Two shooting stars, both near the end?

Question

Shooting stars appear as a Poisson process over a one-hour watch, [0, 60] minutes. You saw exactly 2 shooting stars. Given this, what is the probability that both appeared during the last 20 minutes?

Accepted Answer

Once we fix the count at 2, each star's time is an independent uniform point over the full 60-minute window, with the rate playing no further role. The last 20 minutes make up \frac{20}{60} = \frac13 of the window, so each star independently fell in that final stretch with probability 	frac13. For both to do so: \left(\frac13ight)^2 = \frac19 \approx 0.111. About an 11\% chance. Takeaway. Conditioned on the count, arrival times are independent uniforms; the chance all of them fall in a stretch is (length fraction) raised to the number of arrivals.

Two shooting stars, both near the end?

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