Question
If X has Poisson distribution with $m = 1$, then find $P(X \leq 1)$ given $e^{−1} = 0.3678$

Answer

Given, $m =1$ and $e ^{-1}=0.3678$
$\therefore X \sim P ( m )= X \sim P (1)$
The p.m.f. of $X$ is given by
$ P ( X = x )=\frac{ e ^{- m } m ^x}{x !}$
$\therefore P ( X = x )=\frac{ e ^{-1} 1^x}{x !}, x=0,1,2, \ldots$
$\therefore P ( X \leq 1)= P ( X =0 \text { or } X =1)$
$= P ( X =0)+ P ( X =1)$
$=\frac{ e ^{-1} 1^0}{0 !}+\frac{ e ^{-1} 1^1}{1 !}$
$=\frac{0.3678 \times 1}{1}+\frac{0.3678 \times 1}{1}$
$=0.7356 $

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