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  • Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable Exercise 31.2 Question 21 maths Textbook Solution.

Explain Solution R.D. Sharma Class 12 Chapter 31Mean and Variance of a Random Variable  Exercise 31.2 Question 21 maths Textbook Solution.

Answers (1)

Answer:

\frac{2}{3}

Hint:

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

Given:

A pair of dice are thrown

Solution:

\begin{aligned} &S=\{(1,1),(2,2),(3,3,),(4,4),(5,5),(6,6)\} \\\\ &p=\frac{6}{36}=\frac{1}{6}, q=\frac{5}{6} \\ \end{aligned}

\begin{aligned} &P(X=0)=4 C_{0}\left(\frac{1}{6}\right)^{0} \times\left(\frac{5}{6}\right)^{4}=\frac{625}{1296} \\\\ &P(X=1)=4 C_{1}\left(\frac{1}{6}\right) \times\left(\frac{5}{6}\right)^{3}=\frac{125}{324} \\\\ &P(X=2)=4 C_{2}\left(\frac{1}{6}\right)^{2} \times\left(\frac{5}{6}\right)^{2}=\frac{25}{216} \\ \end{aligned}

\begin{aligned} &P(X=3)=4 C_{3}\left(\frac{1}{6}\right)^{3} \times\left(\frac{5}{6}\right)=\frac{5}{324} \\\\ &P(X=4)=4 C_{4}\left(\frac{1}{6}\right)^{4} \times\left(\frac{5}{6}\right)^{0}=\frac{1}{1296} \end{aligned}

\begin{array}{|c|c|c|c|c|c|} \hline X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{625}{1296} & \frac{125}{324} & \frac{25}{216} & \frac{5}{324} & \frac{1}{1296} \\ \hline X \cdot P(X) & 0 & \frac{125}{324} & \frac{25}{108} & \frac{5}{108} & \frac{1}{324} \\ \hline \end{array}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} X_{i} P_{i} \\ &=0+\frac{125}{324}+\frac{25}{108}+\frac{5}{108}+\frac{1}{324} \\ &=\frac{216}{324} \\ &=\frac{2}{3} \end{aligned}

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