Customers arrive at a service facility according to a Poisson process of rate λ = 2 per half an hour. Given that 5 customers arrived during the first two hours of service, what is the probability that more than 3 customers arrived during the first 90 minutes?
This is what I have so far:
Let A be the event that 5 customers arrive in the first two hours and B the event that more than 3 customers arrive in the first 90 minutes. I want Pr(B|A). This is $\frac{\Pr(B\cap A)}{\Pr(A)}$.
Calculate. The number of arrivals in 2 hours is Poisson with parameter 8 (2 x 120/30 = 8) , so Pr(A) = $e^{-8}\frac{8^{5}}{5!}$=0.0916.
Now I need to calculate $\Pr(B\cap A)$.
Can I write $\Pr(B\cap A)=Pr(B)Pr(A)$? I can't prove that I can or that I cannot.
The events $A$ and $B$ are not independent, because the intervals $[0,90]$ and $[0,120]$ are overlapping. Therefore it is incorrect to write $P(B\cap A)=P(B)P(A)$. However, the event $B\cap A$ can be re-expressed in terms of non-overlapping intervals. Argue that: $$B\cap A=\{ N_1 = 4, N_2=1\} \cup \{N_1=5, N_2 = 0\}$$ where $N_1$ denotes the number of customers arriving in the first 90 minutes and $N_2$ denotes the number arriving between minutes 90 and minutes 120. Then $N_1$ and $N_2$ refer to non-overlapping intervals and so they are independent, and therefore $$P(B\cap A) = P(N_1=4)P(N_2=1) + P(N_1=5) P(N_2=0).$$