If $$X$$ is a random variable with the following probability distribution given below: $$X=x$$ $$0$$ $$1$$ $$2$$ $$3$$ $$P(X=x)$$ $$k$$ $$3k$$ $$3k$$ $$k$$ Then the value of $$k$$ and its variance are:

$$\dfrac{1}{8},\dfrac{22}{27}$$

$$\dfrac{1}{8},\dfrac{23}{27}$$

MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 3

If the variance of the random variable

$X$ is

$4$ , then the variance of the random variable

$5X+10$ is

0%
$100$
0%
$10$
0%
$50$
0%
$25$

Explanation

If a random variable X is adjusted by multiplying by the value b and adding the value a, then the variance is affected as follows:

${ \sigma }_{ a+bX }^{ 2 }={ b }^{ 2 }{ \sigma }^{ 2 }$
If the variance of the random variable X =

${ \sigma }^{ 2 }$ = 4
For the random variable

$5X+10$ ,

${ \sigma }_{ 10+5X }^{ 2 } = {b }^{ 2 }{ \sigma }^{ 2 } = { 5 }^{ 2 }{4} = 100$

${ \sigma }_{ 10+5X }^{ 2 } = 100$

From a lot of 10 items containing 3 defectives, a sample of 4 items is drawn at random without replacement. The expected number of good items is

0%
$3$
0%
$2.8$
0%
$1.2$
0%
$1.8$

Explanation

Probability of good items =

$7/10$
Total number of items drawn =

$4$
Expected number of good items =

$4*7/10 = 2.8$

If the probability distribution of a random variable x is

$X=x_{1}:\quad -2 \quad -1 \quad 0 \quad 1 \quad 2 \quad 3$

$p(X=x_{1}): 0.1 \ \ \ \ \ \ k \ \ \ \ 0.2 \ \ \ 2k \ \ 0.3 \ \ \ k$ then the mean of x is

0%
$0.6$
0%
$0.8$
0%
$1.0$
0%
$0.3$

Explanation

$\sum _{ i=-2 }^{ 3 }{ { p }_{ i } } =1\\ 0.1+k+0.2+2k+0.3+k=1\\ 4k=1-0.6=0.4\\ k=0.1\\ Mean=\sum _{ i=-2 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =-0.2-k+2k+0.6+3k\\ =4k+0.4=0.4+0.4\\ =0.8\\$

The Probability distribution of a random variable

$X$ is given by

$P(X=x) = 0.1 ,0.1, 0.1 ,0.3 ,0.4$ for

$(X=x) = 4,3,2,1,0$ . The variance of

$X$ is

0%
$1. 76$
0%
$2.45$
0%
$3.2$
0%
$4.8$

Explanation

Mean =

$\sum x_{i}p_{i}=1.24$

variance =

$\sum (x_{i}-\bar{x})^{2}p_{i}$

$=1.44*0.4+0.2^{2}*0.3+0.8^{2}*0.1+1.8^{2}*0.1+2.8^{2}*0.1$

$=1.76$

If a random variable

$x$ has the following probability distribution

$X=x_{i} : \quad \quad0 \ \quad\quad 1 \ \quad\quad 2 \ \quad 3$

$P(X=x_{i}): \ 2K^{2},3K^{2},5K^{2},6K^{2}$
then the value of

$K$ is

0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{-1}{4}$
0%
$\displaystyle \pm\frac{1}{4}$
0%
$\displaystyle \frac{1}{2}$

Explanation

Using the fact ,

$\sum P(X) = 1$

$\Rightarrow 2K^2+3K^2+5K^2+6K^2=1$

$\Rightarrow K^2 = \cfrac{1}{16} \Rightarrow K = \pm \cfrac{1}{4}$

The value of

$C$ for which

$P( X=k)= Ck^{2}$ can serve the probability function of a random variable

$X$ that takes values

$0, 1, 2, 3, 4$ is

0%
$\displaystyle \frac{1}{30}$
0%
$\displaystyle \frac{1}{10}$
0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{1}{15}$

Explanation

$\displaystyle \sum _{ k=0 }^{ 4 }{ P\left( X=k \right) } =1\Rightarrow \sum _{ k=0 }^{ 4 }{ c{ k }^{ 2 } } =1$

$\displaystyle \Rightarrow c\left( { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 }+{ 4 }^{ 2 } \right) =1\Rightarrow c=\frac { 1 }{ 30 }$

If the probability distribution of a random variable

$x$ is

$X=x_{1}:\quad -2 \quad -1 \quad 0 \quad 1 \quad 2 \quad 3$

$p(X=x_{1}): 0.1 \ \ \ \ \ \ k \ \ \ \ 0.2 \ \ \ 2k \ \ 0.3 \ \ \ k$ , then the variance of

$x$ is

0%
$2.16$
0%
$2.8$
0%
$\sqrt{2.16}$
0%
$\sqrt{2.8}$

Explanation

$\sum _{ i=-2 }^{ 3 }{ { p }_{ i } } =1\\ 0.1+k+0.2+2k+0.3+k=1\\ 4k=1-0.6=0.4\\ k=0.1\\ E[X]=\sum _{ i=-2 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =-0.2-k+2k+0.6+3k\\ =4k+0.4=0.4+0.4\\ =0.8\\ E[{ X }^{ 2 }]=\sum _{ i=-2 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =0.4+k+2k+1.2+9k=2.6\\ Variance = E[{ X }^{ 2 }]-{ E[X] }^{ 2 }=2.6-{ 0.8 }^{ 2 }= 2.16\\$

If a random variable

$x$ has the following probability distribution

$X=x_{i} : \quad \quad0 \ \quad\quad 1 \ \quad\quad 2 \ \ \quad 3$

$P(X=x_{i}): \ 2K^{2}\ \ 3K^{2}\ \ \ 5K^{2}\ \ 6K^{2}$

Then find the mean.

0%
$\displaystyle \frac{33}{16}$
0%
$\displaystyle \frac{31}{16}$
0%
$\displaystyle \frac{35}{16}$
0%
$\displaystyle \frac{29}{16}$

Explanation

Using

$\sum P(X) = 1\Rightarrow 16 K^2 = 1\Rightarrow K^2 = \cfrac{1}{16}$

Thus mean

$= \sum x_iP(X=x_i) = 0+P(X=1)+2P(X=2)+3P(X=3)$

$0+3K^2+10K^2+18K^2$

$=31K^2 = \cfrac{31}{16}$

The probability distribution of a random variable

$X$ is given below, then

$K =$

$X=x_{1}: \quad 1 \quad , 2, \quad 3, \quad 4$

$p(X=x_{1}):2k , 4k \ \ ,3k , \ \ \ k$

0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{10}$
0%
$\dfrac{1}{15}$

Explanation

$\displaystyle\sum_{1}^{4}P(X=x)=1$

$K(2+4+3+1)=1$

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

A random variable

$X$ takes the values

$-1, 0, +1$ . Its mean is

$0.6$ . If

$P(X=0)=0.2$ , then find

$P(X=1)$ and

$P(X=-1)$

0%
$0.2, 0.8$
0%
$0.3, 0.7$
0%
$0.7, 0.1$
0%
$0.4, 0.2$

Explanation

We know

$\sum P(X)=1\Rightarrow P(X=-1)+P(X=0)+P(X=1)=1\Rightarrow P(X=-1)+P(X=1)=0.8$ -----(i)

Given mean is

$10.6$

$\Rightarrow (-1)(P(X=-1))+(0)(P(X=0))+1(P(X=1))=0.6$

$\Rightarrow -P(X=-1)+P(X=1)=0.6$ ------(ii)

Adding (i) and (ii)

$2P(x=1)=1.4$

$P(x=1)=0.7$

$P(x=-1)=0.8-P(x=1)=0.8-0.7=0.1$

$\therefore\ P(X=-1) =0.1, P(X=1)=0.7$

The value of k, if the probability distribution of a

$X=x:1 , 2 , 3$ random variable X is

$p(X=x):\dfrac{1}{k} , \dfrac{2}{k} , \dfrac{3}{k}$ is

0%
$\dfrac{1}{6}$
0%
$6$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{\sqrt{6}}$

Explanation

concept : { sum of probability distribution against random variable always is equal to 1 }

$\sum_{i=1}^{n}P(x_i)=1$

$\frac{1}{k}+\frac{2}{k}+\frac{3}{k}=1$

$\frac{6}{k}=1$

$k=6$

value of k is 6.

A random variable

$X$ has the following probability distribution, then

$C =$

$X=x: 1 ,2 , 3 , 4$ ,

$P(X=x) :C , 2C, 3C ,4C$

0%
$0.1$
0%
$0.2$
0%
$10$
0%
$20$

Explanation

$\displaystyle\sum_{1}^{4}P(X=x)=1$

$C(2+4+3+1)=1$

$C=0.1$

A random variable

$X$ takes the values

$0,\ 1,\ 2,\ 3$ and its mean is

$1.3$ . If

$P(X=3)=2P(X=1)$ and

$P(X=2)=0.3$ , then

$P(X=0)=$

0%
$0.1$
0%
$0.2$
0%
$0.3$
0%
$0.4$

Explanation

$E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0*P(X=0)+1*P(X=1)+2*P(X=2)+3*P(X=3)=1.3\\ 1*P(X=1)+2*0.3+3*2*P(X=1)=1.3\\ P(X=1)=0.1\\ Now\quad using\\ \sum _{ i=0 }^{ 3 }{ { p }_{ i } } =1=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\ P(X=0)+0.1+0.3+0.2=1\\ P(X=0)=0.4$

The variance of the random variable

$x$ whose probability distribution is given by

$X=x : \quad \quad 0 \quad 1 \quad 2 \quad 3$

$p(X=x): \ \ \ \dfrac{1}{3} \ \ \ \dfrac{1}{2} \ \ \ 0 \ \ \ \dfrac{1}{6}$

0%
$0.5$
0%
$1$
0%
$1.5$
0%
$2.0$

Explanation

$E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0+\dfrac { 1 }{ 2 } +0+\dfrac { 1 }{ 2 } =1\\ E[{ X }^{ 2 }]=\sum _{ i=0 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =0+\dfrac { 1 }{ 2 } +0+\dfrac { 9 }{ 6 } =2\\ Variance= E[{ X }^{ 2 }]-{ E[X] }^{ 2 }= 2-{ 1 }^{ 2 }=1$

$X$ is a random variable with the following probability distribution given below:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$k$	$3k$	$3k$	$k$

Then the value of

$k$ and its variance are:

0%
$\dfrac{1}{8},\dfrac{22}{27}$
0%
$\dfrac{1}{8},\dfrac{23}{27}$
0%
$\dfrac{1}{8},\dfrac{8}{9}$
0%
$\dfrac{1}{8},\dfrac{3}{4}$

Explanation

$\sum _{ i=-2 }^{ 3 }{ { p }_{ i } } =1\\ k+3k+3k+k=1\\ \Rightarrow k=\dfrac { 1 }{ 8 } \\ E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0\times P(X=0)+1\times P(X=1)+2\times P(X=2)+3\times P(X=3)\\ 3k+6k+3k=12k=\dfrac { 12 }{ 8 } \\ E[{ X }^{ 2 }]=\sum _{ i=0 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =3k+12k+9k=24k=3\\ Variance = E[{ X }^{ 2 }]-{ E[X] }^{ 2 } = 3-\left( \dfrac { 3 }{ 2 }\right)^{ 2 } = 3-\dfrac { 9 }{ 4 } =\dfrac { 3 }{ 4 } \\$

A random variable X follows the following distribution

$X=x_{i}: \quad \ \ 1 , \ 2 , \ 3, \ 4$

$p(X=x_{i}):\dfrac{2}{6}, \dfrac{3}{6} , \dfrac{0}{6}, \dfrac{1}{6}$
, then the mean and variance are

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

Explanation

$E[X]=\sum _{ i=1 }^{ 4 }{ { x }_{ i }{ p }_{ i } } =1*P(X=1)+2*P(X=2)+3*P(X=3)+4*P(X=4)\\ =\dfrac { 2 }{ 6 } +\dfrac { 6 }{ 6 } +\dfrac { 0 }{ 6 } +\dfrac { 4 }{ 6 } =2\\ E[{ X }^{ 2 }]=\sum _{ i=0 }^{ 3 }{ { x }^{ 2 }{ p }_{ i } } =\dfrac { 2 }{ 6 } +\dfrac { 12 }{ 6 } +\dfrac { 0 }{ 6 } +\dfrac { 16 }{ 6 } =5\\ Variance\quad =\quad E[{ X }^{ 2 }]-{ E[X] }^{ 2 }\quad =\quad 5-{ 2 }^{ 2 }\quad =\quad 1$

A random variable X has its range

${1, 2, 3}$ with respective probabilities

$P(X=1)=K, \ P(X=2)=2K, P(X=3)=3K,$ then the value of K is

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

Explanation

$\displaystyle\sum_{1}^{3}P(X=x)=1$

$K(1+2+3)=1$

$K=\dfrac{1}{6}$

Let X be the random variable with the probability distribution function

$f(x)=\dfrac{e^{-4}4^{x}}{x!};x=0,1,2,3,....$ then the standard deviation of X is

0%
$2$
0%
$4$
0%
$16$
0%
$\sqrt{2}$

Explanation

The above distribution is Poisson Distribution of type

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

$\lambda=4$ .
Now

$Mean=Variance=\lambda$ in a Poisson Distribution.
Hence

$Variance=\lambda=4$
Or

$S.D^{2}=Variance=4$
Or

$S.D=\sqrt{Variance}=\sqrt{4}$

$=2$ .

A random variable

$X$ has its range

$X = {0, 1, 2}$ with respective probabilities

$P(X=0)=3K^{3}, P(X=1)=4K-10K^{2}, P(X=2)=5K-1$ , then the value of

$K$ is

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

Explanation

Given,

$P(X=0)=3K^{3}, P(X=1)=4K-10K^{2}, P(X=2)=5K-1$

Now,

$P(X=0)+P(X=1)+P(X=2)=1$

$\Rightarrow 3K^{3}-10K^2+9K-2=0$

$\Rightarrow (K-1)(3K^2-7K+2)=0$

$\Rightarrow (K-1)(K-2)(3K-1)=0$

$\Rightarrow K=1,2, \dfrac{1}{3}$

But

$K=1,2$ not possible as at these values of

$K$ , the probabilities

$P(X=0)$ ,

$P(X=1)$ ,

$P(X=2)$ does not lie between

$0$ and

$1$ .

If a random variable X takes values

$(-1)^{k}2^{k}/k;k=1,2,3,....$ with probabilities

$P(X=k)=\dfrac{1}{2^{k}}$ then E(X)=

0%
$log \ 2$
0%
$log \ e$
0%
$log \ \left ( \dfrac{1}{2} \right )$
0%
$log \ \left ( \dfrac{1}{4} \right )$

Explanation

$E[X]=\sum _{ k=1 }^{ \infty }{ \dfrac { { (-1) }^{ k }{ 2 }^{ k } }{ k } \dfrac { 1 }{ { 2 }^{ k } } } =\quad \dfrac { { (-1) }^{ k } }{ k } =\dfrac { -1 }{ 1 } +\dfrac { 1 }{ 2 } +\dfrac { -1 }{ 3 } +\dfrac { 1 }{ 4 } +........\\ =-1+\dfrac { 1 }{ 2 } -\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 4 } +.........\\ =\quad -\log { 2= } \log { \dfrac { 1 }{ 2 } }$

If F(x) is the cumulative distributive function of a random variable x whose range is from

$-\alpha$ to

$+\alpha$ , then

$P(X< -\alpha )$

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

Explanation

We have for a random variable

$X$ ,

$F(x)=P(X\le x)= 0$ for

$x<0$ (since the random variable cannot assume negative values, so its probability is 0)
So,

$P(X<-\alpha)=0$

A random variable

$X$ takes value

$0, 1, 2$ . Its mean is

$1.3$ . If

$P(X=0)=0.2$ , then

$P(X=2)=$

0%
$0.3$
0%
$0.4$
0%
$0.5$
0%
$0.2$

Explanation

Given Mean

$= 1.3,P(x=0)=0.2$

$0\times P(X=0)+ 1\times P(X=1)+ 2\times P(X=2)= 1.3$

Let

$P(X=2)= p$

$P(X=1)= 1- P(X=0)- p= 0.8-p$

Substituting in original equation

$P(X=1)+ 2P(X=2)= 1.3$

$0.8-p+2p= 1.3$

$p=0.5$

Let the discrete random variable

$X = x$ has the probabilities given by

$\displaystyle \frac{x}{6}$ for

$x=0, 1, 2, 3$ , then its mean is

0%
$\displaystyle \frac{1}{3}$
0%
$\displaystyle \frac{5}{3}$
0%
$\displaystyle\frac{7}{3}$
0%
$\displaystyle \frac{9}{3}$

Explanation

$E[X]=\sum _{ i=0 }^{ 3 }{ { x }_{ i }{ p }_{ i } } =0*P(X=0)+1*P(X=1)+2*P(X=2)+3*P(X=3)\\ =0+\dfrac { 1 }{ 6 } +\dfrac { 2*2 }{ 6 } +\dfrac { 3*3 }{ 6 } =\dfrac { 14 }{ 6 }$

If the range of the random variable

$X$ is from

$-\alpha$ to

$+\alpha$ , the limits of

$F(X)$ are

0%
$0$ to $\alpha$
0%
$-\alpha$ to $3$
0%
$-1$ to $+1$
0%
$0$ to $1$

Explanation

We have for a random variable

$X$ ,

$F(x)=P(X\le x)= 0$ for

$x<0$ (since the random variable cannot assume negative values, so its probability is 0)

$F(x)=P(X\le x)$ for

$x \ge 0$
Since, the value of probability lies between 0 and 1.
So,

$0 < F(x) \le 1$
Hence, limits of

$F(x)$ are 0 to 1.

The range of a random variable

$X$ is

$1, 2, 3, 4 ....$
and the probabilities are given by

$\displaystyle p(X=k)=\frac{c^{k}}{k!}k=1,2,3,4....,$ , then the value of

$C$ is

0%
$2 2$
0%
$\log e$
0%
$\log_e 2$
0%
4

Explanation

We know,

$\sum P(X)=1$

$\Rightarrow \displaystyle c+\frac{c^2}{2!}+\frac{c^3}{3!}+........... = 1$

$\Rightarrow \displaystyle 1+c+\frac{c^2}{2!}+\frac{c^3}{3!}+........... =1+1=2$

$\displaystyle \Rightarrow e^c = 2\Rightarrow c = \log_e 2$

A person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score

$X$ is observed, then the range of

$X$ is

0%
${0,3,6}$
0%
${-3,0,3}$
0%
${-3,0,3,6}$
0%
${-3,3,6}$

Explanation

$3$ coins are tossed . There can be

$4$ possible outcomes

$1.\ 3 T , 0 H$ . Score

$= -3$

$2.\ 2 T , 1 H$ . Score

$= -2 + 2 = 0$

$3.\ 1 T , 2 H$ . Score

$= -1 +4 = 3$

$4.\ 0 T, 3 H$ . Score

$= 0 + 6 = 6$
Hence, the range of

$X$ is

$\{ -3, 0,3,6 \}$

A discrete random variable

$X$ , can take all possible integer values from

$1$ to

$K$ , each with a probability

$1/K$ . Its mean is

0%
$K$
0%
$K+1$
0%
$K/2$
0%
$K/4$

Explanation

Mean

$=\mu=\sum_{{i=1}}^{K}x_ip_i$

$=1\times 1+2 \times \dfrac{1}{2}+3 \times \dfrac{1}{3}+.....K \times \dfrac{1}{K}$

$=1+1+....upto\ K\ times$
Mean

$\mu=K$

In a World Cup final match against Srilanka, for six times Sachin Tendulkar hits a six out of 30 balls he plays. What is the probability that in a given throw, the ball does not hit a six?

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

The probability distribution of random variable X: number of heads , when a fair coin is tossed twice is given by

$x$	$0$	$1$	$2$
p(x)	$p_1$	$p_2$	$p_3$

then

0%
$\sum p_i=0$
0%
$\Pi p_i=1$
0%
$\sum p_i=1$
0%
$\sum p_i=3$

Explanation

We know that, the probability distribution of a random variable is a list of probabilities associated with each of its possible values.

Suppose a random variable

$X$ may take

$k$ different values, with the probability that

$X=x_i$ defined to be

$P(X=x_i)=p_i$ .

The probabilities

$p_i$ must satisfy the following:

$0\leq p_i \leq 1$ for each

$i$

$p_1+p_2+...+p_k=1. \:That \:is\: \sum p_i=1$

Thus if the probability distribution of random variable

$X$ : number of heads , when a fair coin is tossed twice is given then

$\sum p_i=1$

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 some demand is refused is

0%
$1.12 \times e^{-1.5}$
0%
$1.2.5 \times e^{-1.5}$
0%
$1-3.625 \times e^{-1.5}$
0%
$3.625 \times e^{-1.5}$

Explanation

let

$X$ be a random variable denoting the number of demands for a car on any day.
So,

$X$ is Poisson distributed with the parameter

$\mu =1.5$
Therefore, the probability mass function is

$\displaystyle P(X=i)=\frac { { e }^{ \mu }{ \mu }^{ t } }{ i! } ={ e }^{ -1.5 }\frac { { \left( 1.5 \right) }^{ 2 } }{ i! } ;i=0,1,2$
Now, the proportion of days in which neither car is used is actually the probability of there being no demand of cars which is given by

$\displaystyle P(X=0)=\frac { { e }^{ -1.5 }{ \left( 1.5 \right) }^{ 2 } }{ 0! } ={ e }^{ -1.5 }$
and the proportion of days on which some demand is refused, is the probability that the number of demands become more than

$2$ and is given by

$P(X>2)=1-P(X\le 2)$

$=1-{P(X=0)+P(X=1)+P(X=2)}$

$\displaystyle=1-{\frac { { e }^{ 1.5 }{ \left( 1.5 \right) }^{ 0 } }{ 0! } +\frac { { e }^{ -1.5 }{ \left( 1.5 \right) }^{ 1 } }{ 1! } +\frac { { e }^{ -1.5 }{ \left( 1.5 \right) }^{ 2 } }{ 2! } }$

$=1-3.625\times { e }^{ -1.5 }$

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 3 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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