Overlapping shifts at a help desk

Question

Two volunteers each begin a help-desk shift at a uniformly random time during a $60$ -minute window, independently. One stays $15$ minutes, the other stays $30$ minutes (neither past the end of the window).

What is the probability they are both on duty at some common moment?

Accepted Answer

Let a, b be start minutes, each uniform on [0, 60]. They fail to overlap iff a - b 30 \quad	ext{or}\quad b - a 15 , triangles with legs 60 - 30 = 30 and 60 - 15 = 45 inside the 60 	imes 60 square: P(	ext{miss}) = \frac{	frac12(30)^2 + 	frac12(45)^2}{60^2} = \frac{450 + 1012.5}{3600} = \frac{1462.5}{3600} = \frac{13}{32} . So P(	ext{overlap}) = 1 - 	frac{13}{32} = 	frac{19}{32} \approx 0.594. Takeaway. The same two-triangle picture works for any window and any pair of durations. Compute each miss triangle from its own leg T - w, add, and subtract from one.

Overlapping shifts at a help desk

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