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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 14  Maths Textbook Solution.

Answers (1)

Answer:

Variance=\frac{1}{2}

Hint:

Var\left ( X \right )=E\left ( X^{2} \right )-\left ( E\left ( X \right ) \right )^{2}

Given:

A die is tossed twice a “success” of getting odd number of toss

Solution:

\begin{aligned} &S=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,1) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\\ \end{aligned}

\begin{aligned} &\begin{array}{|c|c|c|c|} \hline X & 0 & 1 & 2 \\ \hline P(X) & \frac{9}{36} & \frac{18}{36} & \frac{9}{36} \\ \hline \end{array}\\ \end{aligned}

\begin{aligned} &\operatorname{Var}(X)=E\left(X^{2}\right)-(E(X))^{2} \end{aligned}

\begin{aligned} &=\frac{18}{36}+1-\left(\frac{18}{36}+\frac{18}{36}\right)^{2} \\ &=1.5-1=0.5 \\ &=\frac{1}{2} \end{aligned}

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