Let a random variable X take values 0, 1, 2, 3 with P(X=0)=P(X=1)=p, P(X=2)=P(X=3) and E(X2)=2E(X). Then the value of 8p−1 is :
Solution
<p>$$\begin{aligned}
& 2 p+2 q=\frac{1}{2} \\
& p+q \\
& E\left(x^2\right)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 . p+4 \cdot q+9 q \\
& =p+13 q \\
& E(x)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 . p+1 . p+2 q+3 q=p+5 q \\
& p+13 q=2(p+5 q) \\
& p=3 q \\
& \text { So, } q=\frac{1}{8} \& p=\frac{3}{8} \\
& \text { So, } 8 p-1=2
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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