Exactly three at the coffee kiosk

Question

Customers arrive at a coffee kiosk according to a Poisson process at rate \lambda = 12 per hour. What is the probability that exactly 3 customers arrive during the next 10 minutes?

Accepted Answer

The key habit for any Poisson process: first turn the rate into the average number of arrivals for the exact window you care about, then plug into the counting formula. The window is 10 minutes, which is 	frac{10}{60} = 	frac16 of an hour. So the average (mean) number of arrivals in that window is \mu = \lambda t = 12 	imes 	frac16 = 2. In plain words, we expect about 2 customers in 10 minutes on average. The actual count N in such a window follows a Poisson distribution with this mean, and the chance of seeing exactly k of them is P(N = k) = e^{-\mu}\,\frac{\mu^k}{k!}. Here k = 3 and \mu…

Exactly three at the coffee kiosk

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