Assume calls enter a 9-1-1 call center as a Poisson Process. If all operators are on a call, then the caller is placed into a queue that waits for the next available operator.

(a) How many calls are expected to be placed into the 9-1-1 call center from 9:00AM to 9:02AM, if calls are placed at a rate of 2 per minute? 1

(b) Assume at 9:00AM, all operators are available. Additionally, the call center has 5 operators and assume each call lasts exactly 2 minutes. What is the probability that a caller who calls between 9:00AM and 9:02AM will be placed into the queue given that the rate of calls is 2 per minute?

(c) Using the information from part (b), what is the minimum number of operators that need to be working during 9:00AM - 9:02AM to ensure that all calls are handled with at least 96% certainty?

I figured out part a), but I do not know how to approach part b and c. For part b) do I need to do a conditional probability?


For part (a), the calls are coming in at two per minute, not one every two minutes.

For part (b), the wording is somewhat unclear to me, but I assume they mean "what is the probability that one or more callers will be placed into the queue between 9 and 9:02". Since calls last two minutes, any call an operator takes will tie them up for the entire window, so this is the same as the probability that there will be more than five calls. Given that the number of calls $X$ in the interval is Poisson distributed with mean $4$ we have $$ P(X>5) = \sum_{n=6}^\infty P(X=n) = \sum_{n=6}^\infty e^{-4}\frac{4^n}{n!}$$ (Hint for computing this: it might be easier to compute $P(X\le 5)$ and infer the answer from that.)

For part (c), as they say, part (b) will help. Think about how to write the answer for general number of operators $m$ rather than $m=5$ and then you need to make $m$ large enough so that the probability is greater than $0.96.$


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