The chance of a quiet fifteen minutes

Question

Customers walk into a shop as a Poisson process at rate $4$ per hour.

What is the probability that no customer arrives during a particular $15$ -minute stretch? And what is the expected time until the next customer after the shop opens?

Accepted Answer

Quiet stretch. For a Poisson process at rate \lambda, the number of arrivals in a window of length t is Poisson with mean \lambda t. Here \lambda = 4 per hour and t = 15 minutes = 	frac14 hour, so the mean is \lambda t = 4 	imes 	frac14 = 1. The probability of exactly zero arrivals: P(N = 0) = e^{-\lambda t} = e^{-1} \approx 0.368. So about a 37\% chance of a completely quiet quarter-hour. Equivalently, "no arrival in [0, t]" means the waiting time to the first customer exceeds t. That waiting time is exponential with rate \lambda, and P(	ext{wait} t) = e^{-\lambda t}, the same e^{-1}. Th…

The chance of a quiet fifteen minutes

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