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  • Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable Exercise 31.2 Question 18 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable  Exercise 31.2 Question 18 maths Textbook Solution.

Answers (1)

Answer:

\frac{4}{7},0.34

Hint:

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

Given:

An urn contains 5 red 2 black balls. Two balls are randomly drawn, without replacement.

Solution:

\begin{aligned} &\begin{aligned} &P(X=0)=\frac{5 C_{2}}{7 C_{2}}=\frac{10}{21} \\ &P(X=1)=\frac{2 C_{1} \times 5 C_{1}}{7 C_{2}}=\frac{10}{21} \\ &P(X=2)=\frac{2 C_{2}}{7 C_{2}}=\frac{1}{21} \end{aligned}\\ &\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{10}{21} & \frac{10}{21} & \frac{1}{21} \\ \hline \end{array} \end{aligned}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0+\frac{10}{21}+\frac{2}{21} \\ &=\frac{12}{21}=\frac{4}{7} \\ &E\left(X^{2}\right)=\sum_{i=1}^{n} X_{i}^{2} P_{i} \\ &=\frac{10}{21}+\frac{4}{21} \\ &=\frac{14}{21}=\frac{2}{3} \\ &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \\ &=\frac{2}{3}-\frac{16}{49} \\ &=\frac{50}{147}=0.34 \end{aligned}

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