-\left[2\cdot (1{/2})^8\right]$

$=1-\left[2\cdot \displaystyle\frac{1}{16\times 16}\right]$

$=1-\displaystyle\frac{1}{128}$

$=\displaystyle\frac{127}{128}$ .

A random variable X has Poisson distribution with meanThen P

$(x> 1.5)$ equals

0%
$\cfrac{2}{e^{2}}$
0%
$0$
0%
$1-\cfrac{3}{e^{2}}$
0%
$\cfrac{3}{e^{2}}$

Explanation

We know,

$\displaystyle P(x=k)=e^{-\lambda }.\frac{\lambda ^{k}}{K!}\, \,$
Here

$\lambda = 2$

$\therefore \displaystyle P(x>1.5)= P(x\geq 2)$

$=1-p(x=0)-P(x=1)$

$=1-e^{-\lambda }-e^{-\lambda }\left ( \dfrac{\lambda }{1!} \right )$

$=1-\dfrac{3}{e^{2}}$

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

0%
$\dfrac{1}{729}$
0%
$\dfrac{8}{9}$
0%
$\dfrac{8}{729}$
0%
$\dfrac{8}{243}$

Explanation

Probability of getting score

$9$ in a single throw

$\displaystyle=\frac { 4 }{ 36 } =\frac { 1 }{ 9 }$
Probability of getting score

$9$ exactly twice

$\displaystyle=^{ 3 }{ { C }_{ 2 }\times { \left( \frac { 1 }{ 9 } \right) }^{ 2 }\times \frac { 8 }{ 9 } }=\frac { 8 }{ 243 }$

In a binomial distribution

$\mathrm{B} ($ n,p

$=\displaystyle \dfrac{1}{4})$ , if the probability of at least one success is greater than or equal to

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

$\mathrm{n}$ is greater than

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

Explanation

Given

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

$\displaystyle \mathrm{P}(\mathrm{x}\geq 1)\geq\dfrac{9}{10}$

$\Rightarrow 1-\displaystyle \mathrm{P}(\mathrm{x}=0)\geq\dfrac{9}{10}$

$P(x=0)=^nC_0p^0q^n=\left(\dfrac 34\right)^n$

$\Rightarrow \displaystyle \dfrac{1}{10}\geq(\dfrac{3}{4})^{\mathrm{n}}$

$\Rightarrow (\displaystyle \dfrac{3}{4})^{\mathrm{n}}\leq\dfrac{1}{10}$

$\Rightarrow \mathrm{n}[\log_{10}{3}-\log_{10}{4}]\leq-1$

$\Rightarrow \displaystyle \mathrm{n}\geq\dfrac{1}{\log_{10}{4}-\log_{10}{3}}$

$X$ has a binomial distribution,

$B(n, p)$ with parameters

$n$ and

$p$ such that

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

$E(X)$ , the mean of variable

$X$ , is :

0%
$3-p$
0%
$\dfrac {p}{2}$
0%
$\dfrac {p}{3}$
0%
$2-p$

Explanation

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

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

$^{n}C_{3}p^{3}(1-p)^{n-3}$

$\Rightarrow\displaystyle\frac { n! }{ (n-2)!2! } =\frac { n! }{ (n-3)!3! } (\frac { p }{ 1-p } )$

$\Rightarrow 3(1-p)=p(n-2)$

$\Rightarrow np =3-p$

$\Rightarrow E(x)=3-p$

If the mean and the variance of a binomial variate X are 2 and 1 respectively, then probability that X takes a value greater than or equal to one is

0%
$\displaystyle \frac{1}{16}$
0%
$\displaystyle \frac{9}{16}$
0%
$\displaystyle \frac{3}{4}$
0%
$\displaystyle \frac{15}{16}$

Explanation

For binomial distribution,
Mean

$=np=2$ ....(1)
Variance

$=npq=1$

$\Rightarrow 2q=1$

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

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

$\because p+q=1$ )

$\Rightarrow n=4$

Now,

$P(X\ge 1)=1-P(X=0)$

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

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

$=\dfrac{15}{16}$

The mean of a binomial distribution is

$5$ , then its variance has to be :

0%
$> 5$
0%
$= 5$
0%
$< 5$
0%
$= 25$

Explanation

The mean of a binomial distribution is

$np$ , where

$n$ is the number of independent trials and

$p$ being the probability of success in any one of the trials.

The variance of the distribution is

$np(1 - p)$ , which is less than

$n$ since

$p$ has to be less than

$1$ .

Since the mean

$=np=5$ ,
Also,

$np(1-p) < np$
So, the variance would be less than

$5$ .

If in a binomial distribution

$n=4,P(X=0)=\cfrac { 16 }{ 81 }$ , then

$P(X=4)$ equals

0%
$\cfrac { 1 }{ 16 }$
0%
$\cfrac { 1 }{ 81 }$
0%
$\cfrac { 1 }{ 27 }$
0%
$\cfrac { 1 }{ 8 }$

Explanation

Let

$p$ be the probability of success and

$q$ that of failure in trial
Then,

$P(X=0)={ _{ }^{ 4 }{ C } }_{ 0 }{ p }^{ 0 }{ q }^{ 4 }=\cfrac { 16 }{ 81 }$

$\Rightarrow { q }^{ 4 }={ \left( \cfrac { 2 }{ 3 } \right) }^{ 4 }\left( \because { _{ }^{ n }{ C } }_{ 0 }=1 \right)$

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

$\Rightarrow p=1-\dfrac {2}{3}=\cfrac { 1 }{ 3 }$
Therefore,

$P(X=4)={ _{ }^{ 4 }{ C } }_{ 4 }{ p }^{ 4 }{ q }^{ 0 }={ p }^{ 4 }={ \left( \cfrac { 1 }{ 3 } \right) }^{ 4 }=\cfrac { 1 }{ 81 }$

For a binomial distribution

$n = 20,\ q = 0.75$ . The mean of the distribution is

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

Explanation

Mean

$=np$
and

$p=1-q$

$=0.25$
Hence,

Mean

$=np$

$=0.25(20)$

$=5$

If the mean of P.D. is 5, then the variance of the same distribution is

0%
$25$
0%
$10$
0%
$5$
0%
$15$

Explanation

The mean of P.D. is

$5$ .
In P.D, the variance and mean of the same distribution is same.
So, variance =

$5$

If a random variable

$X$ follows Bernoulli Distribution with mean

$2.4$ and variance

$1.44$ , the number of independent trials

$n$ is

0%
$10$
0%
$8$
0%
$6$
0%
$2$

Explanation

Given, mean

$=np = 2.4$ and variance

$=npq = 1.44$

$\because q=1-p$
Solving these we get

$p=\cfrac{2}{5}, q = \cfrac{3}{5}$ and

$n = 6$

In a binomial distribution

$AM=3$ , variance

$=4$ . The statement is

0%
True
0%
False
0%
We cannot say
0%
None

Explanation

Mean of $B(N,P) = np$

variance of $B(N,P) = npq$

So $\dfrac{variance}{mean} \ = q$

Here, $q = \dfrac{4}{3} > 1$

Hence the statement is $False$ .

A symmetrical die is tossed twice. Getting a four is considered to be a success. The mean and variance of number of successes are

0%
$\cfrac { 2 }{ 3 } ,\ \cfrac { 5 }{ 18 }$
0%
$\cfrac { 1 }{ 3 } ,\ \cfrac { 5 }{ 18 }$
0%
$\cfrac { 1 }{ 3 } ,\ \cfrac { 7 }{ 18 }$
0%
$\cfrac { 2 }{ 3 } ,\ \cfrac { 7 }{ 18 }$

Explanation

Here,

$n = 2, p = \cfrac{1}{6}, q = \cfrac{5}{6}$

$\therefore$ mean

$= np = \cfrac{1}{3}$ and variance

$=npq = \cfrac{5}{18}$

The parameter

$\lambda$ of poisson distribution is always

0%
zero
0%
1
0%
-1
0%
a finite positive value

Explanation

In Poisson distribution the parameter

$( > 0)$ , it is always greater than zero or a finite positive value.

The number of parameters of B.D. are

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

Explanation

The number of parameters of B.D. are

$n$ (number of trials) and

$p$ (probability of success).

$q$ is not a parameter as it is equal to

$1-p$ .

Hence there are

$2$ parameters in a binomial distribution.

lf the mean is

$\lambda$ and the variance is

$\sigma^{2}$ in a Poisson distribution, then

0%
$\displaystyle \lambda=\frac{1}{2}\sigma^{2}$
0%
$\displaystyle \sigma^{2}=\frac{1}{2}\lambda$
0%
$\lambda=\sigma^{2}$
0%
$\sigma^{2}=\lambda^{2}$

Explanation

For a Poisson distribution, mean and variance are equal .
i.e.

$\lambda = \sigma^2$

${\overline{x}}$ and

$\sigma^{2}$ are mean and variance of poisson distribution, then

0%
$\overline{x}>\sigma^{2}$
0%
$\overline{x}<\sigma^{2}$
0%
$\overline{x}=\sigma^{2}$
0%
$\overline{x}+\sigma^{2}=1$

Explanation

For a Poisson distribution, mean (

$\mu$ ) and variance (

$\sigma^2$ )are equal .
i.e.

$\mu = \sigma^2$

The S.D. of poisson distribuition whose mean is

$\lambda$ is

0%
$\lambda$
0%
$\sqrt{\lambda}$
0%
$\lambda^{2}$
0%
$\displaystyle \frac{1}{\sqrt{\lambda}}$

Explanation

Mean =

$\lambda$
In Poisson distribution, the expected value of a Poisson-distributed random variable is equal to that is mean and is equal to its variance.
Variance =

$\lambda$
S.D =

$\sqrt {variance} =\sqrt { \lambda }$

The mean of a B.D, is

$15$ and standard deviation is

$5$ , then which one of the following is correct.

0%
$\displaystyle p=\frac{2}{3}$
0%
$\displaystyle q=\frac{5}{3}$
0%
data's are absolutely correct
0%
data's are absolutely wrong

Explanation

We know that mean

$\mu = np=15$ and

$\displaystyle \sigma^{2}=npq=25$
Here variance is greater than mean which can't possible. Hence the statement is wrong.

In a poisson distribution, the probability of

$0$ success is

$10$ %. The mean of the distribution is equal to

0%
$\log_{10}e$
0%
$\log_{e}10$
0%
$0$
0%
$\dfrac{1}{10}$

Explanation

Given

$P(X=0)=.1$

$\Rightarrow e^{-\lambda}=\cfrac{1}{10}\therefore \lambda = \log_e 10$

The probability that a student is not a swimmer is

$\frac{1}{5}.$ Then the probability that out of the five students, four are swimmers, is

0%
$^{5}C_{4}(\frac{4}{5})^{2}(\frac{1}{2})$
0%
$(\frac{4}{5})^{4}(\frac{1}{5})$
0%
$^{5}C_{1}(\frac{1}{5})(\frac{4}{5})^{4}$
0%
None of these

Explanation

Here

$q=p$ (not swimmer )

$=\frac{1}{5}$

$\therefore =p$ (swimmer)

$=1-q$

$=1-\frac{1}{5}=\frac{4}{5}$

$\therefore$ Required probability

$= ^{5}C_{4}p^{4}q^{5-4}$

$= ^{5}C_{1}(\frac{4}{5})^{4}(\frac{1}{5})^{1}$

$= ^{5}C_{1}(\frac{1}{5})(\frac{4}{5})^{4}$

The probability that a student is not a swimmer is

$\displaystyle \frac{1}{5}.$ Then the probability that out of five students, four are swimmers is

0%
$\displaystyle ^{5}C_{4}\left ( \frac{4}{5} \right )^{4}\frac{1}{5}$
0%
$\displaystyle \left ( \frac{4}{5} \right )^{4}\frac{1}{5}$
0%
$\displaystyle ^{5}C_{1} \frac{1}{5}\left ( \frac{4}{5} \right )^{4}$
0%
none of these.

Explanation

Here

$p = \cfrac{4}{5}, n = 5, q = 1-p=\cfrac{1}{5}$
Hence required probability is

$= ^5C_4\left(\cfrac{1}{5}\right)\left(\cfrac{4}{5}\right)^4= ^5C_1\left(\cfrac{1}{5}\right)\left(\cfrac{4}{5}\right)^4$

The standard deviation of P.D. is 1.5, then its mean is

0%
1.5
0%
2
0%
2.25
0%
3.25

Explanation

Given

$\sigma =1.5$
We know that for a Poisson's distribution,mean and variance are equal.

$\mu=\sigma^2$

$\Rightarrow \mu =2.25$

If the mean of poisson distribution is

$16$ , then its S.D. is

0%
$16$
0%
$4$
0%
$10$
0%
$15$

Explanation

$S.D= \sqrt { \lambda }$

Mean

$= \lambda$

Therefore S.D.

$= \sqrt { 16 } \\=4$

If X is a poisson variable with parameter 0.09,then its S.D. is

0%
0.009
0%
0.3
0%
0.03
0%
0.09

Explanation

X is a Poisson variable with parameter 0.09.
In Poisson distribution, Poisson variable(X) is equal to expected value of X and also to its variance.
S.D =

$\sqrt {variance} = \sqrt { X } =\sqrt { 0.09 } =0.3$

10% of tools produced by a certain manufacturing process turn out to be defective. Assuming binomial distribution, the probability of 2 defective in simple of 10 tools chosen at random, is

0%
0.368
0%
0.194
0%
0.271
0%
None of these

Explanation

Let the event of a tool being defective be

$X$ .

$\therefore$ Probability of

$X$ occurring =

$p=\dfrac{1}{10}$

$\therefore q=1-p=\dfrac{9}{10}$

Now, as the process follows the binomial distribution, the probability of 2 tools being defective is

$^{10}C_2p^2q^{10-2}$

$={}^{10}C_2p^2q^8$

$=\dfrac{10!}{2!8!}\left(\dfrac{1}{10}\right)^2 \left(\dfrac{9}{10}\right)^8$

$=0.194$

This is the required solution.

A die is thrown 100 times, getting an even number is considered a success. Variance of number of successe, is

0%
$10$
0%
$20$
0%
$25$
0%
$50$

Explanation

This is an example of a binomial distribution.

Let

$A$ be the event of obtaining an even number when a fair dice is thrown.

$\therefore p_A = \dfrac{1}{2}$

$\therefore p_{A^c} = 1-p_A=\dfrac{1}{2}$

Variance over 100 trials =

$n\ p_A\ p_{A^c}=100\times\dfrac{1}{4}=25$

$16$ throws of a die getting an even number is considered a success, then the variance of the success is

0%
$4$
0%
$6$
0%
$2$
0%
$256$

Explanation

Probability of getting even number on throwing dice is

$p=\dfrac{1}{2}$ , which is probability of success

And probability of failure is

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

Hence variance in

$16$ throws is

$=npq=16\cdot \dfrac{1}{2}\cdot \dfrac{1}{2}=4$

Which one of the following may be the parameter of a binomial distribution?

0%
$np=2$ , $npq = 4$
0%
$n=4$ , $p = \dfrac{3}{2}$
0%
$n=8$ , $p=1$
0%
$np=10$ , $npq = 8$

Explanation

For binomial distribution

$0<p,q<1$ ,

Now in (d) part,

$np=10$ and

$npq=8$ gives

$q=0.8,p=0.2,n=50$ .

So they can be the parameters of a binomial distribution.

Hence, D is correct.

In a binomial distribution, the occurrence and the non-occurrence of an event are equally likely and the mean is

$6$ . The number of trials required is

0%
$15$
0%
$12$
0%
$10$
0%
$6$

Explanation

The occurrence and non occurrence of an event are equally likely.

So, here

$p=0.5$ and

$q=0.5$ .

Mean

$=np$ ...... [Given]

$\Rightarrow 6=n(0.5)$

$\Rightarrow n=12$

So, the required number of trials is

$12$ .

Hence, B is correct.

Let

$X\sim B(n,p)$ , if

$E(X)=5,Var(X)=2.5$ , then

$P(X< 1)=$ ...................

0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 11 }$
0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 10 }$
0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 6 }$
0%
${ \left( \cfrac { 1 }{ 2 } \right) }^{ 9 }$

Explanation

$E(X)=np\\ Var(X)=npq$

where

$n=$ number of trials,

$p=$ probability of a success,

$q=$ probability of a failure

Given

$E(X)=5$

$Var(X)=2.5$

$\Rightarrow n=10,\quad p=\dfrac { 1 }{ 2 } ,q=\dfrac { 1 }{ 2 }$

$P(X<1)={ \left( \dfrac { 1 }{ 2 } \right) }^{ 10 }$

$X$ is a binomial variate with the range {

$0,1,2,3,4,5,6$ } and

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

$p$ of

$X$ is :

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

Explanation

According to the problem

where,

$n=6$

let us consider the problem

$^6C_2P^2Q^4=4.^6C_4P^4C^2$

$Q^2=4P^2$

$(1-P)^2=4P^2$

$3P^2+2P-1=0$

$(P+1)(3P-1)=0$

Hence,

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

$P=-1$

But

$P$ can not be negative

Therefore, the answer is

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

A coin is tossed

$6$ times, find the probability of getting no tails.

0%
$\dfrac{1}{64}$
0%
$\dfrac{5}{64}$
0%
$\dfrac{1}{32}$
0%
$0.25$

Explanation

Given: A coin tossed 6 times.

To find: the probability of getting no tails

Sol: Using Binomial Theorem

$^nC_r (p)^r (q)^{n-r}$

Probability of getting a tail is

$p=\dfrac 12$

And, the probability of NOT getting a tail is

$q=\dfrac 12$

Here

$n = 6, r=0$

$^6C_0\times \left(\dfrac 12 \right)^0\times \left(\dfrac 12\right)^6 = 1\times$

$2^{-6}=\dfrac {1}{64}$

is the required probability of getting no tails.

Four cards are drawn from a deck of

$52$ cards, the probability of only two cards being spade is......

0%
$\dfrac{27}{128}$
0%
$\dfrac{54}{128}$
0%
$\dfrac{81}{128}$
0%
$\dfrac{31}{128}$

Explanation

Let

$X$ : be the number of spade cards.

Drawing is a Bernoulli trial.

So,

$X$ has the binomial distribution,

$P(X=x)=^nC_xq^{n-x}p^x$

where n = number of cards drawn = 4, p = probability of getting a spade card =

$\dfrac {13}{52}=\dfrac 14$

Hence,

$q=1-p=1-\dfrac 14=\dfrac 34$

Hence

$P(X=x)=^4C_x\left(\dfrac 34\right)^{4-x}\left(\dfrac 14\right)^x$

Hence the probability of only two cards being spade is

$P(X=2)=^4C_2\left(\dfrac 34\right)^{4-2}\left(\dfrac 14\right)^2=6\times \left(\dfrac 34\right)^2\left(\dfrac 14\right)^2=6\times \dfrac {9}{16}\times\dfrac {1}{16}=\dfrac {27}{128}$

Four cards are drawn from a deck of

$52$ cards, the probability of only

$1$ card being spade is:

0%
$\dfrac{108}{256}$
0%
$\dfrac{54}{128}$
0%
$\dfrac{81}{256}$
0%
$\dfrac{31}{256}$

Explanation

Let X: be the number of spade cards.

Drawing is a Bernoulli trial.

So, X has the binomial distribution,

$P(X=x)=^nC_xq^{n-x}p^x$

where n = number of cards drawn = 4, p = probability of getting a spade card =

$\dfrac {13}{52}=\dfrac 14$

Hence,

$q=1-p=1-\dfrac 14=\dfrac 34$

Hence

$P(X=x)=^4C_x\left(\dfrac 34\right)^{4-x}\left(\dfrac 14\right)^x$

Hence the probability of only one cards being spade is

$P(X=1)=^4C_1\left(\dfrac 34\right)^{4-1}\left(\dfrac 14\right)^1=4\times \left(\dfrac 34\right)^3\left(\dfrac 14\right)^1=\dfrac {108}{256}$

If the mean and variance of a binomial variable

$X$ are

$2$ and

$1$ respectively, then

$P\left( {X \geq 1} \right) =$

0%
$\dfrac{2}{3}$
0%
$\dfrac{15}{16}$
0%
$\dfrac{7}{8}$
0%
$\dfrac{4}{5}$

Explanation

For Binomial distribution, Mean

$(\mu) =np$ and Variance

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

According to question, Mean = 2 and Variance = 1

Thus,

$np=2.....(1)\quad np(1-p)=1.......(2)$

Putting the value of np from first equation in the second equation.

$2\times(1-p)=1\Rightarrow1-p=\dfrac{1}{2}\Rightarrow p=\dfrac{1}{2}$

Now, putting the value of p in equation(1),

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

$\because \Sigma P(X_i)=1$

$P(X\ge1)=1-P(X<1)=1-P(X=0)=1-{^4C_0p^0(1-p)^{4-0}}$

$=1-\dfrac{4!}{0!(4-0)!}\times \left(\dfrac{1}{2}\right)^0\times \left(\dfrac{1}{2}\right)^4=1-\dfrac{1}{16}=\dfrac{15}{16}$

CBSE Questions for Class 12 Commerce Applied Mathematics Standard Probability Distributions Quiz 1 - 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 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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