Question

If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$, then $P(X=4)$ is equal to

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

Answer 1

Answer: (C)

Given that, $P(X-1)=P(X=2)$
$\frac{e^{-\lambda }{\lambda }^1}{1!}=\frac{e^{-\lambda }{\lambda }^2}{2!}\Rightarrow \lambda =2$
$\therefore P(X=4)=\frac{e^{-2}(2{)}^4}{4!}=\frac{e^{-2}\times 16}{24}=\frac{2}{3e^2}$