Poisson process. A problem about customer arrival at different time intervals. Wackerly 3.125 I was solving exercise 3.125 of Wackerly's Probability book and i did not understand the solution given in the solutions manual.
The problem says:
Customers arrive at a shop following a poisson distribution for an average of 7 customers per hour. What is the probability that exactly two clients arrive in 2 hours between :
(a). 2:00pm and 4:00pm (a continous period of two hours)?
(b). between 1:00pm and 2:00pm or between 3:00pm and 4:00pm (two periods of 1 hour that in total add to two hours)?
For part (a) i responded correctly. That is $P(Y=2) = \frac{14^2 e^{-14} }{2!}$ since initially $\lambda$=7 (per hour), but because we are considering a continous period of 2 hours, our random variable Y has $\lambda_1$ = 14.
However for part (b) the solutions manual says the result is the same as (a), even though we are considering separate 1 hour periods. I would have said that the answer was $2P(Y=2) = 2 \frac{7^2 e^{-7} }{2!}$. Why is this not the answer? Does it have something to do with the "Poisson process"?  The book does not dive much into this, and wikipedia's article was overwhelming.
Thank you
 A: Simple answer is that for both problems, you have a span of two hours, so the calculation is identical for both of them. On average 7 customers arrive in 1 hour so 14 customers arrive in 2 hours. Using $X\sim\text{Poisson}(14)$ to be the number of customer arrivals in a 2 hour time interval, we have $$\Pr(X=2)=\frac{e^{-14}(14)^2}{2!}$$
In terms of a Poisson process, if 7 customers arrive per hour, then the Poisson process has intensity/ rate 7 customers per hour. Generally this is given in units of time, which is one hour here, so it's just 7. And then for a two hour period, the number of arrivals would be modeled by the Poisson distribution/ random variable having parameter $7(2)$.
If we assume the customers arrive according to a Poisson process with rate $r$, i.e. for every interval of length $t$ the distribution of arrivals is $\text{Poisson}(rt)$ and arrivals in separate intervals are independent of one another, then the answers to the two parts are the same.
Probably the problem should have said customers arrive according to a Poisson process.
A: The question is asking that the probability that exactly two people arrive in the combined two-hour period. In particular, it is ok for one person to arrive at 1:30 and one other to arrive at 2:30.
You computed the probability that exactly two people arrived between 1:00 and 2:00 and added it to the probability that exactly two people arrived between 2:00 and 3:00, which is not what the question is asking.
