Car accidents and the Poisson distribution
Answer
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.
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