If X has a Poisson distribution such that $P(X=1)=4P(X=2),$ then $P(X=0)$ is :
Answer & explanation
Correct answer: option 1
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,
$P(X=1)=4P(X=2)$
$\frac{λ^1e^{-λ}}{1!}=\frac{λ^2e^{-λ}}{2!}⇒λ=\frac{1}{2}$
$P(X=0)=\frac{λ^0e^{-λ}}{0!}=e^{-λ}=e^{-\frac{1}{2}}$