Question

If $X$ is a random poission variate such that $2P(X=0)+P(X=2)=2P(X=1)$ then $E\left(X\right)=$

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

Poisson's distribution implies
$P(X=n)=\frac{{\lambda }^n{e}^{-\lambda }}{n!}$
Where mean=variance=$\lambda$
Hence $2.P(x=0)+P(x=2)=2P(x=1)$
$2\frac{{\lambda }^0{e}^{-\lambda }}{0!}+\frac{{\lambda }^2{e}^{-\lambda }}{2!}=2\frac{{\lambda }^1{e}^{-\lambda }}{1!}$

$2e^{-\lambda }+\frac{{\lambda }^2{e}^{-\lambda }}{2!}=2\lambda e^{-\lambda }$

$2\lambda e^{-\lambda }-2e^{-\lambda }=\frac{{\lambda }^2{e}^{-\lambda }}{2!}$
$4\lambda -4={\lambda }^2$
${\lambda }^2-4\lambda +4=0$
$(\lambda -2{)}^2=0$
$\lambda =2$
Hence mean=variance=$2$