The random  variable X can take only the values 1 and 2. Given that P(X=1)=p and that $E(X^2)=E(X)+1.$ The value of p is :

$\frac{1}{2}$

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

$E(X)=P(X=1).1+P(X=2).2$

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

$=p+2-2p$

$=2-p$

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

$=p+(1-p).4$

$=4-4p+p$

$=4-3p$

$E(X^2)=E(X)+1$  (given)

$4-3p=2-p+1$

$2p=1$

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