RD Sharma Class 12 Exercise 31.1 Mean and Variance of a Random variable Solutions Maths-Download PDF Free Online

RD Sharma Class 12 Exercise 31.1 Mean and Variance of a Random variable Solutions Maths

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 04:38 PM IST

The students of the CBSE board schools prefer using the RD Sharma books for their reference. Every student comes across doubts when they try to solve homework, especially in mathematics. A chapter like Mean and Variance of a Random Variable is challenging to solve. The RD Sharma Class 12th Exercise 31.1 solution book is used by the students to clarify their doubts.

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RD Sharma Class 12 Solutions Chapter31 Mean and Variance of a Random Variable- Other Exercise
Mean and Variance of a Random Variable Excercise: 31.1
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter31 Mean and Variance of a Random Variable- Other Exercise

Mean and Variance of a Random Variable Excercise: 31.1

Mean and Variance of a Random Variable exercise 31 point 1 question 1(i)

Answer: The given distribution of probability is not a probability distribution.
Hint: $\sum P\left ( x=n \right )=1$
Given:

$X$	$3$	$2$	$1$	$0$	$-1$
$P\left ( X \right )$	$0.3$	$0.2$	$0.4$	$0.1$	$0.05$

Solution:
$\begin{aligned} &P(x=3)+P(x=2)+P(x=1)+P(x=0)+P(x=-1) \\ &=0.3+0.2+0.4+0.1+0.05 \\ &=1.05 \neq 1 \end{aligned}$
So, the given distribution of probabilities is not a probability distribution.

Mean and Variance of a Random Variable exercise 31 point 1 question 1(ii)

Answer: The given distribution of probability is not a probability distribution.
Hint: $\sum P\left ( x=n \right )=1$
Given:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$0.6$	$0.4$	$0.2$

Solution:
$\begin{aligned} &P(x=0)+P(x=1)+P(x=2) \\ &=0.6+0.4+0.2 \\ &=1.2 \neq 1 \end{aligned}$
So, the given distribution of probabilities is not a probability distribution.

Mean and Variance of a Random Variable exercise 31.1 question 1(iii)

Answer: The given distribution of probability is a probability distribution

Hint: $\sum P\left ( x=n \right )=1$

Given:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$0.1$	$0.5$	$0.2$	$0.1$	$0.1$

Solution:

$P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)$

$=0.1+0.5+0.2+0.1+0.1$

$=1$

So, the given distribution of probability is a probability distribution.

Mean and Variance of a Random Variable exercise 31.1 question 1(iv)

Answer: The given distribution of probability is a probability distribution

Hint: $\sum P\left ( x=n \right )=1$

Given:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$0.3$	$0.2$	$0.4$	$0.1$

Solution:

$P(x=0)+P(x=1)+P(x=2)+P(x=3)$

$=0.3+0.2+0.4+0.1$

$=1$

So, the given distribution of probability is a probability distribution

Mean and Variance of a Random Variable exercise 31.1 question 2

Answer: So the required value of $K=0.1$

Hint: we will use the formula $\sum P\left ( x=n \right )=1$

Given:

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$0.1$	$K$	$0.2$	$2K$	$0.3$	$K$

Solution:
$P(-2)+P(-1)+P(0)+P(1)+P(2)+P(3)=1$
$\Rightarrow 0.1+K+0.2+2 K+0.3+K=1$
$\Rightarrow 4 K+0.6=1$
$\Rightarrow 4 K=1-0.6$
$\Rightarrow 4 K=0.4$
$\Rightarrow K=\frac{0.4}{4}$
$\Rightarrow K=\frac{1}{10}$
$\Rightarrow K=0.1$

Mean and Variance of a Random Variable exercise 31.1 question 3(i)

Answer: $a=\frac{1}{81}$

Hint: $\sum P\left ( x=n \right )=1$

Given:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P\left ( X \right )$	$a$	$3a$	$5a$	$7a$	$9a$	$11a$	$13a$	$15a$	$17a$

Solution: Since $\sum P\left ( x \right )=1$

$\Rightarrow P(0)+P(1)+P(2)+P(3)+P(4)+P(5)+P(6)+P(7)+P(8)=1$

$\Rightarrow a+3 a+5 a+7 a+9 a+11 a+13 a+15 a+17 a=1$

$\Rightarrow 81 a=1$

Mean and Variance of a Random Variable exercise 31.1 question 3(ii)

Answer: $P(x<3)=\left(\frac{1}{9}\right), P(x \geq 3)=\frac{8}{9}, P(0<x<5)=\frac{8}{27}$

Hint: $\sum P\left ( x=n \right )=1$

Given:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P\left ( X \right )$	$A$	$3a$	$5a$	$7a$	$9a$	$11a$	$13a$	$15a$	$17a$

Solution:

$P(x<3)=P(0)+P(1)+P(2)$

$P(x<3)=a+3 a+5 a$

$P(x<3)=9 a$

$P(x<3)=9\left(\frac{1}{81}\right)$ [ using the previous solution ]

$\therefore P(x<3)=\left(\frac{1}{9}\right)$

$\therefore P(x \geq 3)=1-P(x<3)=1-\frac{1}{9}=\frac{8}{9}$

$P(0<x<5)=P(1)+P(2)+P(3)+P(4)$

$P(0<x<5)=3 a+5 a+7 a+9 a$

$P(0<x<5)=24 a$

$P(0<x<5)=24\left(\frac{1}{81}\right)$

Mean and Variance of a Random Variable exercise 31.1 question 4(i)

Answer: So, the possible value of $c=\frac{1}{3}$

Hint: $\sum P\left ( x=n \right )=1$

Given:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$3c^{3}$	$4c-10c^{2}$	$5c-1$

where $c> 0$

Solution :
Since $\sum P\left ( x \right )=1$

$\Rightarrow P(0)+P(1)+P(2)=1$

$\Rightarrow 3 c^{3}+4 c-10 c^{2}+5 c-1=1$

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

$\Rightarrow 3 c^{3}-3 c^{2}-7 c^{2}+7 c+2 c-2=0$

$\Rightarrow 3 c^{2}(c-1)-7 c(c-1)+2(c-1)=0$

$\Rightarrow(c-1)\left(3 c^{2}-7 c+2\right)=0$

$\Rightarrow(c-1)\left(3 c^{2}-6 c-c+2\right)=0$

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

$c=1, c=2, \quad c=\frac{1}{3}$

Only $c=\frac{1}{3}$ is possible, because if $c=1$ or $c=2$ then $P\left ( 1 \right )$ will become negative.

Mean and Variance of a Random Variable exercise 31.1 question 4(ii)

Answer: $P\left ( x< 2 \right )=\frac{1}{3}$
Hint: $P\left ( x< 2 \right )=P\left ( 0 \right )+P\left ( 1 \right )$

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$3c^{3}$	$4c-10c^{2}$	$5c-1$

where $c< 2$
Solution: $P\left ( x< 2 \right )=P\left ( 0 \right )+P\left ( 1 \right )$
$P(x<2)=3 c^{3}+4 c-10 c^{2}$
$P(x<2)=3\left(\frac{1}{3}\right)^{3}+4\left(\frac{1}{3}\right)-10\left(\frac{1}{3}\right)^{2}$
$P(x<2)=\frac{3}{27}+\frac{4}{3}-\frac{10}{9}$
$P(x<2)=\frac{1}{9}+\frac{4}{3}-\frac{10}{9}$
$P(x<2)=\frac{3}{9}$
$\therefore P(x<2)=\frac{1}{3}$

Mean and Variance of a Random Variable exercise 31.1 question 4(iii)

Answer: So, the required value of : $P\left ( x< 2 \right )=\frac{2}{3}$
Hint: $P\left ( 1<x\leq 2 \right )=P\left ( 2 \right )$
Given:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$3c^{3}$	$4c-10c^{2}$	$5c-1$

Solution: $P\left ( 1<x\leq 2 \right )=P\left ( 2 \right )$
$P\left ( 1<x\leq 2 \right )=5c-1$
$P\left ( 1<x\leq 2 \right )=5\left ( \frac{1}{3} \right )-1$
$\therefore P\left ( 1<x\leq 2 \right )=\frac{2}{3}$

Mean and Variance of a Random Variable exercise 31.1 question 5

Answer: The required probability distribution of x is.

$X$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$P\left ( X \right )$	$\frac{15}{61}$	$\frac{10}{61}$	$\frac{30}{61}$	$\frac{6}{61}$

Hint: $\sum P\left ( x=n \right )=1$
Given: $2 P\left(X=x_{1}\right)=3 P\left(X=x_{2}\right)=P\left(X=x_{3}\right)=5 P\left(X=x_{4}\right)$
Solution: $2 P\left(x_{1}\right)=3 P\left(x_{2}\right)=P\left(x_{3}\right)=5 P\left(x_{4}\right)$
$Let P\left(x_{3}\right)=a$
$\begin{aligned} &2 P\left(x_{1}\right)=P\left(x_{3}\right) \Rightarrow P\left(x_{1}\right)=\frac{a}{2} \\ &3 P\left(x_{2}\right)=P\left(x_{3}\right) \Rightarrow P\left(x_{2}\right)=\frac{a}{3} \\ &5 P\left(x_{4}\right)=P\left(x_{3}\right) \Rightarrow P\left(x_{4}\right)=\frac{a}{5} \end{aligned}$
$Since, P\left(x_{1}\right)+P\left(x_{2}\right)+P\left(x_{3}\right)+P\left(x_{4}\right)=1$
$\begin{aligned} &\Rightarrow \frac{a}{2}+\frac{a}{3}+\frac{a}{1}+\frac{a}{5}=1 \\ &\Rightarrow \frac{15 a+10 a+30 a+6 a}{30}=1 \\ &\Rightarrow 61 a=30 \\ &\Rightarrow a=\frac{30}{61} \end{aligned}$
So,

$X$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$
$P\left ( X \right )$	$\frac{15}{61}$	$\frac{10}{61}$	$\frac{30}{61}$	$\frac{6}{61}$

Mean and Variance of a Random Variable exercise 31.1 question 6

Answer: The probability distribution of x is

$X$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{3}$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{9}$

Hint: $\sum P\left ( x=n\right )=1$
Given:
$\begin{aligned} &P(x=0)=P(x>0)=P(x<0) \\ &P(x=-3)=P(x=-2)=P(x=-1) \\ &P(x=1)=P(x=2)=P(x=3) \end{aligned}$
Solution: Here $\begin{aligned} &P(x=0)=P(x>0)=P(x<0) \\ \end{aligned}$
Let $\begin{aligned} &P(x=0)=k \end{aligned}$
$\begin{aligned} \Rightarrow P\left ( x> 0 \right )=k=P\left ( x< 0 \right ) \end{aligned}$
Since: $\sum p\left ( x \right )=1$
$\begin{aligned} &\Rightarrow P(x<0)+P(x=0)+P(x<0)=1 \\ &\Rightarrow k+k+k=1 \\ &\Rightarrow 3 k=1 \\ &\Rightarrow k=\frac{1}{3} \end{aligned}$
So, $P(x<0)$
$\begin{aligned} &\Rightarrow P(x=-1)=P(x=-2)=P(x=-3)=\frac{1}{3} \\ &\Rightarrow 3 P(x=-1)=\frac{1}{3} \end{aligned}$ $\begin{aligned} \quad[\text { as }, P(x=-1)=P(x=-2)=P(x=-3)] \end{aligned}$
$\begin{aligned} &\Rightarrow P(x=-1)=\frac{1}{9} \\ \end{aligned}$
$\begin{aligned} &\Rightarrow P(x=-1)=P(x=-2)=P(x=-3)=\frac{1}{9} \\ \end{aligned}$ ....(i)
$\begin{aligned} &\Rightarrow P(x=0)=\frac{1}{3} \end{aligned}$ .....(ii)
$\begin{aligned} And P(x>0)=k \end{aligned}$
$P(x=1)+P(x=2)+P(x=3)=\frac{1}{3}$
$\begin{aligned} &\Rightarrow 3 P(x=1)=\frac{1}{3} \\ \end{aligned}$ $[\text { as, } P(x=1)=P(x=2)=P(x=3)]$
$\begin{aligned} &\Rightarrow P(x=1)=\frac{1}{9} \\ &\Rightarrow P(x=1)=P(x=2)=P(x=3)=\frac{1}{9} \end{aligned}\\$ .....(iii)

From equation $\left ( i \right ),\left ( ii \right )\left ( iii \right )$

$X$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{3}$	$\frac{1}{9}$	$\frac{1}{9}$	$\frac{1}{9}$

Mean and Variance of a Random Variable exercise 31.1 question 7

Answer: The probability distribution of the number of aces is

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

Hint: Let x denotes numbers of aces in a sample of 2 cards drawn, there are four aces in a pack of 52 cards so, x can have values 0,1,2.
Given: Two cards are drawn from a well shuffled pack of 52 cards.
Solution:
Now,
$\begin{aligned} &P(x=0)=\frac{48_{C_{2}}}{52_{C_{2}}}=\frac{48 \times 47}{2}=\frac{2}{52 \times 51}=\frac{188}{221} \\ &P(x=1)=\frac{48 c_{1} \times 4 c_{1}}{52 c_{2}}=\frac{48 \times 4 \times 2}{52 \times 51}=\frac{32}{221} \\ &P(x=2)=\frac{4 c_{2}}{52 c_{2}}=\frac{4 \times 3}{2}=\frac{2}{52 \times 51}=\frac{1}{221} \end{aligned}$
So,

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

Mean and Variance of a Random Variable exercise 31.1 question 8

Answer: Required probability distribution is

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Hint: Probability of getting a head in one throws of a coin= $\frac{1}{2}$
$\begin{aligned} &P(H)=\frac{1}{2} \\ &P(T)=1-\frac{1}{2} \\ &P(T)=\frac{1}{2} \end{aligned}$
Given: Probability distribution of the number of heads, when three coins are tossed.
Solution: Let x denotes the numbers of heads obtained is 3 throws of a coin than $x=0,1,2,3$
$\begin{aligned} &P(x=0)=P(T) P(T) P(T)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8} \\ &P(x=1)=P(H) P(T) P(T)+P(T) P(H) P(T)+P(T) P(T) P(H) \\ &P(x=1)=\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right) \\ &P(x=1)=\frac{3}{8} \\ &P(x=2)=P(H) P(H) P(T)+P(H) P(T) P(H)+P(T) P(H) P(H) \\ &P(x=2)=\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right)+\left(\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}\right) \end{aligned}$
$\begin{aligned} &P(x=2)=\frac{3}{8} \\ &P(x=3)=P(H) P(H) P(H) \\ &P(x=3)=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \\ &P(x=3)=\frac{1}{8} \end{aligned}$
So, the required probability distribution is

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{8}$	$\frac{3}{8}$	$\frac{3}{8}$	$\frac{1}{8}$

Mean and Variance of a Random Variable exercise 31.1 question 9

Answer: Required probability distribution is

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{48c_{4}}{52c_{4}}$	$\frac{48c_{3}\times 4c_{1}}{52c_{4}}$	$\frac{48c_{2}\times 4c_{2}}{52c_{4}}$	$\frac{48c_{1}\times 4c_{3}}{52c_{4}}$	$\frac{ 4c_{4}}{52c_{4}}$

Hint: Let x denotes numbers of aces drawn out of 4 cards drawn, there are four aces in a pack of 52.
So, $x: 0,1,2,3,4$
Given: Cards are drawn simultaneously from a well shuffled pack of 52 playing cards.
Solution:
$\begin{aligned} &P(x=0)=\frac{48 c_{4}}{52 c_{4}} \\ &P(x=1)=\frac{48 c_{3} \times 4 c_{1}}{52 c_{4}} \\ &P(x=2)=\frac{48 c_{2} \times 4 c_{2}}{52 c_{4}} \\ &P(x=3)=\frac{48 c_{1} \times 4 c_{3}}{52 c_{4}} \\ &P(x=4)=\frac{4 c_{4}}{52 c_{4}} \end{aligned}$
So, required probability distribution is

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{48c_{4}}{52c_{4}}$	$\frac{48c_{3}\times 4c_{1}}{52c_{4}}$	$\frac{48c_{2}\times 4c_{2}}{52c_{4}}$	$\frac{48c_{1}\times 4c_{3}}{52c_{4}}$	$\frac{ 4c_{4}}{52c_{4}}$

Mean and Variance of a Random Variable exercise 31.1 question 10

Answer: Required probability distribution is

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Hint: Let x denotes number of red balls out of 3 drawn. Then $x=0,1,2,3$
Given: A bag has 4 red and 6 black balls, three balls are drawn
Solution:
$P( no red balls )=P(x=0)=\frac{6 C 3}{10 c_{3}}=\frac{6 \times 5 \times 4}{3 \times 2}=\frac{3 \times 2}{10 \times 9 \times 8}=\frac{1}{6}$
$P( one red balls )=P(x=1)=\frac{4 C 1 \times 6 C 2}{10 C 3}=\frac{4 \times 6 \times 5}{2}=\frac{3 \times 2}{10 \times 9 \times 8}$ $=\frac{1}{2}$
$P( two red balls )=P(x=2)=\frac{4 C 2 \times 6 C 1}{10 C 3}=\frac{4 \times 3 \times 6}{2}=\frac{3 \times 2}{10 \times 9 \times 8}$ $=\frac{3}{10}$
$P( all three red )=P(x=3)=\frac{4 C 3}{10 C 3}=\frac{4 \times 3 \times 2}{10 \times 9 \times 8}=\frac{1}{30}$
The required probability distribution is

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Mean and Variance of a Random Variable exercise 31.1 question 11

Answer:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{91}{323}$	$\frac{455}{969}$	$\frac{70}{323}$	$\frac{10}{323}$	$\frac{1}{969}$

Hint: Use permutation and combination.
Given: Five defective mangoes are accidentally mixed with 15 good ones. 4 mangoes are drawn at random. Find the probability of the no. of defective mangoes.
Solution: x represents the number of defective mangoes drawn.
$\therefore x can take values 0,1,2 or 3$
$\therefore there are total 20 mangoes (15 good +5 defective ) mangoes$
$n(s)= total possible ways of selecting 5 mangoes n(s)=20 C_{4}$
$n(s)= total possible ways of selecting 5 mangoes$
$n(s)=20 C_{4} P(x=0)=P( selecting no defective mango )$
$P(x=0)=\frac{15 C 4}{20 C 4}=\frac{15 \times 14 \times 13 \times 12}{20 \times 19 \times 18 \times 17}=\frac{91}{323}$
$P(x=1)=P( selecting 1 defective mango and 3 good mangoes )$
$P(x=1)=\frac{5 C _{1} \times 15 C _{3}}{20 C_{4}}=\frac{455}{969}$
$P(x=2)=P( selecting 2 defective mangoes and 2 good mangoes )$
$P(x=2)=\frac{5 C_{2}\times 15 C _{3}}{20 C _{4}}=\frac{70}{323}$
$P(x=3)=P( selecting 3 defective mangoes and 1 good mango )$
$=\frac{5 C _{3} \times 15 C_{1}}{20 C _{4}}=\frac{10}{323}$
$P(x=4)=P (selecting 4 defective mangoes and no good mango)$
$=\frac{5C_{4}}{20C_{4}}=\frac{1}{969}$
Now, we have
Following table represents the probability distribution of x:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{91}{323}$	$\frac{455}{969}$	$\frac{70}{323}$	$\frac{10}{323}$	$\frac{1}{969}$

Mean and Variance of a Random Variable exercise 31.1 question 12

Answer:

$X$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$P\left ( X \right )$	$\frac{1}{36}$	$\frac{1}{18}$	$\frac{1}{12}$	$\frac{1}{9}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{1}{9}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$

Hint: Use probability distribution formula
Given: 2 dice are thrown together and the no. was noted x, denotes the sum of the 2 nos. Assuming that all the 36 outcomes are equally likely, we have to find the probability distribution of x
Solution: When two fair dice are thrown there are total 36 possible outcomes.
$\therefore \lambda$ denotes the sum of 2 numbers appearing on dice.
$\therefore x$ can take values 2,3,4,5,6,7,8,9,10,11,12 As appearance of a number on a fair die is equally likely
$\begin{aligned} &\text { i. e. } P(\text { appearing of } 1)=P(\text { appearing of } 2)=P(\text { appearing of } 3)=P(\text { appearing of } 4) \\ &=P(\text { appearing of } 5)=P(\text { appearing of } 6)=\frac{1}{6} \end{aligned}$
And also the appearance of numbers on two different dice is an independent event: so two find conditions like P(1 in the first dice and 2 in the second dice) can be given using multiplication rule of probability.
Note: $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$ where A and B are independent events.
$P(x=2)=\frac{1}{36}\{\therefore(1,1)$ is the only combination resulting sum $=2\}$
$P(x=3)=\frac{2}{36}=\frac{1}{18}$ $\quad\{\therefore(1,2)$ and $(2,1)$ are the combination resulting in sum $=3\}$
$P(x=4)=\frac{3}{36}=\frac{1}{12}$ $\quad\{\therefore(1,3)(3,1)$ and $(2,2)$ are the combination resulting in sum $=4\}$
$P(x=5)=\frac{4}{36}=\frac{1}{9}$ $\quad\{\therefore(3,2)(2,3)(1,4)$ and $(4,1)$ are the combination resulting in sum $=5\}$
$P(x=6)=\frac{5}{36}$ $\{\therefore(1,5)(5,1)(2,4)(4,2)(3,3)$ are the combination resulting in sum $=6\}$
$P(x=7)=\frac{6}{36}=\frac{1}{6}$
$\{\therefore(1,6)(6,1)(2,5)(5,2)(3,4)(4,3) are the combination resulting in sum$ $=7\}$
$P(x=9)=\frac{4}{36}=\frac{1}{9}$ $\quad\{\therefore(3,6)(6,3)(5,4)(4,5)$ are the combination resulting in sum $=9\}$
$P(x=10)=\frac{3}{36}=\frac{1}{12}$ $\quad\{\therefore(6,4)(4,6)$ and $(5,5)$ are the combination resulting in sum $=10\}$
$P(x=11)=\frac{2}{36}=\frac{1}{18}$ $\quad\{\therefore(5,6)$ and $(6,5)$ are the combination resulting in sum $=11\}$
$P(x=12)=\frac{1}{36} \quad\{\therefore(6,6)$ is the only combination resulting in sum $=2\}$
Now we have X and P(X)
$\therefore$ Required probablity distribution is

$X$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$
$P\left ( X \right )$	$\frac{1}{36}$	$\frac{1}{18}$	$\frac{1}{12}$	$\frac{1}{9}$	$\frac{5}{36}$	$\frac{1}{6}$	$\frac{5}{36}$	$\frac{1}{9}$	$\frac{1}{12}$	$\frac{1}{18}$	$\frac{1}{36}$

Mean and Variance of a Random Variable exercise 31.1 question 13

Answer:

$X$	$14$	$15$	$16$	$17$	$18$	$19$	$20$	$21$
$P\left ( X \right )$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

Hint: Use frequency table
Given: A class has 15 students whose ages are 14,17,15,14,21,19,20,16,18,17,20,17,16,19 and 20 years respectively. One student is selected in such a manner that each has the same chance of being selected and the age x of the selected student is recorded. We have to find the probability distribution of the random variable x
Solution:
Note: many of the time while solving such simple problems we make a mistake in counting so at first, we should make a frequency table which tells us no of students in the class of the same age.
This makes our interpretation easier.
$\therefore$ Frequency distribution table for age and number of student is:

$Age$	$14$	$15$	$16$	$17$	$18$	$19$	$20$	$21$
$Number$ $of$ $student$	$2$	$1$	$2$	$3$	$1$	$2$	$3$	$1$

X represents the age of a random student
$\therefore$ can take value 14, 15, 16,17,18,19,20 or 21
Total no. of students in class = 15
Using the above frequency table, we can easily see a total number of students of a particular age, and hence we can find probability easily.
$\begin{aligned} &\left\{\text { As probability }=\frac{\text { no of favourable outcome }}{\text { total no of outcome }}\right\} \\ &P(x=14)=\frac{2}{15} \end{aligned}$
Similarly,
$\begin{aligned} &P(x=15)=\frac{1}{15} ; \\ &P(x=16)=\frac{2}{15} \\ &P(x=17)=\frac{3}{15}=\frac{1}{5} \\ &P(x=18)=\frac{1}{15} \\ \end{aligned}$
$\begin{aligned} &P(x=19)=\frac{2}{15} \\ &P(x=20)=\frac{3}{15}=\frac{1}{5} \\ &P(x=21)=\frac{1}{15} \end{aligned}$
$\therefore$ Required Probability distribution is

$X$	$14$	$15$	$16$	$17$	$18$	$19$	$20$	$21$
$P\left ( X \right )$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

Mean and Variance of a Random Variable exercise 31.1 question 14

Answer:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{969}{2530}$	$\frac{455}{969}$	$\frac{38}{253}$	$\frac{2}{253}$	$\frac{1}{2530}$

Given: defective bolts are accidently mixed with 20 good ones. If 4 bolts are drawn at random, find the probability of defective bolts drawn.Hint: we combination formula.
$\therefore X$ can take values 0,1,2, 3 or 4
$\therefore$ there are total 25 bolts ( 20 good + 5 defective) bolts
N(s) = total possible ways of selecting 5 bolts = 25 $c_{4}$
P (x= 0) = P (selecting no defective bolt)
$=\frac{20 c_{4}}{25 c_{4}}=\frac{20 \times 19 \times 18 \times 17}{25 \times 24 \times 23 \times 22}=\frac{969}{2530}$
P (x=1) =P (selecting 1 defective bolt and 3 good bolts)
$=\frac{5 c_{1} \times 20 c_{3}}{25 c_{4}}=\frac{114}{969}$
P (x=2)= P (selecting 2 defective bolts and 2 good bolt)
$=\frac{5 c_{2} \times 20 c_{2}}{25 c_{4}}=\frac{38}{253}$
P (x=3)= P (selecting 3 defective bolts and 1 good bolt)
$=\frac{5 c_{3} \times 20 c_{1}}{25 c_{4}}=\frac{4}{253}$
P (x=4) =P (selecting 4 defective bolts and no good bolt)
$=\frac{5 c_{4}}{25 c_{4}}=\frac{1}{2530}$
Now we have X and P(X)
Following table represents the probability distribution of x:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{969}{2530}$	$\frac{455}{969}$	$\frac{38}{253}$	$\frac{2}{253}$	$\frac{1}{2530}$

Mean and Variance of a Random Variable exercise 31.1 question 15

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Hint: use permutation formula
Given: 2 cards are drawn successively with replacement from well shuffled pack of 52 cards find the probability distribution of aces
Solution: Let solve the problem now:
Note 1 in our problem, it is given that after every draw we are replacing the card.
So our sample space will be not change
In a deck of 52 cards, there are y aces each of our suit respectively.
Let x be the random variable denoting the number of aces for an event when 2 cards are drawn successfully.
$\therefore x$ can take values 0,1 or 2
$P(x=0)=\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}$
{ as we have to select 1 card at a time such that no ace is there so the first probability is $\frac{48}{52}$ and as the drawn card is replaced , next time again probability is $\frac{48}{52}$ }
Similarly, we proceed for other cases
$P(x=1)=\frac{4}{52} \times \frac{48}{52}=\frac{24}{169}$
{we might get are in the first card or second card, so both probabilities are added}
Similarly,
$P(x=1)=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}$
Now we have X and P(X)
∴ now we are ready to write the probability distribution of x
The following table gives probability Distributions:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Mean and Variance of a Random Variable exercise 31.1 question 16

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Given: two cards are drawn successively with replacement from a well shuffled pack of 52 cards, find the probability distribution of the number of kingsHint: use permutation formula
Solution: Lets us solve the problem now:
Note: in our problem, it is given that after every draw, we are replacing the card so our sample will not change
In a deck 52 cards, there are 4 kings each of our suit respectively
Let x be the random variable denoting the number of kings for an event when 2 cards are drawn successively.
∴ X can take value 0,1 or 2
$P(x=0)=\frac{48}{52} \times \frac{48}{52}=\frac{144}{169}$
{As we have to select 1 card at a time such that no king is there so that first probability is $\frac{48}{52}$ and as the drawn card is replaced, used time again probability is $\frac{48}{52}$ }
Similarly we proceed for other cases
$P(x=1)=\frac{4}{52} \times \frac{48}{52}=\frac{24}{169}$
{We might get a king in the first card or second card, so both probabilities are added}
Similarly
$P(x=2)=\frac{4}{52} \times \frac{4}{52}=\frac{1}{169}$
Now we have X and P(X)
∴ Now we have are ready to write the probability distribution of x+
The following table gives probability

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{144}{169}$	$\frac{24}{169}$	$\frac{1}{169}$

Mean and Variance of a Random Variable exercise 31.1 question 17

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

Given: two cards are drawn successively without replacement from a well shuffled pack of 52 cards. Find the probability distribution of the number of acesHint: use probability formula
Solution: while reading this question you might have observed that cards are being drawn successively and sometimes in other question you might have get cards are drawn simultaneously.
You have to be careful regarding these two words, while solving the question. Both have a different meaning
Let’s solve the problem now
Note: in our problem, it is given that after every draw we are not replacing the card so our sample space will change in the case
In a deck of 52 cards, there are 4 aces each of one suit respectively
Let x be the random variable denoting the number of aces for an event when 2 cards are drawn respectively
∴ x can take value 0,1 or 2
$P(x=0)=\frac{48}{51} \times \frac{47}{51}=\frac{188}{221}$
{As we have to select 1 card at a time such that no ace is there so that first probability is $\frac{48}{52}$ and as the drawn card is not replaced, used time again probability is $\frac{47}{51}$ because now our card is our sample space and that card is not from group of ace, so now are card are 47}
Similarly we proceed for other cases
$P(x=1)=\frac{4}{52} \times \frac{48}{51}+\frac{48}{52} \times \frac{4}{51}=\frac{32}{221}$
{We might get are in the first card in the second card, so both probabilities are added}
Similarly,
$P(x=2)=\frac{4}{52} \times \frac{3}{51}=\frac{1}{221}$
Now we have X and P(X)
∴ Now we are ready to write the probability distribution of x
The following table gives probability

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{188}{221}$	$\frac{32}{221}$	$\frac{1}{221}$

Mean and Variance of a Random Variable exercise 31.1 question 18

Answer:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Hint: use combination formula
Given: find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement from a bag containing 4 white and 6 red balls.
Solution: Let x denote the number balls drawn in a random draw of 3 balls
∴x can take values 0, 1, 2 or 3
Since bag contains 6 red and 4 white ball, i.e. total of 10 balls.
∴ Total no of ways of selecting 3 balls put of 10 = 10 $c_{3}$
For selecting 0 white balls, we will select all 3 balls from red
$\therefore P\left ( x=0 \right )=P$ (not selecting any white ball)
$=\frac{6 c_{3}}{10 c_{3}}=\frac{6 \times 5 \times 4}{10 \times 9 \times 8}=\frac{1}{6}$
For selecting 1 white ball we will select all 2 balls from red and 1 from white
$\therefore P(x=1)=\frac{6 c_{2} \times 4 c_{1}}{10 c_{3}}=\frac{15}{30}=\frac{1}{2}$
For selecting 2 white balls, we will select and 1 ball from red and 2 from white
$\therefore P(x=2)=\frac{6 c_{1} \times 4 c_{2}}{10 c_{3}}=\frac{9}{30}=\frac{3}{10}$
For selecting 3 white balls, we will select all 0 balls from red and 3 from white
$\therefore P(x=3)=\frac{6 c_{0} \times 4 c_{3}}{10 c_{3}}=\frac{1}{30}$
Now we have X and P(X)
∴ Now we have are ready to write the probability distribution of x ÷
The following table gives probability Distributions:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{3}{10}$	$\frac{1}{30}$

Mean and Variance of a Random Variable exercise 31.1 question 19

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{64}{81}$	$\frac{16}{81}$	$\frac{1}{81}$

Given: find the probability of y in two throws of two dice, where y represents the number of times a total of 9 appears when two fair dice are drawn.
Solution: total 36 possible outcome
In our question, we are throwing two dice 2 times
∴y Denotes the number of times a sum of two numbers appearing on dice is equal to 9
∴y Can take values 0,1 or 2
Means in both the throw sum of 9 is not obtained and both the throw of two die a sum of 9 is obtained
∴(3,6) (6,3) (5,4) and (4,5) are the combinations resulting in sum = 9
Let A denotes the event of getting a sum of 9 a throw of 2 dice
$\therefore P\left ( A \right )$ =probability of getting the sum of 9 in a throw of 2 dice is
$\frac{4}{36}=\frac{1}{9}$
And probability of not getting the sum of 9 in the throw of 2 dice $=\frac{32}{36}=\frac{8}{9}$
Note: $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$ where A and B are independent events
$P\left ( Y=0 \right )=P\left ( {A}' \right )P\left ( {A}' \right )$
$\begin{aligned} &=\frac{8}{9} \cdot \frac{8}{9}=\frac{64}{81} \\ &P(Y=1)=P(A) P\left(A^{\prime}\right)+P\left(A^{\prime}\right) P(A) \\ &=\frac{8}{9} \times \frac{1}{9}+\frac{1}{9} \times \frac{8}{9}=\frac{16}{81} \\ &P(Y=2)=P(A) P(A) \\ &=\frac{1}{9} \times \frac{1}{9}=\frac{1}{81} \end{aligned}$
Now we have X and P(X)
∴ the required probability distribution of x
The following table gives probability
Distributions:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{64}{81}$	$\frac{16}{81}$	$\frac{1}{81}$

Mean and Variance of a Random Variable exercise 31.1 question 20

Answer:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{969}{2530}$	$\frac{114}{253}$	$\frac{38}{253}$	$\frac{4}{253}$	$\frac{1}{2530}$

Given: from a lot containing 25 items, 5 of which are defective, 4 are chosen at random.Hint: use probability formula and combination formula
Let X be the number of defectives found obtains the probability distribution of x if the items are chose without replacement
Solution: X represents the number of defective items drawn.
∴x can take value 0,1,2,3,4
∴ there are total 25 items ( 20 good + 5 defective) items
N(s) = total possible ways of selling 5 items 25 $c_{4}$
P (x=0)= P (selecting no defective item)
$\frac{20 c_{4}}{25 c_{4}}=\frac{969}{2530}$
P (x=1)= P (selecting 1 defective item and 3 good item)
$=\frac{5 c_{2} \times 20 c_{2}}{25 c_{4}}=\frac{38}{253}$
P (x=2) = P ( selecting 2 defective item and 2 good item)
$=\frac{5 c_{2} \times 20 c_{2}}{25 c_{4}}=\frac{38}{253}$
P (x=3) = P ( selecting 3 defective item and 1 good item)
$=\frac{5 c_{3} \times 20 c_{1}}{25 c_{4}}=\frac{4}{253}$
P (x=4) =P (selecting 4 defective items and no good item)
$=\frac{5 c_{4}}{25 c_{4}}=\frac{1}{2530}$
Now we have $P_{1}$ and $X_{1}$
The following table gives probability Distribution of x:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{969}{2530}$	$\frac{114}{253}$	$\frac{38}{253}$	$\frac{4}{253}$	$\frac{1}{2530}$

Mean and Variance of a Random Variable exercise 31.1 question 21

Answer:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{27}{64}$	$\frac{27}{64}$	$\frac{9}{64}$	$\frac{1}{64}$

Hint: use probability formula
Given: Three cards are drawn successively with replacement from a well shuffled deck of 52 cards.
A random variable x denotes the number of hearts in the three cards drawn. Determine the probability distribution of x
Solution: let us solve the problem now:
Note: in our problem, it is given that after every draw, we are replacing the card, so our sample spaces will no change.
In a desk of 52 cards, there are 13 hearts let x be the random variable denoting the number of hearts drawn for an event when 3 cards are drawn successively with replacement
∴x Can take value 0,1,2,3
$P(x=0)=\frac{39}{52} \times \frac{39}{52} \times \frac{39}{52}=\frac{27}{64}$
{As we have to select 1 card at a time such that no heart is there so the first probability is $\frac{39}{52}$ and as the drawn card is replaced time again probability is $\frac{39}{52}$ and again the same thing}
Similarly we proceed for other cases.
$P(x=1)=\frac{13}{52} \times \frac{39}{52} \times \frac{39}{52}+\frac{39}{52} \times \frac{13}{52} \times \frac{39}{52}+\frac{39}{52} \times \frac{39}{52} \times \frac{13}{52}=\frac{27}{64}$
{We figure might get a heart is the first card or second card or third so probabilities in all 3 cases are added as they are mutually exclusive events}
Similarly
$P(x=2)=\frac{13}{52} \times \frac{13}{52} \times \frac{39}{52}+\frac{39}{52} \times \frac{13}{52} \times \frac{13}{52}+\frac{13}{52} \times \frac{39}{52} \times \frac{13}{52}=\frac{9}{64}$
{first two cards are heart and 3rd is non-heart last 2 are hearts and so on cases as these cases are mutually exclusive hence they are added }
Similarly
$P(x=3)=\frac{13}{52} \times \frac{13}{52} \times \frac{13}{52}=\frac{1}{64}$
Now we have X and P(X)
∴ Now we are ready to write the probability distribution of x.
The following table gives probability distribution

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{27}{64}$	$\frac{27}{64}$	$\frac{9}{64}$	$\frac{1}{64}$

Mean and Variance of a Random Variable exercise 31.1 question 22

Answer:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{64}{343}$	$\frac{144}{343}$	$\frac{108}{343}$	$\frac{27}{343}$

Hint: use probability formula
Given: An contains 4 red and 3 blue balls, find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement
Solution: Let X denote the number of blue balls drawn in a random draw of 3 balls.
∴X can take value 0,1,2,3
Since bag contains 4 red and 3 blue balls i.e. at a total of 7 balls
∴ Balls are drawn with replacement for selecting 0 blue balls, we will select all 3 red balls.
$=\frac{4}{7} \times \frac{4}{7} \times \frac{4}{7}=\frac{64}{343}$
For selecting 1 blue ball, we will select all 2 red balls and 1 blue
$\therefore P(x=1)=\frac{3}{7} \times \frac{4}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{4}{7} \times \frac{3}{7}=\frac{144}{343}$
For selecting 2 blue balls, we will select all 1 red balls and 2 blue.
$\therefore P(x=2)=\frac{3}{7} \times \frac{3}{7} \times \frac{4}{7}+\frac{4}{7} \times \frac{3}{7} \times \frac{3}{7}+\frac{3}{7} \times \frac{4}{7} \times \frac{3}{7}=\frac{108}{343}$
For selecting 3 blue balls, we will select all 3 blue and 0 red balls
$\therefore P(x=3)=P(\text { selecting all blue balls })=\frac{3}{7} \times \frac{3}{7} \times \frac{3}{7}=\frac{27}{343}$
Now we have X and P(X)
∴ Now we are ready to write the probability distribution of x.
The following table gives probability distribution

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$\frac{64}{343}$	$\frac{144}{343}$	$\frac{108}{343}$	$\frac{27}{343}$

Mean and Variance of a Random Variable exercise 31.1 question 23

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{19}{34}$	$\frac{13}{34}$	$\frac{1}{17}$

Hint: use combination formula
Given: two cards are drawn simultaneously from a well shuffled deck of 52 cards. Find the probability distribution of the number of successes when getting a spade is considered a success
Solution: in a deck of 52 cards, there are 13 spades, let x be the random variable denoting the number of success and success here is getting a spade for an event when two cards are drawn simultaneously
∴ We can say the number of successes is equal to some spades obtained in each draw.
∴ X can take value 0, 1 or 2
[For selecting 0 spades, we removed all 13 spade from the deck and selected out of 39]
$\therefore P(x=0)=\frac{39 c_{2}}{52 c_{2}}=\frac{39 \times 38}{52 \times 51}=\frac{19}{34}$
[For selecting 1 spades, we need to select and 1 out of 13 spade and not any other spade]
$P(x=1)=\frac{13 c_{1} \times 39 c_{1}}{52 c_{2}}=\frac{13 \times 39 \times 38}{52 \times 51}=\frac{13}{34}$
[For selecting 2 spade, we need to select and 2 out of 13 spade]
$P(x=2)=\frac{13 c_{2}}{52 c_{2}}=\frac{13 \times 12}{52 \times 51}=\frac{1}{17}$
Now we have X and P(X)
∴ Now we are ready to write the probability distribution of x
The following probability distribution is given in this table

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{19}{34}$	$\frac{13}{34}$	$\frac{1}{17}$

Mean and Variance of a Random Variable exercise 31.1 question 24

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$

Hint: use probability formula
Given: A fair die is tossed twice if the number appearing on the top less than 3 it is successes find the probability distribution of number of successes
Solution: A fair dice is tossed twice. Every time a throwing dice is an independent event.
Note: $P\left ( A\cap B \right )=P\left ( A \right )P\left ( B \right )$ where A and B are independent events.
Let A denote the event of getting a number less than 3 on a single throw of dice
$\therefore P\left ( A \right )=\frac{2}{6}=\frac{1}{3}$ { out of 6 outcomes 1 and 2 are favourable}
$P(\operatorname{not} A)=P\left(A^{\prime}\right)=1-\frac{1}{3}=\frac{2}{3}$
As success is considered when number appearing on dice is less than 3
Let x denotes the success
As we are throwing two dice so that we can or we may even not get success
∴x can take value 0,1 or 2
$\begin{aligned} &P(x=0)=P(\text { not success }) \times \mathrm{P}(\text { not success })=P\left(A^{\prime}\right) \times P\left(A^{\prime}\right)=\frac{2}{3} \times \frac{2}{3}=\frac{4}{9} \\ &P(x=0)=P\left(A^{\prime}\right) \times P(A)+P\left(A^{\prime}\right) \times P(A) \\ &=\frac{2}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{3}=\frac{4}{9} \\ &P(x=2)=P(\text { success }) \times P(\text { success })=P(A) \times P(A)=\frac{1}{3} \times \frac{1}{3}=\frac{1}{9} \end{aligned}$
Now we have X and P(X)
∴ Now we are ready to write the probability distribution of x.
The following table gives probability distribution

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{4}{9}$	$\frac{4}{9}$	$\frac{1}{9}$

Mean and Variance of a Random Variable exercise 31.1 question 25

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{10}{21}$	$\frac{10}{21}$	$\frac{1}{21}$

Given: An urn contains 5 red and 2 black balls, two balls are randomly selected. let X represent the number of black balls. What are the possible values of x? is x a random variable?
Solution: the key point to solve the problem: A variable X is said to be a random variable if the sum of probabilities associated with each value of X gets equal to 1
i.e. $\varepsilon \left ( p_{1} \right )=1$ where $p_{1}$ is probability associated with $x_{1}$
x represents the number of black balls drawn
∴ x can take value 0,1 or 2 as both balls drawn can be black which corresponds to x = 2, either of 2 balls can be black which corresponds to x = 1 and if neither of balls drawn is black it corresponds to x = 0
∴ there are the of 7 balls.
$\mathrm{n}(\mathrm{s})=$ total possible ways of selecting 2 balls $=7 c_{2}$
$P(x=0)=P($ selecting no blank balls $)$
$\frac{5 c_{2}}{7 c_{2}}=\frac{5 \times 4}{7 \times 6}=\frac{10}{21}$
$P(x=1)=P($ selecting 1 black ball and 1 red $)$
$=\frac{5 c_{1} \times 2 c_{1}}{7 c_{2}}=\frac{5 \times 2 \times 2}{7 \times 6}=\frac{10}{21}$
$P(x=2)=P($ selecting all black balls $)$ $=\frac{2 c_{2}}{7 c_{2}}=\frac{2}{7 \times 6}=\frac{1}{21}$

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{10}{21}$	$\frac{10}{21}$	$\frac{1}{21}$

Clearly, $\varepsilon\left(P_{1}\right)=\frac{10}{21}+\frac{10}{21}+\frac{1}{21}=1$
∴ x is a random variable, and the above table represents its probability distribution.

Mean and Variance of a Random Variable exercise 31.1 question 26

Answer: X can take values -4, -2, 0, 2, 4 or 6

Hint: use probability concept

Given: Let X represent the difference between the no. of heads and the no. of tails when a coil is tossed 6 times. What are the possible values of X?

Solution: Since X represents the difference between some heads and number of tails when a coin is tossed 6 times.

As the coin is tossed 6 times following outcome are possible

Let H denoted heads and T denotes tails.

Outcomes: 6H 0T (6 Heads 0 tails), 5H 1T, 4H 2T, 3H 3T, 2H 4T and 1H 5T

∴ Difference can be |6-0|,|5- 1|, |4-2|,|3-3| ,| 2-4| and |1-5|

∴X can take value 0, 2, 4 & 6

Mean and Variance of a Random Variable exercise 31.1 question 27

Answer:

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{7}{15}$	$\frac{7}{15}$	$\frac{1}{15}$

Hint: Use probability & combination formula
Given: from a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the number of defective bulbs.
Solution: X represents the number of defective bulbs drawn
∴X can take values 0, 1 or 2 (as maximum bulbs drawn are 2)
∴ there are total 10 bulbs (7 good + 3 defectives)
n(s) = total possible ways of selecting ways of selecting 2 items from sample = 10 $c_{2}$
P(x=0) = P (selecting no defective bulb)
$=\frac{7 c_{2}}{10 c_{2}}=\frac{7 \times 6}{10 \times 9}=\frac{7}{15}$
P(x=1) = P (selecting 1 defective bulb and 1 good bulb)
$=\frac{7 c_{1} \times 3 c_{1}}{10 c_{2}}=\frac{7}{15}$
P(x=2) = P (selecting 2 defective bulb and 0 good bulb) $=\frac{7 c_{0} \times 3 c_{2}}{10 c_{2}}=\frac{3 \times 2}{10 \times 9}=\frac{1}{15}$

$X$	$0$	$1$	$2$
$P\left ( X \right )$	$\frac{7}{15}$	$\frac{7}{15}$	$\frac{1}{15}$

Mean and Variance of a Random Variable exercise 31.1 question 28

Answer:

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{1}{495}$	$\frac{32}{495}$	$\frac{168}{495}$	$\frac{224}{495}$	$\frac{70}{495}$

Given: 4 balls are drawn without replacement from box containing 8 red and 4 white balls. If X denotes the no. of red balls drawn, find the probability distribution of X.Hint: Use combination formula
Solution: Let X denote the no of red balls drawn in a random draw of 4 balls
∴X Can take values 0, 1, 2, 3 or 4
Since bag contains 8 red and 6 white balls i.e. total of 12 balls.
∴ Total no. of ways of selecting 4 balls and of 12 =12 $c_{1}$ for selecting 0 red balls are will select all 3 balls from red
$\therefore P(x=0)=P$ (not selecting any red ball)
$=\frac{4 c_{4}}{12 c_{4}}=\frac{4 \times 3 \times 2 \times 1}{12 \times 11 \times 10 \times 9}=\frac{1}{495}$
For selecting 1 red ball, we will select 1 ball from 8 red and 3 balls from 4 white
$\therefore P(x=1)=\frac{4 c_{3} \times 8 c_{1}}{12 c_{4}}=\frac{4 \times 8}{495}=\frac{32}{495}$
For selecting 2 red ball, we will select 2 balls from 8 red and 2 balls from 4 white
$\therefore P(x=2)=\frac{4 c_{2} \times 8 c_{2}}{12 c_{4}}=6 \times \frac{28}{495}=\frac{168}{495}$
For selecting 3 red ball, we will select 3 balls from 8 red and 1 ball from 4 white
$\therefore P(x=3)=\frac{4 c_{1} \times 8 c_{3}}{12 c_{4}}=4 \times \frac{56}{495}=\frac{224}{495}$
$\therefore P(x=4)=P($ selecting all red ball $)$
$=\frac{8 c_{4}}{12 c_{4}}=\frac{8 \times 7 \times 6 \times 5}{12 \times 11 \times 10 \times 9}=\frac{70}{495}$
Now we have P(x) and X

$X$	$0$	$1$	$2$	$3$	$4$
$P\left ( X \right )$	$\frac{1}{495}$	$\frac{32}{495}$	$\frac{168}{495}$	$\frac{224}{495}$	$\frac{70}{495}$

Mean and Variance of a Random Variable exercise 31.1 question 29

Answer: $(i) K=\frac{8}{15}$ $(i i) P(x \leq 2)=\frac{14}{15}, P(x>2)=\frac{1}{15}=1$ $(iii) P(x \leq 2)+P(x>2)=1$
Hint: Use probability formula
Given: the probability distribution of a random variable x is given below:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$K$	$\frac{k}{2}$	$\frac{k}{4}$	$\frac{k}{8}$

i }Determine the value of k
ii} determine $P\left ( x\leq 2 \right )$ and $P\left ( x> 2 \right )$
ii} Find $P\left ( x\leq 2 \right )+P\left ( x> 2 \right )$
Solution: the key point to solve the problem is, if a probability distribution is given the as per definition sum of probability associated with each value of a random variable of given distribution is equal to 1
i.e. $\varepsilon \left ( p_{1} \right )=1$
Given distribution is:

$X$	$0$	$1$	$2$	$3$
$P\left ( X \right )$	$K$	$\frac{k}{2}$	$\frac{k}{4}$	$\frac{k}{8}$

$\begin{aligned} &\therefore k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8}=1 \\ &\Rightarrow 8 k+4 k+2 k+k=8 \\ &\Rightarrow k=\frac{8}{15}(\text { ans }) \end{aligned}$
$ii \}P(x=2)=\frac{k}{4}$ and $P(x=3)=\frac{k}{8}$
$\begin{aligned} &\therefore P(x>2)=P(x=3)=\frac{k}{8}=\frac{\frac{8}{15}}{8}=\frac{1}{15} \\ &P(x \leq 2)=P(x=0)+P(x=1)+P(x=2) \\ &\Rightarrow k+\frac{k}{2}+\frac{k}{4} \\ &\Rightarrow \frac{7 k}{4}=\frac{7}{4} \times \frac{8}{15}=\frac{14}{15} \end{aligned}$
$iii \} \quad \therefore P(x \leq 2)+P(x>2)=\frac{14}{15}+\frac{1}{15}=1$

Mean and Variance of a Random Variable exercise 31.1 question 30

Answer: $k=\frac{1}{8}$ $\left ( i \right )\frac{1}{8}$ $\left ( ii \right )\frac{5}{8}$ $\left ( iii \right )\frac{7}{8}$
Hint: Let P(X=0) = 0
$P(X=x)=\left\{\begin{aligned} k x, & \text { if } x=0,1 \\ 2 k x, & \text { if } x=2 \\ k(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { if } x>4 \end{aligned}\right.$
Where k is a positive constant
find the value of k. also find the probability that you will get admission in
$\left ( i \right )$ Exactly on colleges $\left ( ii \right )$ at most 2 colleges $\left ( iii \right )$ at least 2 colleges
Solution:
$P(X=x)=\left\{\begin{aligned} k x, & \text { if } x=0,1 \\ 2 k x, & \text { if } x=2 \\ k(5-x), & \text { if } x=3 \text { or } 4 \\ 0, & \text { if } x>4 \end{aligned}\right.$
Our variable is X and from equation we see that it is taking values X = 0,1,2,3,4….( any whole number )
And it represents the no. of colleges in which you are going to get admission
According to equation given we have:
$\begin{aligned} &\therefore P(x=0)=k \times 0=0 \\ &P(x=1)=k \times 1=k \\ &P(x=2)=2 k \times 2=4 k \\ &P(x=3)=k(5-3)=2 k \\ &P(x=4)=k(5-4)=k \\ &P(x>4)=0 \end{aligned}$
As in question it is not given either x is random variable or the given distribution is a probability distribution, we can’t apply anything directly.
But we have a hint in the question
As $P(x=0)=0$
It means the chance of not getting admission in any college 0
∴admission is sure
Hence the sum of all probability must be equal to as getting admission has become a sure event.
This makes the given distribution a probability distribution an X a random variable also.
$\begin{aligned} &\therefore k+4 k+2 k+k=1 \Rightarrow 8 k=1 \\ &k=\frac{1}{8} \end{aligned}$
i} P (getting admission in exactly one colleges )= $P\left ( x=1 \right )$
$=k.1=\frac{1}{8}$
ii} P (getting admission in almost 2 colleges) = $P\left ( x\leq 2 \right )$
$\begin{aligned} &=P(x=1)+P(x=2) \\ &=k+4 k=5 k=\frac{5}{8} \end{aligned}$
iii} P (getting admission in at least 2 colleges) = $P\left ( x\geq 2 \right )$
$\begin{aligned} &=P(x=2)+P(x=3)+P(x=4) \\ &=4 k+2 k+k=7 k=\frac{7}{8} \end{aligned}$

Mean and Variance of a Random Variable exercise 31.1 question 31

Answer: $\frac{4}{19}$

Hint: use probability concept.

Given: a bag contains 19 tickets numbered from 1 to 19. A ticket is drawn and then another ticket is drawn without replacement. The probability that both the tickets will show even number on the ticket is

Solution: the probability of getting an even number in first draw $=\frac{9}{19}$ . The probability of getting an even number in second draw $=\frac{8}{18}$

Both are independent event and so required probability $=\frac{9}{19}\times \frac{8}{18}==\frac{4}{19}$

The portions in the class 12 mathematics, chapter 21, Mean and Variance of a Random Variable, consist of two exercises, ex 31.1 and ex 31.2. The first exercise, ex 31.1, consists of 37 questions. The concepts in this exercise are finding value from the probability distributions and finding the probability of the given combinations. It includes balls, cards, dice, coins and so on. When students doubt the method or formula for solving these sums, they can refer to the RD Sharma Class 12 Chapter 31 Exercise 31.1 reference book.

This exercise is related to the previous Chapter, Probability. The students must concentrate on understanding the formulas used in the last Chapter and the present one. The sums in the RD Sharma Class 12th Exercise 31.1 are based on the NCERT pattern that makes the CBSE students follow easily. They can use this solution book to prepare for their exams, complete their homework and work on assignments. The students will soon witness them scoring above their benchmark.

Each answer in the Class 12 RD Sharma Chapter 31 Exercise 31.1 Solution is solved and verified by the mathematical experts. The solutions are provided in every possible method using formulas and shortcuts. The students must understand the concept in this exercise before jumping to ex 31.2. With the help of various practice questions given in the RD Sharma Class 12 Solutions Mean and Variance of a Random Variable Ex 31.1, they can work out any additional sums.

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