Two probability distributions of the discrete random variable $$X$$ and $$Y$$ are given below. $$X$$ $$0$$ $$1$$ $$2$$ 3 $$P\left( X \right)$$ $$\dfrac { 1 }{ 5 } $$ $$\dfrac { 2 }{ 5 } $$ $$\dfrac { 1 }{ 5 }$$ $$\dfrac { 1 }{ 5 }$$ $$Y$$ $$0$$ $$1$$ $$2$$ $$3$$ $$P\left( Y \right)$$ $$\dfrac { 1 }{ 5 }$$ $$\dfrac { 3 }{ 10 } $$ $$\dfrac { 2 }{ 5 } $$ $$\dfrac { 1 }{ 10 }$$ Then

$$E\left( { Y }^{ 2 } \right) =2E\left( X \right)$$

$$E\left( { Y }^{ 2 } \right) =E\left( X \right)$$

$$E\left( Y \right) =E\left( X \right)$$

$$E\left( { X }^{ 2 } \right) =2E\left( Y \right) $$

MCQ Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 7

The probability distribution of a discrete random variable X is given below.

The value of K is

0%
8
0%
16
0%
32
0%
48

Explanation

We know that,

$\sum { p }(x) =1$

$\Rightarrow \dfrac{5}{k}+\dfrac{7}{k}+\dfrac{9}{k}+\dfrac{11}{k}=1$

$\Rightarrow \dfrac{32}{k}=1$

$\therefore K = 32$

For the following probability distribution

$E(X^2)$

0%
3
0%
5
0%
7
0%
10

Explanation

$E(X^2)=\sum X^2P(X)=1\cdot\dfrac{1}{10}+4\cdot\dfrac{1}{5}+9\cdot\dfrac{3}{10}+16\cdot\dfrac{2}{5}$

$=\dfrac{1}{10}+\dfrac{4}{5}+\dfrac{27}{10}+\dfrac{32}{5}$

$\dfrac{1+8+27+64}{10}=10$

Two cards are drawn randomly from a well-shuffled deck of

$52$ cards. If

$X$ denotes the number of aces, then find mean of

$X$ :

0%
$\cfrac{5}{13}$
0%
$\cfrac{1}{13}$
0%
$\cfrac{37}{221}$
0%
$\cfrac{2}{13}$

Explanation

Two cards are drawn randomly from

$52$ cards, firstly number of total ways that two cards are not an ace.

$={ _{ }^{ 48 }{ C } }_{ 2 }=\cfrac { 48\times 47 }{ 2 } =1128$
Number of ways drawing

$2$ cards out of

$52$ cards

$={ _{ }^{ 52 }{ C } }_{ 2 }=\cfrac { 52\times 51 }{ 2 } =1326$

$P(X=0)$ . Probably that no ace is not drawn

$=\cfrac { 1128 }{ 1326 }$
Secondly number of ways that there is an ace and there is no ace out of

$={ _{ }^{ 4 }{ C } }_{ 1 }\times { _{ }^{ 48 }{ C } }_{ 1 }=4\times 48=192$

$P(X=1)$ , Probably that there is an ace and there is no ace

$=\cfrac { 192 }{ 1326 }$

$P(X=2)$ , Probably of getting two aces

$=\cfrac { 6 }{ 1326 }$
Probability distribution is:

$X$	0	1	2
$P(X)$	$\cfrac { 1128 }{ 1326 }$	$\cfrac { 192 }{ 1326 }$	$\cfrac { 6 }{ 1326 }$

$\sum { { X }_{ i } } =\sum { { p }_{ i }{ x }_{ i } }$

$=\cfrac { 1128 }{ 1326 } \times 0+\cfrac { 192 }{ 1326 } \times 1+\cfrac { 6 }{ 1326 } \times 2=\cfrac { 192 }{ 1326 } +\cfrac { 12 }{ 1326 } =\cfrac { 2 }{ 13 }$

A random variable

$X$ , takes the values

$0,1,2$ and

$3$ . Mean of

$X$ is

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

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

$P(x=0)$ is{

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

Explanation

Let

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

$P(X=3)=p$
and

$P(X=1)=\cfrac{p}{2}$
Also

$P(X=2)=0.3$ (given)
Let

$P(X=0)=x$
Probability distribution is:

$X$	0	1	2	3
$P(X)$	$x$	$\cfrac{p}{2}$	$0.3$	$p$

From table

$\sum { { X }_{ i } } =Mean=\cfrac { \sum { { p }_{ i }{ x }_{ i } } }{ \sum { { p }_{ i } } } \Rightarrow 1.3=x\times 0+1\times \cfrac { p }{ 2 } +2\times 0.3+3\times p$

$\Rightarrow 1.3=\cfrac { p }{ 2 } +0.6+3p\Rightarrow 2.6=p+6p+1.2\Rightarrow 7p=1.4\Rightarrow p=\cfrac { 1.4 }{ 7 } =0.2$
Sum of probabilities

$=1$

$x+\cfrac { p }{ 2 } +0.3+p=1\Rightarrow x+0.1+0.3+0.2=1\Rightarrow x=1-0.6\Rightarrow x=0.4$

$P(X=0)=0.4$

A random variable

$X$ has the probability distribution

$X$	1	2	3	4	5	6	7	8
$P(X)$	0.15	0.23	0.12	0.10	0.20	0.08	0.07	0.05

For the events,

$E = \{X$ is a prime number

$\}$ and

$F = {X < 4}$ , the probability

$\displaystyle P\left ( E\cup F \right )$ is

0%
$0.87$
0%
$0.77$
0%
$0.35$
0%
$0.50$

Explanation

From the given values, prime numbers and the values less than

$4$ are

$1,2,3,5,7$

So,

$P(E \cup F)$ will be the sum of probabilities of each of the above selected values of

$X$

So,

$P( E \cup F) = 0.15 + 0.23 + 0.12 + 0.20 + 0.07 = 0.77$

Suppose

$X$ is a random variable which takes values

$0, 1, 2, 3, ...$ and

$P(X=r)=pq^r$ where

$0 < p < 1, q=1-p$ and

$r=0, 1, 2, ...$ , Then

0%
$P(X\geq n)=q^n$
0%
$P(X\geq m+n|X\geq m+n|X\geq m)=P(X\geq n)$
0%
$P(X= m+n|X\geq m+n|X\geq m)=P(X= n)$
0%
None of these

Explanation

$\displaystyle P(X\geq n)=\sum_{k=n}^{\infty}P(X=k)=\frac {pq^n}{1-q}=q^n$

$P(X\geq m+n|X\geq m)=\dfrac {P(X\geq m+n)}{P(X\geq m)}=\dfrac {q^{m+n}}{q^m}=q^n$

$P(X=m+n|X\geq m)=\dfrac {P(X=m+n)}{P(X\geq m)}=pq^n$ .

In a series of 3 independent trials, the probability of exactly 2 success is 12 times as large as the probability of 3 successes. The probability of a success in each trail is

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

Explanation

For a binomial experiment consisting of n trials, the probability of exactly

k successes is

$P(k \text{ successes}) = ^nC_kp^k(1-p)^{n-k}$

where the probability of success on each trial is p.

According to the question,

$n=3,, P(\text{2 successes})=12\times P(\text{3 successes})$

To find:

The probability of a success in each trail, p

$P(\text{2 successes})=12\times P(\text{3 successes})\\\implies ^3C_2p^2(1-p)^{3-2}=12 \times ^3C_3p^3(1-p)^{3-3}\\\implies 3\times p^2(1-p)^1=12\times 1\times p^3(1-p)^0\\\implies 3p^2-3p^3=12p^3$

Divide throughout by

$3p^2$ , we get

$1-p=4p\\\implies 5p=1\\\implies p=\dfrac 15$

A random variable

$x$ assumes values which are numbers of the form

$\displaystyle \frac{n}{n+1}$ and

$\displaystyle \frac{n+1}{n}$ , where

$n=1, 2, 3,...$ . lf

$P\left(x=\displaystyle \frac{n}{n+1}\right)=P\left(x=\dfrac{n+1}{n}\right)=\left(\dfrac{1}{2} \right)^{n+1}$ , then

0%
$\mathrm{P}(\mathrm{x}<1)=\mathrm{P}(\mathrm{x}>1)$
0%
$\mathrm{P}(1/2<\mathrm{x}<1)<\mathrm{P}(\mathrm{x}>1)$
0%
$\mathrm{P}(\mathrm{x}>3/2)<\mathrm{P}(\mathrm{x}<1)$
0%
$\mathrm{P}(\mathrm{x}>3/2)=0$

Explanation

The numbers of the form

$\dfrac{n}{n+1}$ are less than 1 and

$\dfrac{n+1}{n}$ are greater than 1.

$P(x<1) = P(\dfrac{1}{2}) +P(\dfrac{2}{3}) + ...$

$\Rightarrow P(x<1) = \dfrac{1}{4} +\dfrac{1}{8} + ...$

$\Rightarrow P(x<1) = \dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}}$

$\Rightarrow P(x<1) = \dfrac{1}{2}$

$P(x>1) = \dfrac{1}{2}$

$P\left(\dfrac{1}{2} <x<1\right) = P\left(\dfrac{2}{3} \right) + P\left(\dfrac{3}{4} \right) + ...$

$P\left(\dfrac{1}{2} <x<1\right) = \dfrac18 +\dfrac{1}{16} + ...$

$P\left(\dfrac{1}{2} <x<1\right) =\dfrac{1}{4}$

$P\left(x> \dfrac32\right) =P(x=2)$

$P\left(x> \dfrac32\right) =\dfrac{1}{4}$

$P\left(x> \dfrac32\right) = \dfrac{1}{4}$ ]

$\therefore$

$\mathrm{P}(\mathrm{x}<1)=\mathrm{P}(\mathrm{x}>1)$

$\mathrm{P}(1/2<\mathrm{x}<1)<\mathrm{P}(\mathrm{x}>1)$

$\mathrm{P}(\mathrm{x}>3/2)<\mathrm{P}(\mathrm{x}<1)$

If the range of a random variable

$X$ is

$\{0, 1, 2, 3, \ldots\ldots\}$ with

$\displaystyle P(X=k)=\dfrac{(k+1)(a)}{3^{k}}$ for sample of

$4$ items is drawn at random without

$\mathrm{k}\geq 0$ , then

$\mathrm{a}=$

0%
$2/3$
0%
$4/9$
0%
$8/27$
0%
$16/81$

Explanation

Use the concept of probability distribution function,

$\sum_{k=0}^{\infty}(P(X=k))=1$

So,

$\sum_{k=0}^{\infty}\dfrac{(k+1)a}{3^k}=a\sum_{k=0}^{\infty}(\dfrac{k}{3^k}+\dfrac{1}{3^k})$

$\sum_{k=0}^{\infty}\dfrac{k}{3^k}=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+.......=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+.......=\dfrac{1}{3}(1+\dfrac{2}{3}+\dfrac{3}{3^2}+\dfrac{4}{3^3}+.....)$

Use the Taylor series

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

Let us assume that

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

$\sum_{k=0}^{\infty}\dfrac{k}{3^k}=\dfrac{\dfrac{1}{3}}{(1-\dfrac{1}{3})^2}=\dfrac{3}{4}$

The second Summation will be infinite Geometric Sequence where

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

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

$\sum_{k=0}^{\infty}\dfrac{1}{3^k}=1+\dfrac{\dfrac{1}{3}}{1-\dfrac{1}{3}}=\dfrac{3}{2}$

Therefore,

$\sum_{k=0}^{\infty}(P(X=k))=1$

$(\dfrac{3}{4}+\dfrac{3}{2})\times a=1$

Hence,

$a=\dfrac{4}{9}$

$\mathrm{P}(\mathrm{u}_{\mathrm{i}})\propto \mathrm{i}$ , where

$\mathrm{i}=1,2,3,\ \ldots \mathrm{n}$ , then

$\displaystyle \lim_{\mathrm{n}\rightarrow\infty}\mathrm{P}(\mathrm{w})$ is equal to

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

Explanation

Here

$\mathrm{P}(\mathrm{u}_{\mathrm{i}})=\mathrm{k}\mathrm{i}$

$\Sigma \mathrm{P}(\mathrm{u}_{\mathrm{i}})=1$

$\displaystyle \Rightarrow \mathrm{k}=\frac{2}{\mathrm{n}(\mathrm{n}+1)}$

$\displaystyle \lim_{\mathrm{n}\rightarrow\infty}\mathrm{P}(\mathrm{w})=\lim_{n\rightarrow\infty}\sum_{{i}=1}^{n}P(\frac{w}{u_i})p(u_i)=\lim_{\mathrm{n}\rightarrow\infty}\sum_{\mathrm{i}=1}^{\mathrm{n}}\frac{2\mathrm{i}^{2}}{\mathrm{n}(\mathrm{n}+1)^{2}}=\lim_{\mathrm{n}\rightarrow\infty}\frac{2\mathrm{n}(\mathrm{n}+1)(2\mathrm{n}+1)}{\mathrm{n}(\mathrm{n}+1)^{2}6}=\frac{2}{3}$

A fair coin is tossed

$99$ times. If X is the number of times heads occur then P(X = r) is maximum when r is

0%
$49$
0%
$50$
0%
$51$
0%
none of these

Explanation

Notice that x has binomial distribution with n = 99, p = q = 0.5.

Therefore the probability

$P(x=r)=^nC_rP^rq^{n-r}\\\implies P(x=r)=^nC_r(0.5)^r(0.5)^{n-r}=^{n}C_r(0.5)^n$

Thus maximum of P(x = r) corresponds to the value of r with maximal

$^nC_r=\dfrac {n!}{r!(n-r)!}$

Notice that binomial coefficients are symmetric:

$^nC_r=^{n-r}C_r=\dfrac {n!}{r!(n-r)!}$

hence the maximum corresponds to the points most closest to

$\dfrac n2$

Since n is odd, there are two such values:

$r_1=\dfrac {n-1}2=\dfrac {98}2=49, r_2=\dfrac {n+1}2=\dfrac {100}2=50$

For these numbers the function P(x = r) has maximum.

Let numbers

$1,2,3...4n$ be pasted on

$4n$ blocks. The probability of drawing a number is proportional to

$r$ , then the probability of drawing an even number in one draw is

$\left ( n\in N \right )$

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

Explanation

$P(r)$ be the probability that a number

$r$ is drawn in one draw, it is given that

$P(r) = kr;$
where

$k$ is a constant.

$\therefore P(1) + P(2) + P(3) + \cdots + p(4n) = k \cdot 1 + k \cdot 2 + ...+ k\cdot (4n) = k\left [ 1 + 2 + 3 + \cdots+4n\right ]$

$1= \displaystyle \frac{4n(4n + 1)k}{2}$

$\displaystyle\therefore \sum_{r=0}^{n}P(r)=1$ or

$P(S)= 1$

$\displaystyle\Rightarrow K=\frac{1}{2n\left ( 4n+1 \right )}$
Now

$'A'$ event of drawing

$2, 4, 6, ...4n$

$P(A)= P ( 2$ or

$4$ or

$6 ....$ or

$4n )$

$\therefore P(A)= P(2) + P(4) + \cdots + P(4n)$

$=2k\left (1 + 2 + ... + 2n \right )$

$P(A)= k(2n) \left (2n + I \right )$

$P(A)= (2n)\left (2n + I \right )$

$\therefore P(A)= \displaystyle \frac{2n+1}{4n+1}$

Let

$p$ be the probability that a man aged

$x$ years will die in a year time. The probability that out of

$n$ men

$\displaystyle A_{1},A_{2},A_{3},....,A_{n}$ each aged

$x$ years,

$\displaystyle A_{1}$ will die & will be the first to die, is

0%
$\displaystyle \frac{1-p^{n}}{n}$
0%
$\displaystyle \frac{p}{n}$
0%
$\displaystyle \frac{p\left ( 1-p \right )^{n-1}}{n}$
0%
$\displaystyle \frac{1-\left ( 1-p \right )^{n}}{n}$

Explanation

Let

$E_i$ be the event that

$A_i$ will die in a year where i=1, 2, 3, 4, ..., n

The probability that none of

$A_1, A_2, A_3, ..., A_n$ dies in a year

$(1-p_{A_1}).(1-p_{A_2}), ..., (1-p_{A_n})$

Where

$p_{A_1}, p_{A_2}, ..., p_{A_n}$ are the probabilities that

$A_1, A_2, A_3, ..., A_n$ dies in a year respectively.

$(1-p).(1-p), ...(1-p)=(1-p)^n$

Now the probability that at least one of

$A_1, A_2, A_3, ..., A_n$

$1-(1-p)^n$

The probability that among n men

$A_1$ is the first one to die is

$\dfrac 1n$

Therefore, the required probability is

$=\dfrac 1n[1-(1-p)^n]$

A continuous random variable

$X$ has p.d.f

$f(x)$ , then:

0%
$0\le f(x)\le 1$
0%
$f(x)\ge 0$
0%
$f(x)\le 1$
0%
$0< f(x)< 1$

Explanation

Let

$X$ be a continuous random variable.

Let

$f(x)$ be the p.d.f of

$X$ .

We know that the p.d.f

$f(x)$ of a continuous random variable

$X$ is always positive. That is

$f(x) \geq 0, \forall x\in \mathbb{R}$

Thus a continuous random variable

$X$ has p.d.f

$f(x)$ then

$f(x)\geq 0$

A boy has 20% chance of hitting at a target. Let p denote the probability of hitting the target for the first time at the nth trial. If

$p$ satisfies the inequality

$625p^{2} - 175p + 12 < 0$ then value of

$n$ is

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

Explanation

Let p(n)=Hitting for the 1st time at the n-th trial

$\implies p(n)=0.8^{n-1}\times0.2$

The boundary of the inequality

$62p^2-175p+12=0$

is the solution of a quadratic equation in p :

or,

$175^2-4\times12\times625=625=25^2$

$\implies p=\dfrac {175\pm25}{1250}=\dfrac 3{25}$ or

$\dfrac 4{25}$

So p(n) is negative between those two values.

$p(n)=\dfrac 3{25}=0.8^{n-1}\times 0.2\\\implies \dfrac 35=0.8^{n-1}\\\implies \log\left(\dfrac 35\right)=(n-1)\log (0.8)\\\implies n=1+\dfrac {\log \left(\dfrac 35\right)}{\log (0.8)}=3.289$

$p(n)=\dfrac 4{25}=0.8^{n-1}\times 0.2\\\implies \dfrac 45=0.8^{n-1}\\\implies \log\left(\dfrac 45\right)=(n-1)\log (0.8)\\\implies n=1+\dfrac {\log \left(\dfrac 45\right)}{\log (0.8)}=2$

Hence,

$2<n<3.289\implies n=3$

$X$ is a discrete random variable then which of the following is correct?

0%
$0\le F(x)<1$
0%
$F(-\infty )=0;F(\infty )\le 1$
0%
$P\left[ X={ x }_{ n } \right] =F({ x }_{ n })-F({ x }_{ n-1 })$
0%
$F(x)$ is a constant function

Two cards are drawn simultaneously (without replacement) from a well-shuffled pack of

$52$ cards. Find the mean and variance of the number of red cards.

0%
Mean $= 0.1$ and Variance $= 0.7$
0%
Mean $= 0.6$ and Variance $= 0.3$
0%
Mean $= 0.49$ and Variance $= 0.37$
0%
Mean $= 0$ and Variance $= 0.45$

Explanation

Since we have to draw two red cards(without replacement) from

$52$ cards, therefore the probability is:

$p=\dfrac { 26 }{ 52 } \times \dfrac { 25 }{ 51 } =\dfrac { 25 }{ 102 }$

And,

$q=1-\dfrac { 25 }{ 102 } =\dfrac { 77 }{ 102 }$

Since

$n=2$ , therefore,

$Mean=np=2\times \dfrac { 25 }{ 102 } =\dfrac { 50 }{ 102 } =0.49\\$

$Variance=npq=2\times \dfrac { 25 }{ 102 } \times \dfrac { 77 }{ 102 } =0.37$

$f(x)=\dfrac { A }{ \pi } .\dfrac { 1 }{ 16+{ x }^{ 2 } } ,-\infty <x<\infty$ is a p.d.f of a continuous random variable

$X$ , then the value of

$A$ is:

0%
$16$
0%
$8$
0%
$4$
0%
$1$

Explanation

For a probability distribution function

$f(x), a<x<b$ of a continuous random variable

$X$ , it is required that

$\displaystyle \int_a^b f(x)dx = 1$

$\therefore \displaystyle \int_{-\infty}^{\infty} \cfrac{A}{\pi}\cfrac{1}{16+x^2}dx = 1$

$\Rightarrow \displaystyle \cfrac{A}{\pi}\int_{-\infty}^{\infty}\cfrac{1}{4^2 + x^2}dx = 1$

$\Rightarrow \displaystyle \cfrac{A}{\pi} \cfrac14\tan^{-1}\left(\cfrac x4\right)\big|_{-\infty}^{\infty} =1$

$\Rightarrow \cfrac {A}{4\pi}\left[\cfrac{\pi}2 - \left(-\cfrac{\pi}2\right) \right] = 1$

$\Rightarrow \cfrac{A}{4} =1$

$\Rightarrow A = 4$

What is P(

$Z$ is the product of two prime numbers) equal to ?

0%
$0$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{6}$
0%
$\dfrac{1}{12}$

Explanation

Given set

$=\{1,2,3,4,5,6,7\}$

$Z=X+Y$

Where X is randomly selected from set of odd numbers and

Y is randomly selected from set of even numbers

Which gives Z as odd and greater than one

$X=\{1,3,5,7\}$

$Y=\{2,4,6\}$

$Z=\{3,5,7,9,11,13\}$

To find probability of Z being product of two prime numbers-

Among the above obtained values of Z, none can be expressed in the form of product of two different prime numbers

$\therefore P=0$

$NOTE$ - If we consider that Z is to be expressed as product of two different prime numbers, then the answer is 0

But if we assume that the prime numbers used in the products can be same, then

9 can be expressed as product of 2 prime numbers

$Z=X+Y=9$

$\{X,Y\}=\{\{7,2\},\{3,6\},\{5,4\}\}$

$P=\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } \times \cfrac { 1 }{ 3 }$

$P=\cfrac{1}{4}$

The distribution of a random variable X is given below:

X = x	2	-1	0	1	2	3
P(X = x)	$\frac{1}{10}$	k	$\frac{1}{5}$	2k	$\frac{3}{10}$	k

0%
$\frac{1}{10}$
0%
$\frac{2}{10}$
0%
$\frac{3}{10}$
0%
$\frac{7}{10}$

The probability that the bag contains 2 balls of each colour, is

0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{5}$
0%
$\dfrac{1}{10}$
0%
$\dfrac{1}{4}$

Let the p.m.f. of a random variable

$X$ be -

$P(x) = \dfrac {3 - x}{10}$ for

$x = -1, 0, 1, 2$ otherwise
Then

$E(X)$ is _________.

0%
$1$
0%
$2$
0%
$0$
0%
$-1$

Explanation

$x$	$-1$	$0$	$1$	$2$
$P(x)$	$\dfrac{4}{10}$	$\dfrac{3}{10}$	$\dfrac{2}{10}$	$\dfrac{1}{10}$
$x\cdot P(x)$	$-\dfrac{4}{10}$	$0$	$\dfrac{2}{10}$	$\dfrac{2}{10}$

We have to find

$E(X)$

We know that

$E(X)=\sum x\cdot P(x)$

$\therefore E(X)=-\dfrac{4}{10}+0+\dfrac{2}{10}+\dfrac{2}{10}=0$

Hence

$E(X)=0$

The range of a random variable X =

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

$P (X = k) = \frac{C^k}{ k! }$ ;

$k = 1,2,3,4...,$ then the value of C is

0%
$2$
0%
$log_2 e$
0%
$log_e 2$
0%
4

A random variable

$X$ has the following probability distribution:

$X$	0	1	2	3	4	5
$P(X=x)$	$\cfrac{1}{4}$	$2a$	$3a$	$4a$	$5a$	$\cfrac{1}{4}$

Then

$P(1\le X\le 4)$ is:

0%
$\cfrac { 10 }{ 21 }$
0%
$\cfrac { 2 }{ 7 }$
0%
$\cfrac { 1 }{ 14 }$
0%
$\cfrac { 1 }{ 2 }$

Explanation

We know that

$\sum P(X=x)=1$

$\Rightarrow \dfrac{1}{4}+2a+3a+4a+5a+\dfrac{1}{4}=1$

$\Rightarrow \dfrac{1}{2}+14a=1$

$\Rightarrow 14a=1-\dfrac{1}{2}$

$\Rightarrow 14a=\dfrac{1}{2}$

$\Rightarrow a=\dfrac{1}{28}$

Now we have to find

$P(1\leq x \leq 4)$

$P(1\leq x \leq 4)=P(x=1)+P(x=2)+P(x=3)+P(x=4)$

$=2a+3a+4a+5a$

$=14a$

$=14 \left(\dfrac{1}{28}\right) \quad \left[\because a=\dfrac{1}{28}\right]$

$=\dfrac{1}{2}$

$\therefore P(1\leq x \leq 4)=\dfrac{1}{2}$

The random variable

$X$ follows normal distribution

$f(x)=c{ e }^{ \cfrac { -\cfrac { 1 }{ 2 } { \left( x-100 \right) }^{ 2 } }{ 25 } }$ . Then the value of

$c$ is:

0%
$\sqrt { 2\pi }$
0%
$\cfrac { 1 }{ \sqrt { 2\pi } }$
0%
$5\sqrt { 2\pi }$
0%
$\cfrac { 1 }{5 \sqrt { 2\pi } }$

Explanation

The normal density function of the random variable

$X$ is given by

$f(x)=ce^{\dfrac{-\dfrac{1}{2}(x-100)^2}{25}}$

$...(1)$

We have to find the value of

$c$ .

We know that, if

$X \sim N(\mu ,\sigma^2)$ , the normal density function is

$f(x;\mu,\sigma^2)=\dfrac{1}{\sigma\sqrt{2\pi}}e^{\dfrac{-(x-\mu)^2}{2\sigma^2}}$

$...(2)$

Comparing

$(1)$ and

$(2)$ we get

$c=\dfrac{1}{\sigma\sqrt{2\pi}}, \mu=100, \sigma^2=25 \Rightarrow \sigma=5$

$\therefore c=\dfrac{1}{5\sqrt{2\pi}}$

If the range of random variable X is {0, 1, 2, 3, 4...} with p(X = k) =

$\frac{(k + 1)a}{3^k}$ for

$k>=$ 0, then a =

0%
$\frac{2}{3}$
0%
$\frac{4}{9}$
0%
$\frac{8}{27}$
0%
$\frac{16}{81}$

There are

$5$ men and

$5$ women in a party. In how many can

$5$ dancing pairs be selected ? (each dancing pair consists of

$1$ man and

$1$ women )

0%
$120$
0%
$44$
0%
$32$
0%
$10$

Two probability distributions of the discrete random variable

$X$ and

$Y$ are given below.

$X$	$0$	$1$	$2$	3
$P\left( X \right)$	$\dfrac { 1 }{ 5 }$	$\dfrac { 2 }{ 5 }$	$\dfrac { 1 }{ 5 }$	$\dfrac { 1 }{ 5 }$

$Y$	$0$	$1$	$2$	$3$
$P\left( Y \right)$	$\dfrac { 1 }{ 5 }$	$\dfrac { 3 }{ 10 }$	$\dfrac { 2 }{ 5 }$	$\dfrac { 1 }{ 10 }$

Then

0%
$E\left( { Y }^{ 2 } \right) =2E\left( X \right)$
0%
$E\left( { Y }^{ 2 } \right) =E\left( X \right)$
0%
$E\left( Y \right) =E\left( X \right)$
0%
$E\left( { X }^{ 2 } \right) =2E\left( Y \right)$

Explanation

$E(X)=\sum { XP( } X)=0*\frac { 1 }{ 5 } +1*\frac { 2 }{ 5 } +2*\frac { 1 }{ 5 } +3*\frac { 1 }{ 5 } =7/5\\ E({ Y }^{ 2 })=\sum { { y }^{ 2 }P(Y) } ={ 0 }^{ 2 }*\frac { 1 }{ 5 } +1^{ 2 }*\frac { 3 }{ 10 } +{ 2 }^{ 2 }*\frac { 4 }{ 10 } +{ 3 }^{ 2 }*\frac { 1 }{ 10 } =28/10=14/5\\ E({ Y }^{ 2 })=2E(X)$

The total numbers of outcomes when three coins tossed once is .....

0%
$2$
0%
$8$
0%
$6$
0%
$4$

Explanation

When three coins tossed ones then total number of outcomes is :

$8$

$\{ TTT,HHH,TTH,THT,HTT,HHT,HTH,THH\}$

The random variable X has the following probability massfunction

$P[x=x]=k.\dfrac{2x}{x!}, x=0,1,2,3=0$ , otherwise, then the value of K is

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

CBSE Questions for Class 12 Commerce Applied Mathematics Probability Distribution And Its Mean And Variance Quiz 7 - MCQExams.com

0:0:1

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 7 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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