Poisson process - calls arriving Already posted on MSE.  Had no answer, so will post here.
Assume the number of calls per hour arriving at an answering service follows a Poisson process with $\lambda = 4$.   
Question: If it is know that $8$ calls came in the first two hours.  What is the probability that exactly five arrived in the first hour?
Attempt: Isn't this just a combinatorial question?  So the answer is ${8 \choose 5}/2^8$
 A: Thinking this through, I believe this should be calculated with a binomial distribution with $n = 8$ and $p = 0.5$ as follows:
$P = \binom{8}{5} \cdot 0.5^{5} \cdot (1-0.5)^{3} $
Let me try to proof this:
Let
$X_1$ = number of calls that arrive in the first hour 
$X_2$ = number of calls that arrive in the second hour
$X_3$ = number of calls that arrive in the two hours 
What you want to calculate is the conditional probability of 5 calls arriving in the first hour given that 8 calls arrived in two hours:
$P(X_1 = 5 | X_3 = 8) = \frac {P[(X_1 = 5) \cap (X_3 = 8)]} {P(X_3 = 8)}$
This would be equivalent to : $\frac {P[(X_1 = 5) \cap (X_2 = 3)]} {P(X_3 = 8)}$, however now the events occur over non overlapping time frames which allow us to use the independent increment property of the poisson processes.
$\frac {P[(X_1 = 5) \cap (X_2 = 3)]} {P(X_3 = 8)} = \frac {P(X_1 = 5) \cdot P(X_2 = 3)]} {P(X_3 = 8)}$
$           =\frac {\left[\frac {e^{-4} \cdot 4^5} {5!} \right] \cdot \left[\frac {e^{-4} \cdot 4^3} {3!} \right]} {\frac {e^{-(4 \cdot 2)} \cdot {(4 \cdot 2)}^8} {8!}} $
$=\frac{8!} {5! \cdot 3!} \frac {(4^5) \cdot (4^3)} {8^8} $
$=\frac{8!} {5! \cdot 3!} \frac {(4^5) \cdot (4^3)} {(8^5) \cdot (8^3)} $
$=\frac{8!} {5! \cdot 3!} \cdot \left(\frac {4} {8}\right)^5 \cdot \left(\frac {4} {8}\right)^3$
$= \binom{8}{5} \cdot 0.5^{5} \cdot (0.5)^{3} $
A: Judging from the comments there appears to be a lot of confusion and lack of intuition over this question. A trivial Monte Carlo simulation will give (roughly) the correct answer that can be used to gauge the validity of the solutions. Here it is in R:
> firstHour <- rpois(n=10000000,lambda=4) ; secondHour <- rpois(n=10000000,lambda=4)
> mean(firstHour[firstHour + secondHour == 8]==5)
[1] 0.2181712

Compare this to the OP's attempt:
> choose(8,5)/2^8
[1] 0.21875

Personally the combinatorial approach is not obvious to me. I would have followed @Orlando Mezquita's solution. As you can see, they arrive at the same answer.
