A quiet 45 minutes of texts

Question

Text messages land on your phone as a Poisson process at rate \lambda = 4 per hour. What is the probability that you receive no texts at all during the next 45 minutes?

Accepted Answer

First convert the rate to the mean count for a 45-minute window. Forty-five minutes is 	frac{45}{60} = 	frac34 of an hour, so \mu = \lambda t = 4 	imes 	frac34 = 3. We expect about three texts in that window on average. The chance of exactly k texts is P(N=k) = e^{-\mu}\mu^k/k!. For "no texts" we set k = 0, and since \mu^0 = 1 and 0! = 1 the formula collapses nicely: P(N = 0) = e^{-\mu} = e^{-3} \approx 0.0498. So only about a 5\% chance of complete silence. The probability of an empty window is always just e^{-\mu}. Takeaway. The chance of zero arrivals in a window is simply e^{-\lambda …

A quiet 45 minutes of texts

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