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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise 31.2 Question 5 Maths Textbook Solution.

Provide Solution For R.D. Sharma Maths Class 12 Chapter 31 Mean and Variance of a random Exercise 31.2 Question 5 Maths Textbook Solution.

Answers (1)

Answer:

Mean=\frac{34}{221} , variance =\frac{400}{2873}, standard deviation=0.37

Hint:

E\left ( X \right )=\sum_{i-1}^{n}X_{i}P_{i}

Standard deviation=\sqrt{var\left ( X \right )}

Given:

Number of kings in 52 cards

Solution:

Let number of kings be X

P (X=0) = P (Number of kings)

=\frac{48C_{2}}{52C_{2}}=\frac{48\times 47}{52\times 51}=\frac{188}{221}

\begin{aligned} &P(X=1)=P(1 \mathrm{King}) \\\\ &=\frac{4 C_{1} 48 C_{1}}{52 C_{2}}=\frac{4 \times 48 \times 2}{52}=\frac{32}{221} \\\\ &P(X=2)=P(2 \mathrm{King}) \\\\ &=\frac{4 C_{2}}{52 C_{2}}=\frac{4 \times 3}{52 \times 51}=\frac{1}{221} \end{aligned}

\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{188}{221} & \frac{32}{221} & \frac{1}{221} \\ \hline \end{array}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0 \times \frac{188}{221}+1 \times \frac{32}{221}+2 \times \frac{1}{221} \\ &=\frac{32}{221}+\frac{2}{221} \\ &=\frac{34}{221} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=0^{2} \times \frac{188}{221}+1^{2} \times \frac{32}{221}+2^{2} \times \frac{1}{221} \\ &=\frac{36}{221} \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}

\begin{aligned} &=\frac{36}{221}-\left(\frac{34}{221}\right)^{2} \\ &=\frac{400}{2873} \\ &\text { Standard deviation }=\sqrt{\operatorname{var}(X)}=\sqrt{\frac{400}{2873}}=0.37 \end{aligned}

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