# Poisson Process

Customers arrive at a certain facility according to a Poisson process of rate lambda. Suppose that it is known that five customers arrived in the first hour. Each customer spends a time in the store that is a random variable, exponentially distributed with parameter alpha and independent of other customer times and then departs. What is the probability that the store is empty at the end of the first hour?

I have no clue how to model this situation or proceed. Help would be much appreciated. Thanks so much.

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I suggest you go over the material a second time and then try solving your questions again. You won't learn anything if we do all the questions for you. – Yuval Filmus Mar 24 '11 at 2:44
Would you at least be able to help me set up this situation as a probability statement? My textbook has no examples in it so it is very hard for me to apply the theory to these practical problems! – icobes Mar 24 '11 at 2:49
Here's a start. Suppose you knew that only one customer arrived at the store during the hour. Could you solve that problem? What does the Poisson process tell you about the relationship between different customers? – Carl Brannen Mar 24 '11 at 3:07
I am still unsure what to do! My teacher posted the solution [1 - (1-e^-alpha / alpha)]^5 but I do not have any clue how she got it after trying to work out a few scenarios. Please help! – icobes Mar 24 '11 at 4:38
1. Work out Carl's suggestion. 2. Figure out how to phrase P(nobody is in the store after 1hr) in terms of the probabilities for each of the people. 3. Profit!! – Aaron Mazel-Gee Mar 24 '11 at 7:59

First: You probably meant to write $[1 - (1 - e^{ - \alpha } )/\alpha ]^5$ as the solution to this exercise, right?
Hint 1: The arrival times of the five customers can be simulated as five i.i.d. uniform$[0,1]$ rv's. That is, if you place five i.i.d. uniform rv's on $[0,1]$, the points correspond to arrival times of five customers.
Hint 2: If $Z_1,\ldots,Z_5$ are i.i.d. rv's with distribution function $F$, how can you express the distribution function of $\max \{ Z_1 , \ldots ,Z_5 \}$ in terms of $F$?
Hint 3: You'll have to use the law of total probability, in order to calculate a certain probability. In this context, it may be useful to note that $1-U$ and $U$ are identically distributed, for $U$ a uniform$[0,1]$ random variable.
EDIT (further hints, in response to the OP's request). Suppose that $Z = U + Y$, where $U$ and $Y$ are independent uniform$[0,1]$ and exponential$(\alpha)$ random variables, respectively. First note that $${\rm P}(Z \le 1) = {\rm P}(U + Y \le 1) = {\rm P}(Y \le 1 - U) = {\rm P}(Y \le U),$$ where the last equality (which is actually not essential) follows from the fact that $U$ and $1-U$ are identically distributed. Now, since $U$ has constant density $f(u)=1$ for $u \in [0,1]$, the law of total probability gives $${\rm P}(Y \le U) = \int_0^1 {{\rm P}(Y \le U|U = u)1\,{\rm d}u} = \int_0^1 {{\rm P}(Y \le u)\,{\rm d}u} = \int_0^1 {(1 - e^{ - \alpha u} )\,{\rm d}u} .$$ Calculate the integral on the right-hand side, and recall Hint 2.