The random variable x can take values 0, 1, 2, 3, Given that P(x=0)=P…

The correct answer is Option (2) → $\frac{3}{8}$

Given,

$P(x=0)=P(x-1)=P$

$P(x-2)=P(x-3)$

$E(x^2)=2E(x)$

Since the total probability must be 1,

$p+p+q+=1$

$⇒p+q=\frac{1}{2}$   ....(1)

$E(x)=0.p+1.p+2.q+3.q=p+5q=\frac{1}{2}+4q$

$E(x^2)=0^2.p+1^2.p+2^2.q+3^2.q=p+13q=\frac{1}{2}+12q$

and,

$E(x)=E(x^2)$

$\frac{1}{2}+4q=\frac{1}{2}+12q$

$q=\frac{1}{8}$

$⇒p=\frac{3}{8}$