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Explain Solution R.D.Sharma Class 12 Chapter 31Mean and Variance of a Random Variable  Exercise 31.2 Question 16 maths Textbook Solution.

Answers (1)

Answer:

-\frac{30}{13}

Hint:

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

Given:

Person loss 10 rupees and get 90 rupees.

Solution:

P (Wheels/Getting same slot) =\frac{1}{13}=p

\begin{aligned} &p^{\prime}=q=1-p=1-\frac{1}{13}=\frac{12}{13}\\ &\begin{array}{|c|c|c|} \hline x_{i} & 90 & -10 \\ \hline p\left(x_{i}\right) & \frac{1}{13} & \frac{12}{13} \\ \hline \end{array} \end{aligned}

\begin{aligned} &\text { Mean } E(X)=\sum_{i=1}^{n} x_{i} p_{i} \\ &=90 \times \frac{1}{13}-10 \times \frac{12}{13} \\ &=\frac{90}{13}-\frac{120}{13} \\ &=-\frac{30}{13} \end{aligned}

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