Two vending machines, one sale

Question

Two vending machines sit in a hallway. One records sales as a Poisson process at $6$ per hour and the other, independently, at $9$ per hour.

What is the probability of exactly $1$ sale from the two machines combined in the next $4$ minutes?

Accepted Answer

Independent Poisson processes superpose into a single Poisson process with the summed rate: \lambda = 6 + 9 = 15 	ext{ per hour}. Four minutes is 	frac{4}{60} = 	frac{1}{15} of an hour, so the mean number of sales in the window is \mu = \lambda t = 15 	imes 	frac{1}{15} = 1. The chance of exactly one sale is P(N = 1) = e^{-1}\,\frac{1^1}{1!} = e^{-1} \approx 0.368. So about a 37\% chance. When the mean count works out to exactly 1, the probability of exactly one event is always e^{-1}, a handy landmark. Takeaway. After merging, everything reduces to a single mean count \mu = \lambda t; w…

Two vending machines, one sale

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