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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise 31.2 Question 13 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 31 Mean and Variance of a Random Variable Exercise 31.2  Question 13  Maths Textbook Solution.

Answers (1)

Answer:

Mean = 2.3

Variance = 5

Hint:

E(X)=\sum_{i=1}^{n} X_{i} P_{i} \text { and } \operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2}

Given:

There are 5 cards numbered 1, 1, 2, 2 and 3

Solution:

There are 5 cards numbered 1, 1, 2, 2 and 3

Let, x = sum of two numbers on card =2, 3, 4, 5

Y = Maximum of two numbers = 1, 2, 3

Thus the probability distribution of x is given by

\begin{array}{|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 \\ \hline P(x) & \frac{1}{10} & \frac{4}{10} & \frac{3}{10} & \frac{2}{10} \\ \hline \end{array}

Total number of cards = 5 (2 cards with “1” + 2 cards with “2” + 1 card with “3”

Total number of possible choice (in drawing the two cards) = number of ways in which threw two cards can be drawn from the total 5

\begin{aligned} &=5 C_{2} \\ &=\frac{5 !}{2 ! 3 !}=10 \end{aligned}

Probability that the two car drawn would be having the number 1 and 1 on them is P (11)

P(1,1)=\frac{2 C_{2} \times 2 C_{0} \times 1 C_{0}}{5 C_{2}}=\frac{1 \times 1 \times 1}{10}=\frac{1}{10}

Probability that two cards drawn would be having the number 1 and 2 on them is P (11)

\begin{aligned} &P(1,1)=\frac{2 C_{2} \times 2 C_{0} \times 1 C_{0}}{5 C_{2}} \\ &=\frac{1 \times 1 \times 1}{10}=\frac{1}{10} \end{aligned}

"1" and "2" on them

P(1,2)=\frac{2 C_{1} \times 2 C_{1} \times 1 C_{0}}{5 C_{2}}

\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 1 & 1 & 0 & 2 \\ \hline \text { Choices } & 2 C_{1} & 2 C_{1} & 1 C_{0} & 5 C_{2} \\ \hline \end{array}\\ &\begin{aligned} &=\frac{\frac{2}{1} \times \frac{2}{1} \times 1}{10} \\ &=\frac{2 \times 2 \times 1}{10}=\frac{4}{10}=\frac{2}{5} \end{aligned} \end{aligned}

"1" and "3" on them

P(1,3)=\frac{2 C_{1} \times 2 C_{0} \times 1 C_{1}}{5 C_{2}}

\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 1 & 0 & 1 & 2 \\ \hline \text { Choices } & 2 C_{1} & 2 C_{0} & 1 C_{1} & 5 C_{2} \\ \hline \end{array}\\ &=\frac{\frac{2}{1} \times 1 \times \frac{1}{1}}{10}=\frac{2 \times 1 \times 1}{10}=\frac{2}{10}=\frac{1}{5} \end{aligned}

"2" and "2" on them

P(2,2)=\frac{2 C_{1} \times 2 C_{2} \times 1 C_{1}}{5 C_{2}}

\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 0 & 2 & 0 & 2 \\ \hline \text { Choices } & 2 C_{0} & 2 C_{2} & 1 C_{0} & 5 C_{2} \\ \hline \end{array}\\ &=\frac{1 \times 1 \times 1}{10}=\frac{1}{10} \end{aligned}

"2" and "3" on them

P(2,3)=\frac{2 C_{0} \times 2 C_{1} \times 1 C_{1}}{5 C_{2}}

\begin{aligned} &\begin{array}{|c|c|c|c|c|} \hline & 1^{\prime} s & 2^{\prime} s & 3^{\prime} s & \text { Total } \\ \hline \text { Available } & 2 & 2 & 1 & 5 \\ \hline \text { To choose } & 0 & 1 & 1 & 2 \\ \hline \text { Choices } & 2 C_{0} & 2 C_{1} & 1 C_{1} & 5 C_{2} \\ \hline \end{array}\\ &=\frac{1 \times \frac{2}{1} \times 1}{10}=\frac{1 \times 2 \times 1}{10}=\frac{2}{10}=\frac{1}{5} \end{aligned}

Probability for the sum of the members on the cards drawn to be

\begin{aligned} &2 \Rightarrow P(x=2)=P(1,1)=\frac{1}{10} \\\\ &3 \Rightarrow P(x=3)=P(1,2)=\frac{4}{10} \\\\ &4 \Rightarrow P(x=4)=P(1,3 \text { or } 2,2) \\\\ &\text { i. } e, P(1,3 \text { or } 2,2)=P(1,3)+P(2,2) \\\\ \end{aligned}

\begin{aligned} &=\frac{2}{10}+\frac{1}{10} \\\\ &=\frac{2+1}{10}=\frac{3}{10} \\\\ &5 \Rightarrow P(x=2)=P(2,3)=\frac{2}{10} \end{aligned}

The probability distribution of x would be

\begin{array}{|c|c|c|c|c|} \hline x & 2 & 3 & 4 & 5 \\ \hline P(X=x) & \frac{1}{10} & \frac{4}{10} & \frac{3}{10} & \frac{2}{10} \\ \hline \end{array}

Calculations for mean and standard deviation

\begin{array}{|c|c|c|c|c|} \hline x & P(X=x) & P x=x \cdot P(X=x) & x^{2} & P x^{2}=x^{2} \cdot P(X=x) \\ \hline 2 & \frac{1}{10} & \frac{2}{10} & 4 & \frac{4}{10} \\ \hline 3 & \frac{4}{10} & \frac{12}{10} & 9 & \frac{36}{10} \\ \hline 4 & \frac{3}{10} & \frac{12}{10} & 16 & \frac{48}{10} \\ \hline 5 & \frac{2}{10} & \frac{10}{10} & 25 & \frac{50}{10} \\ \hline \text { Total } & 1 & \frac{36}{10}=3.6 & & \frac{138}{10}=13.6 \\ \hline \end{array}

The expected value of the sum

Expectation of x

\begin{aligned} &E(x)=\sum P x=3.6 \\ \end{aligned}

Variance of the sum of the numbers on the cards

\begin{aligned} &\operatorname{Var}(x)=E\left(x^{2}\right)-(E(x))^{2} \\ &=\sum P x-\left(\sum P x\right)^{2} \\ &=13.8-3.6^{2} \\ &=13.8-12.96 \\ &=0.84 \end{aligned}

Standard deviation of the sum of the numbers on the cards

\begin{aligned} &\text { S.D }(x)=+\sqrt{\operatorname{Var}(x)} \\ &=+\sqrt{0.84} \\ &=+0.917 \end{aligned}

Computation of mean and variance

\begin{array}{|c|c|c|c|} \hline x_{i} & p_{i} & p_{i} x_{i} & p_{i} x_{i}^{2} \\ \hline 2 & \frac{1}{10} & \frac{2}{10} & \frac{4}{10} \\ \hline 3 & \frac{4}{10} & \frac{12}{10} & \frac{36}{10} \\ \hline 4 & \frac{3}{10} & \frac{12}{10} & \frac{48}{10} \\ \hline 5 & \frac{2}{10} & 1 & \frac{50}{10} \\ \hline \end{array}

\begin{aligned} &\sum p_{i} x_{i}=\frac{36}{10}=3.6 \\ &\sum p_{i} x_{i}^{2}=\frac{138}{10}=13.8 \\ &\text { Mean }=\sum p_{i} x_{i}=3.6 \\ &\text { Variance }=\sum p_{i} x_{i}^{2}-(\text { Mean })^{2}=1 \end{aligned}

Thus the probability distribution of Y is given by

\begin{array}{|c|c|c|c|} \hline Y & 1 & 2 & 3 \\ \hline P(Y) & 0.1 & 0.5 & 0.4 \\ \hline \end{array}

Computation of Mean and Variance

\begin{array}{|c|c|c|c|} \hline y_{i} & p_{i} & p_{i} y_{i} & p_{i} y_{i}{ }^{2} \\ \hline 1 & 0.1 & 0.1 & 0.1 \\ \hline 2 & 0.5 & 1 & 2 \\ \hline 3 & 0.4 & 1.2 & 3.6 \\ \hline \end{array}

\begin{aligned} &\sum p_{i} y_{i}=2.3 \\\\ &\sum p_{i} y_{i}^{2}=5.7 \\\\ &\text { Mean }=\sum p_{i} y_{i}=2.3 \\\\ &\text { Variance }=\sum p_{i} y_{i}^{2}-(\text { mean })^{2}=5.7-5.29=0.41 \end{aligned}

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