Question
If X has Poisson distribution with parameter m and $P(X = 2) = P(X = 3)$, then find $P(X ≥ 2)$. Use $e^{−3} = 0.0497$

Answer

The p.m.f. of $X$ is given by
$P ( X = x )=\frac{ e ^{- m } m ^x}{x !}$
Given, $P(X=2)=P(X=3)$
$ \therefore \frac{ e ^{- m ^2} m ^2}{2 !}=\frac{ e ^{- m ^3} m ^3}{3 !}$
$\therefore \frac{ m ^2}{2}=\frac{ m ^3}{6}$
$\therefore \frac{1}{2}=\frac{m}{6}$
$\therefore \frac{6}{2}=m$
$\therefore m =3$
$\therefore P ( X \geq 2)=1- P ( X <2)$
$=1- P ( X =0 \text { or } X =1) $
$=1-[P(X=0)+P(X=1)]$
$=1-\left[\frac{ e ^{-3}(3)^0}{0 !}+\frac{ e ^{-3}(3)^1}{1 !}\right]$
$=1-[0.0497+3 \times 0.0497]$
$=0.8012$

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