Two counters feeding one waiting area

Question

At a government office, arrivals to counter A form a Poisson process at rate $8$ per hour and arrivals to counter B form an independent Poisson process at rate $12$ per hour. Everyone who arrives at either counter then steps into a shared waiting area.

What is the probability that exactly $2$ people enter the waiting area in the next $6$ minutes?

Accepted Answer

When you overlay two (or more) independent Poisson processes, the combined stream is again a Poisson process, and its rate is simply the sum of the individual rates. This is called superposition, and it is one of the most useful shortcuts with Poisson arrivals. So the waiting area receives arrivals at \lambda = 8 + 12 = 20 	ext{ per hour}. Now treat it as one process. Six minutes is 	frac{6}{60} = 	frac{1}{10} of an hour, so the mean count is \mu = \lambda t = 20 	imes 	frac{1}{10} = 2. The chance of exactly 2 arrivals is P(N = 2) = e^{-2}\,\frac{2^2}{2!} = 2\,e^{-2} \approx 0.271. About…

Two counters feeding one waiting area

Your answer

Discussion

Learn the concepts

Related questions