# "Time until arrival/departure" in a Poisson process...

Customers are served at a bank with the following process. While there is at most one customer in the bank, there will be only one person teller at a window. If a second customer comes into the bank while the first is still in service, a second teller will come to another window and begin to serve the second customer. Likewise, if a third customer enters while the first two are still being served, a third teller comes to a window and begins to serve the third customer. But, there are only three windows, so any further customers have to wait in line. Assume that the times between customer arrivals are independent and identically distributed Exponential random variables with mean $$1/λ$$, and that the service times for each customer are independent and identically distributed Exponential random variables with mean $$1/μ$$.

If $$x$$ customers are in the bank, what is the expected length of time until an arrival?

So the reason I'm struggling with this question is that I first don't really know what is meant by an "arrival". Is that just another person entering the bank? Because if so, that is just independent and exponentially distributed so regardless of how many $$x$$ customers are in the bank I would assume it would be $$1/\lambda$$.

Perhaps (if that's the case) a follow up question would be, if there were $$10$$ customers inside the bank what would be the expected time until a departure? Would this depend on how many people were inside? I'm unsure how I would calculate this.

As far as I would say, an "arrival" describes the time until the next person enters the bank. As you pointed out, this is independent of the number of customers in the bank and therefore the expected time until an arrival is $1/\lambda$.
For your second question, you have to consider the number of customers being served. If all three employees are working (as it would be the case if 10 costumers are inside the bank), then the time until the next customer is leaving is described by $\min(Exp(\mu),Exp(\mu),Exp(\mu))$, where all three random variables are independent. This random variable is $Exp(3\mu)$-distributed and therefore the expected time until a departure is $1/3\mu$. Observe, this depends on the number of employees working and therefore indirectly on the number of customers inside the bank.
• Thanks for your clarification! A follow up a friend told me I wouldn't be able to get, is if all three tellers are busy, what is the probability that the customer at Teller #1 would depart first? And I have an idea that it would be $1-\text{probability that customer at teller 1 is maximum}$, but I'm not really sure where to go from there. Any ideas? Commented Jul 30, 2015 at 15:15
• Are they equivalent? So it just becomes $1/3$? Commented Jul 30, 2015 at 15:26