Past exam of the options I course of the University of Oxford / 2013 / M / 1

A Poisson process has a Poisson distribution for the count in an interval of length $t$, with mean equal to rate times length. Fifteen minutes is $1/4$ hour, so

The first arrival time $T_1$, measured in hours after opening, has exponential distribution of rate twelve. ThereforeEquivalently, there must be no arrivals in the first quarter hour and at least one in the next; independent increments give $e^{-3}(1-e^{-3})$.

Let $I_2=T_2-T_1$ and $I_3=T_3-T_2$ be the interarrival times adjacent to the second arrival. They are independent exponential random variables of rate twelve per hour. The second customer meets the first if $I_2<1/12$ hour, and meets a later customer if $I_3<1/12$ hour. If the third arrives too late, every subsequent customer does also. Thus the only way to meet nobody is for both gaps to be at least five minutes:Hence $\boxed{\mathbb{P}(\text{meeting})=1-e^{-2}}$. Exact equality of arrival and departure times has probability zero. The calculation includes both customers already present and customers arriving during the visit; it is a two-sided interarrival test for overlap.

Independent marking with probability $5/6$ is Poisson thinning and gives a Poisson process of chip-buying arrivals, of rate $12(5/6)=10$ per hour. Its first arrival time therefore satisfiesFor example, its survival probability is $\mathbb{E}[(1/6{)}^{N(t)}]=\exp (12t(1/6-1))=e^{-10t}$, which also derives the thinning conclusion directly.