I froze for a bit because I kept trying to think of it as a discrete problem.
Model the problem geometrically by representing arrival times as points in a unit square. The meeting condition defines a region within the square, so compute its area to find the probability.
Pro tip: After solving, mention that this geometric method generalizes to any meeting window, and that the complement (not meeting) is often easier to compute.
Let X and Y be the arrival times in minutes after the start of the hour, each uniformly distributed in [0, 60]. The sample space is the square [0,60] x [0,60] with area 3600.
They meet if |X - Y| <= 15. This inequality defines a band around the line X = Y within the square.
The complement (not meeting) consists of two right triangles where X > Y + 15 and Y > X + 15. Each triangle has legs of length 45, so total non-meeting area is 2 * (1/2 * 45 * 45) = 2025. Thus meeting area = 3600 - 2025 = 1575.
Probability = meeting area / total area = 1575 / 3600 = 7/16 = 0.4375.
Check that the probability is between 0 and 1. Note that for a window of w minutes, the probability is 1 - (1 - w/60)^2, which for w=15 gives 7/16.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.