The random variable X can take only values 0, 1, 2. Given that P(X=0)…

The correct answer is option (1) → $\frac{1}{4}$

$P(X=0)=P(X=1)=ρ$ [Given]

let $P(X=2)=ρ_2$, since the total probability must be 1,

$ρ+ρ+ρ_2=1$

$⇒ρ_2=1-2ρ$

and,

$E(X^2)=E(X)+1$ [Given]

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

$=0.ρ+1.ρ+2(1-2ρ)$

$=2-3ρ$

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

$=0.ρ+1.ρ+4(1-2ρ)$

$=4-7ρ$

$∴4-7ρ=(2-3ρ)+1$

$⇒1=4ρ$

$⇒ρ=\frac{1}{4}$