An old bus breaks down an average of 3 times per month. Using the Poisson probability distribution formula, find the probability that during the next month this bus will have

i) exactly 2 breakdowns;
ii) at most one breakdown.

Solution

Let $X$ denote the number of break downs per months. The average number of break downs is 3, i.e., $E(X) = \lambda = 3$.

$X\sim P(3)$.

The probability mass function of Poisson distribution with $\lambda =3$ is

$$ \begin{aligned} P(X=x) &= \frac{e^{-3}(3)^x}{x!},\; x=0,1,2,\cdots \end{aligned} $$

i) The probability that during the next month this bus will have exactly 2 breakdowns is $P(X =2)$

$$ \begin{aligned} P(X = 2) &= \frac{e^{-3}3^{2}}{2!}\\ &= 0.224 \end{aligned} $$

ii) The probability that during the next month this bus will have at most one breakdown is $P(X\leq 1)$.

$$ \begin{aligned} P(X\leq1) &= P(X=0)+ P(X=1)\\ &= \frac{e^{-3}3^{0}}{0!}+\frac{e^{-3}3^{1}}{1!}\\ &= 0.0498+0.1494\\ &= 0.1992 \end{aligned} $$

Further Reading