Poisson emails in the inbox

Question

Emails arrive according to a Poisson process with rate $\lambda = 3$ per hour.

(a) What is the probability that exactly $5$ emails arrive in the next $2$ hours? (b) What is the probability that none arrive in the next $20$ minutes?

Accepted Answer

Scale the rate to each window first: the count in a window of length t is Poisson with mean \mu = \lambda t. (a) Over 2 hours, \mu = 3 	imes 2 = 6: P(N = 5) = e^{-6}\frac{6^5}{5!} = e^{-6}\frac{7776}{120} = 64.8\,e^{-6} \approx 64.8 	imes 0.00248 \approx 0.161. (b) Over 20 minutes = 	frac13 hour, \mu = 3 	imes 	frac13 = 1: P(N = 0) = e^{-1} \approx 0.368. Equivalently, the wait to the next email is exponential with rate 3/hour, so P(	ext{wait} 	frac13 	ext{ hr}) = e^{-3 	imes 1/3} = e^{-1}, the same 0.368 seen from the waiting-time side. Takeaway. Always convert to the mean count \mu…

Poisson emails in the inbox

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