Suppose that X has Poisson distribution. If $P(X=2)=\frac{2}{3} P(X=1)$, then $P(X=0)$ is equal to

$e^{\frac{-4}{3}}$

The correct answer is Option (3) → $e^{\frac{-4}{3}}$

$P(X=k) = \frac{e^{-\lambda}\lambda^k}{k!}$

$P(X=2) = \frac{e^{-\lambda}\lambda^2}{2}, \;\; P(X=1) = e^{-\lambda}\lambda$

$\frac{P(X=2)}{P(X=1)} = \frac{\lambda}{2}$

$\frac{\lambda}{2} = \frac{2}{3}$

$\lambda = \frac{4}{3}$

$P(X=0) = e^{-\lambda} = e^{-\frac{4}{3}}$

$P(X=0) = e^{-\frac{4}{3}}$