If X has a Poisson distribution such that $P(X=1)=4P(X=2),$ then $P(X=0)$ is :

$e^{-\frac{1}{2}}$

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

The probability mass function (PMF) is,

$P(X=k)=\frac{λ^ke^{-λ}}{k!}$

and,

$P(X=1)=4P(X=2)$

$\frac{λ^1e^{-λ}}{1!}=\frac{λ^2e^{-λ}}{2!}⇒λ=\frac{1}{2}$

$P(X=0)=\frac{λ^0e^{-λ}}{0!}=e^{-λ}=e^{-\frac{1}{2}}$