A random variable X follow Poisson distribution such that $P(X=1) = 2P(X=2)$, then $P(X=0)$ is

$\frac{1}{e}$

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

$\text{Let }X\sim \text{Poisson}(\lambda).$

$P(X=1)=\frac{\lambda^1 e^{-\lambda}}{1!}=\lambda e^{-\lambda}$

$P(X=2)=\frac{\lambda^2 e^{-\lambda}}{2!}=\frac{\lambda^2}{2}e^{-\lambda}$

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

$\lambda e^{-\lambda}=2\left(\frac{\lambda^2}{2}e^{-\lambda}\right)$

$\lambda e^{-\lambda}=\lambda^2 e^{-\lambda}$

$\lambda=\lambda^2$

$\lambda(\,1-\lambda\,)=0$

$\lambda=1$ (non-zero parameter)

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

$P(X=0)=\frac{1}{e}$