# Car accidents and the Poisson distribution

In a busy highway on the average 5 accidents happen per week. What is the probability that there will be no accidents next week? Assume that the number of accidents has a Poisson distribution.

Let $X$ be the number of accidents per week. Then $X$ has the Poisson distribution
$p(x)=\frac{\lambda^{x}e^{-\lambda}}{x!}, x=0,1,2, \dots .$
Since the average number of accidents happen per week is $5$, $\lambda=5$. Hence
$p(x)=\frac{5^{x}e^{-5}}{x!}, x=0,1,2, \dots .$
The probabality of no accident is
$P[X=0]=\frac{5^{0}e^{-5}}{0!}=\frac{e^{-5}}{1}=e^{-5}=\frac{1}{e^{5}}=0.0067=6.7\%.$
Note that  $0! =1$.

The answer is accepted.