If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.

Solution

Let $X$ denote the number of electricity failures per week. The average number of electricity failures per twenty weeks is 3. Thus the average number of electricity failures per week $E(X) = \lambda = 3/20 = 0.15$.

$X\sim P(0.15)$.

The probability mass function of Poisson distribution with $\lambda =0.15$ is

$$ \begin{aligned} P(X=x) &= \frac{e^{-0.15}(0.15)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$

The probability that there will not be more than one failure during a particular week is $P(X\leq 1)$

$$ \begin{aligned} P(X\leq1) &= P(X=0)+ P(X=1)\\ &= \frac{e^{-0.15}0.15^{0}}{0!}+\frac{e^{-0.15}0.15^{1}}{1!}\\ &= 0.8607+0.1291\\ &= 0.9898 \end{aligned} $$