Discussion
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
Answer: C
The probability that they meet is 7/16.
Let their arrival times be x and y minutes after 1 PM, each uniform on 0 to 60. They meet when |x - y| ≤ 15. In the 60 by 60 square of possible arrivals, the region where they miss is two corner triangles, each with legs 60 - 15 = 45:
\[ \begin{aligned} P(\text{miss}) &= \frac{2 \times \tfrac{1}{2} \times 45^{2}}{60^{2}} = \frac{2025}{3600} = \frac{9}{16} \\ P(\text{meet}) &= 1 - \frac{9}{16} = \frac{7}{16} \end{aligned} \]
9/16 is the probability that they do not meet. 1/4 is 15/60, the waiting time as a fraction of the hour, which ignores that either may arrive first, and 1/16 is its square.
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