Question

If $X$ is a poisson variate and $P(X=1)=2P(X=2)$, then $P(X=3)=X$

• A. $\frac{e^{-1}}{6}$
• B. $\frac{e^{-2}}{2}$
• C. $\frac{e^{-1}}{2}$
• D. $\frac{e^{-1}}{3}$

Answer 1

Answer: (A)

$\frac{e^{-1}}{6}$

$p\left(k\right)=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$

Now it is given that
$P\left(1\right)=2P\left(2\right)$
$\frac{\lambda .e^{-\lambda }}{1}=2\frac{{\lambda }^2.e^{-\lambda }}{2!}$

$\lambda =2\frac{{\lambda }^2}{2}$
$\lambda (1-\lambda )=0$
$\lambda =0$ or $\lambda =1$

If $\lambda =1$, then $P\left(3\right)=\frac{1^3.e^{-1}}{3!}$

$=\frac{e^{-1}}{6}$