A random variable X takes only 3 values {0, 1, 2} with $P(X=0)=P(X=1).$ If $E(X^2)=1+E(X)$ then the var(X) is :

$\frac{11}{16}$

The correct answer is Option (3) → $\frac{11}{16}$

$P(X=0)=P(X=1)=p$

$E(X^2)=1+E(X)$

$⇒P(X=2)=1-2p$

$E(X)=0.p(X=0)+1.p(X=1)+2.p(X=2)$

$=0p+1p+2(1-2p)$

$=2-3p$

$E(X^2)=0^2.p(X=0)+1^2.p(X=1)+2^2.p(X=2)$

$=0p+1p+4(1-2p)$

$=4-7p$

$E(X^2)=1+E(X)$

$4-7p=3-3p$

$p=\frac{1}{4}$

$Var X=E(X^2)-[E(X)]^2$

$=\frac{9}{4}-\frac{25}{16}=\frac{11}{16}$