Question

If $X$ has Poisson distribution with parameter $m=1$. Find $P[X\le 1]$(use $e^{-1}=0.367879)$.

Answer 1

$P(X=x)=\frac{m^x{e}^{-x}}{x!}$

$P(X=0)=\frac{e^{-0}}{0!}=1$

$P(X=1)=\frac{e^{-1}}{1!}=e^{-1}=0.367879$

$P(X\le 1)=P(X=0)+P(X=1)=1+0.367879=1.367879$