Question

If a random variable $X$ follows a P.D. such that $P(X=1)=P(X=2)$, then $P(X=0)=$

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

Answer 1

Answer: (B)

$\frac{1}{e^2}$

$P(X=1)=P(X=2)$
$P$$(1;\mu )=\frac{e^{-\mu }{\mu }^1}{1!}=P(2;\mu )=\frac{e^{-\mu }{\mu }^2}{2!}$
$e^{-\mu }\left[\mu \right(2-\mu \left)\right]$ $=0$

either $\mu =0or\mu =2$

$P(X=0)=$$\frac{e^{-\mu }{\mu }^0}{0!}=e^{-\mu }$

So, either $e^{-0}$or $e^{-2}$
that is 1 or $\frac{1}{e^2}$