The time between arrivals of small aircraft at a county airport is exponentially distributed with a mean of one hour.

What is the probability that more than three aircraft arrive within an hour?

Solution

If the inter-arrival time is exponential with mean $\mu= 1$ hour, the arrivals must form a Poisson process with rate $\lambda = 1$.

Let $X$ denote the number of arrivals in 1 hour.

Therefore $X\sim P(1)$.

The pmf of Poisson distribution with $\lambda =1$ is

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

The probability that more than three aircraft arrive within an hour is

$$ \begin{aligned} P(X\geq 3) &= 1- P(X\leq 2)\\ &= P(X=0) + P(X=1) +P(X=2)\\ &= \frac{e^{-1}1^{0}}{0!}+\frac{e^{-1}1^{1}}{1!}+\frac{e^{-1}1^{2}}{2!}\\ &= 0.3679+0.3679+0.1839\\ &= 0.9197 \end{aligned} $$

Further Reading