A slow forty minutes for support tickets

Question

Support tickets arrive as a Poisson process at rate \lambda = 3 per hour. What is the probability that the next ticket takes more than 40 minutes to arrive?

Accepted Answer

The wait until the next Poisson event is exponential, so the chance it exceeds a time t is P(T t) = e^{-\lambda t}. Forty minutes is 	frac{40}{60} = 	frac23 of an hour, and \lambda = 3 per hour, so P(T 	frac23) = e^{-3 	imes 2/3} = e^{-2} \approx 0.135. About a 13.5\% chance of a gap longer than forty minutes. Takeaway. Convert the wait to the same time units as the rate, then read off P(T t) = e^{-\lambda t}.

A slow forty minutes for support tickets

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