Question

Suppose $X$ is a poisson variable such that $P(X=2)=\frac{2}{3}P(X=1)$, then $P(x=0)$ is

• A. $\frac{3}{4}$
• B. $e^{\frac{4}{3}}$
• C. $e^{\frac{-4}{3}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

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

Poisson's distribution implies
$P(X=n)=\frac{{\lambda }^n{e}^{-\lambda }}{n!}$

Where mean=variance=$\lambda$

Hence, $P(X=2)=\frac{2}{3}P(x=1)$

$\frac{{\lambda }^2{e}^{-\lambda }}{2!}=\frac{2{\lambda }^1{e}^{-\lambda }}{3.1!}$
$\frac{\lambda }{2}=\frac{2}{3}$

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