Waiting for the next shooting star

Question

Shooting stars appear as a Poisson process at rate \lambda = 5 per hour. What is the probability that you wait more than 30 minutes for the next shooting star?

Accepted Answer

In a Poisson process, the time until the next event follows an exponential distribution. The single fact you need is that the chance of waiting longer than a time t is P(T t) = e^{-\lambda t}. Here waiting more than 30 minutes means t = 0.5 hours, and \lambda = 5 per hour, so P(T 0.5) = e^{-5 	imes 0.5} = e^{-2.5} \approx 0.0821. So only about an 8\% chance of a wait longer than half an hour. This lines up with the "no arrivals" view: waiting more than t for the next star is the same event as getting zero stars in the window of length t, which also has probability e^{-\lambda t}. Takeaway. T…

Waiting for the next shooting star

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