Question

If X is a poisson variate such that $P(X=2)=9p(X=4)+90p(X=6)$ , then the mean of x is

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

Answer: (C)

$1$

For Poisson's distribution, $P\left(X\right)$ is given by
$P\left(X\right)=\frac{e^{-\lambda }.{\lambda }^x}{x!}$
It is given that $P(X=2)=9P(X=4)+90P(X=6)$
$\Rightarrow \frac{e^{-\lambda }.{\lambda }^2}{2!}=9.\frac{e^{-\lambda }.{\lambda }^4}{4!}+90.\frac{e^{-\lambda }.{\lambda }^6}{6!}$
Cancelling $e^{-\lambda }$ from both sides, we get
$\frac{{\lambda }^2}{2!}=9.\frac{{\lambda }^4}{4!}+90.\frac{{\lambda }^6}{6!}$
Since $\lambda$ can not be zero, cancel out ${\lambda }^2$ from both sides to obtain
${\lambda }^4+3{\lambda }^2-4=0$.
Solving , we get $\lambda =1$ as a valid solution.