= 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:
$$
\mathbb P(\text{no meeting})=\mathbb P(I_2\geq1/12)\mathbb P(I_3\geq1/12)=e^{-1}e^{-1}=e^{-2}.
$$
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>.
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