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(X…

Question

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 :

Accepted Answer

$$\begin{aligned}
& 2 p+2 q=\frac{1}{2} \
& p+q \
& E\left(x^2\right)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 \cdot p+1 . p+4 \cdot q+9 q \
& =p+13 q \
& E(x)=\sum_{i=0}^3 x_i^2 p\left(x_i\right)=0 . p+1 . p+2 q+3 q=p+5 q \
& p+13 q=2(p+5 q) \
& p=3 q \
& \text { So, } q=\frac{1}{8} & p=\frac{3}{8} \
& \text { So, } 8 p-1=2\quad\text{Option (2)}
\end{aligned}$$

Solutions · 1