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 :

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Answer: A. 2 \begin{aligned} & 2 p+2 q=\frac{1}{2} \ & p+q \ & E\left(x^2ight)=\sum_{i=0}^3 x_i^2 p\left(x_iight)=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ight)=0 . p+1 . p+2 q+3 q=p+5 q \ & p+13 q=2(p+5 q) \ & p=3 q \ & 	ext { So, } q=\frac{1}{8} \& p=\frac{3}{8} \ & 	ext { So, } 8 p-1=2\quad	ext{Option (2)} \end{aligned}

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 :

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