I knew this problem from school but blanked on how to set it up cleanly under pressure.
Model the problem geometrically by representing each person's arrival time as a point in a unit square, then identify the region where the absolute time difference is at most 15 minutes. Calculate the area of that region and divide by the total area to get the probability.
Pro tip: After solving, mention that this is a classic 'meeting problem' and that the same geometric approach generalizes to any waiting time w, with probability 1 - (1 - w)^2 for a 1-hour window. This shows you can abstract the solution and apply it to similar problems.
Let X and Y be the arrival times of the two people, each uniformly distributed on [0, 60] minutes. The sample space is the square [0,60] x [0,60] with area 3600.
The condition 'within 15 minutes of each other' means |X - Y| ≤ 15. This region is a diagonal band of width 15√2 around the line X = Y.
The area of the band is the total square area minus the areas of the two triangles where |X - Y| > 15. Each triangle has legs of length 45, so area = 0.5 * 45 * 45 = 1012.5. Total unfavorable area = 2025. Favorable area = 3600 - 2025 = 1575.
Divide the favorable area by the total area: 1575 / 3600 = 0.4375, or 43.75%.
Note that for a waiting time w (in minutes) within a 60-minute window, the probability is 1 - (1 - w/60)^2. For w=15, this gives 1 - (0.75)^2 = 0.4375. Discuss how this applies to real-world scenarios like meeting up or customer arrivals.
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