Let X∼Bin (3, p) be a binomial random variable. If $P(X=3)=\frac{1}{12}P(X=1)$, then p is equal to :

$\frac{1}{3}$

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

for a binomial distribution,

$P(X=k)=\left({^nC}_k\right)p^k(1-p)^{n-k}$

and,

$P(X=3)=\frac{1}{12}P(X=1)$

$p^3=\frac{1}{12}3p(1-p)^2$

$⇒3p^2+2p-1=0$

$⇒(p+1)(p-\frac{1}{3})=0$

∴ Probability (p) = $\frac{1}{3}$