 Solution

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.