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A and B are friends. They decide to meet between 1 PM and 2 PM on a given day. There is a condition that whoever arrives first will not wait for the other for more than 15 minutes. The probability that they will meet on that day is

1. 1/4
2. 11/16
3. 7/16
4. 9/16

Correct answer: 7/16

Solution

Let the arrival times of A and B be uniformly distributed over the 60-minute interval. They meet if the absolute difference between their arrival times is at most 15 minutes, which corresponds to the region between the lines $|x-y|\le 15$ in a $60\times 60$ square. The favorable area is $60^2-2\cdot\frac{45^2}{2}=1575$, so probability $=1575/3600=7/16$.

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