The number of planes arriving per day at a small private airport is a random variable having a Poisson distribution with $\lambda = 28.8$. What is the probability that the time between two such arrivals is at least 1 hour?

#### Solution

Let $X$ denote the number of planes arriving per day at a small private airport. $X\sim P(28.8)$.

The interarrival time $t$ follows exponential distribution with parameter $\theta =\lambda=28.8$.

The pdf of $t$ is

`$$ \begin{aligned} f(t) &= \theta e^{-\theta t},\; t>0. \end{aligned} $$`

The probability that the time between two such arrivals is at least 1 hour is

`$$ \begin{aligned} P(t\geq 1 \text{hour}) &= 1-P(t<1 \text{hour})\\ &= 1-P(t < \frac{1}{24})\\ &= 1- \int_0^{1/24}\theta e^{-\theta t}\; dt\\ &= 1- 28.8\int_0^{1/24}e^{-28.8 t}\; dt\\ &= 1-28.8\bigg[-\frac{e^{-28.8 t}}{28.8}\bigg]_0^{1/24}\\ &= 1- \bigg[-e^{-28.8 (1/24)} +1\bigg]\\ &= e^{-28.8 (1/24)}\\ &=0.30119 \end{aligned} $$`