Solution
Three for Coffee
Show the problem again
Three friends each arrive at a cafe at an independent, uniformly random time between 12:00 and 13:00. Each stays for exactly 15 minutes after arriving (the cafe stays open as long as needed) and then leaves. What is the probability that at some moment all three are in the cafe at the same time? Give a reduced fraction.
Worked solution
The answer is 5/32. Measure time in hours. All three are present together exactly when the last arrival comes no more than 1/4 hour after the first arrival, because then at the moment of the last arrival nobody has left yet; otherwise the first to arrive has left before the last appears. So we need the range R = max − min of three uniforms to satisfy R ≤ 1/4. For n uniforms, P(R ≤ a) is found by choosing which point is the minimum (n ways), placing it at m, and putting the other n − 1 points in [m, m + a]: P(R ≤ a) = n ∫₀^{1−a} a^{n−1} dm + n ∫_{1−a}^1 (1 − m)^{n−1} dm = n a^{n−1} − (n − 1) a^n. With n = 3 and a = 1/4: 3/16 − 2/64 = 5/32. Compare the two-person version, where the answer is 7/16: requiring a common moment for three people is far more demanding than pairwise overlaps.
Source: Classic geometric-probability exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.