A discrete random variable X can take values 0, 1, 2, 3. Given that P…

The correct answer is Option (3) → $\frac{1}{2}$

$P(X=0)=P(X=2)=k,\quad P(X=1)=P(X=3)=p$

$2k+2p=1 \Rightarrow k+p=\frac{1}{2}$

$E(X)=0\cdot k+1\cdot p+2\cdot k+3\cdot p=2k+4p$

$E(X^2)=0+1\cdot p+4\cdot k+9\cdot p=4k+10p$

$E(X^2)=2E(X)$

$4k+10p=2(2k+4p)=4k+8p$

$10p=8p \Rightarrow 2p=0 \Rightarrow p=0$

$k+\;0=\frac{1}{2} \Rightarrow k=\frac{1}{2}$

$k=\frac{1}{2}$