In airport A, the time between the arrivals of flights coming from the UK is exponentially distributed, with a mean of an hour.
- What is the probability that more than three flights arrive within an hour?
- If we choose 30 separate 1 hour intervals, what is the probability that no interval is a siege of more than 3 arrivals?
- If the probability that no flights arrive during a time interval is 0.1, what is its length?
The mean is $1/\lambda=1\Leftrightarrow\lambda=1$ per hour.
Number 3 is straightforward. I want to find $t$ such that $\int_0^te^{-x}dx=0.1$
I am stuck with the rest, though. What I can do is transform this into a Poisson distribution problem so as to get$$P(X>3)=1-P(X\leq3)=1-\sum_{i=0}^3\frac{e^{-1}}{i!}$$
and then for 2) do $(P(X\leq3))^{30}$, but I am not sure if this is what should be done..
Any guidance? Thank you!
Best Answer
If the time between arrivals is i.i.d. exponential with rate $\lambda$, then the number of arrivals within a fixed time interval of length $t$ is $Poisson\left(\lambda t\right)$. Here $\lambda=1$.
Thus, you are first looking for $P\left(N_1 > 3\right)$, which is indeed $1 - \sum_{i\leq 3} \frac{e^{-1}}{i!}$. For the second part, you use the fact that any two non-overlapping increments of a Poisson process are independent, which justifies $\left(1-P\left(N_1 > 3\right)\right)^{30}$ because you are looking at $30$ "separate" (I presume this means "disjoint") intervals. For the last part, let $T$ be the time of the first arrival; observe that
$$P\left(N_t = 0\right) = P\left(T > t\right) = e^{-t},$$
so you are looking for $t = -\log\left(0.1\right)$.