# 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 ${8 \choose 5}/2^8$

• @Christos I don't think you are correct. – user32442 Nov 7 '13 at 9:28
• @users32442 check the answer of Orlando Mezquita he explains in a more formulated way. – Christos Nov 7 '13 at 10:03
• @Christos I can only think you are missing the point that it is $8$ calls in total across both of the first two hours, rather than $8$ calls in the second hour only – M. Berk Nov 7 '13 at 10:07
• Please don't cross post to different SE sites. The right thing to do if you think your post belongs somewhere other than where you posted it is to flag it for moderator attention and ask them to move it. – Glen_b -Reinstate Monica Nov 7 '13 at 10:29
• @Glen_b could you expand on why the attempt is wrong. I think we all know the definition of the Poisson distribution so your link is not at all insightful – M. Berk Nov 7 '13 at 11:14

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.

$\frac {P[(X_1 = 5) \cap (X_2 = 3)]} {P(X_3 = 8)} = \frac {P(X_1 = 5) \cdot P(X_2 = 3)]} {P(X_3 = 8)}$

$=\frac {\left[\frac {e^{-4} \cdot 4^5} {5!} \right] \cdot \left[\frac {e^{-4} \cdot 4^3} {3!} \right]} {\frac {e^{-(4 \cdot 2)} \cdot {(4 \cdot 2)}^8} {8!}}$

$=\frac{8!} {5! \cdot 3!} \frac {(4^5) \cdot (4^3)} {8^8}$ $=\frac{8!} {5! \cdot 3!} \frac {(4^5) \cdot (4^3)} {(8^5) \cdot (8^3)}$ $=\frac{8!} {5! \cdot 3!} \cdot \left(\frac {4} {8}\right)^5 \cdot \left(\frac {4} {8}\right)^3$ $= \binom{8}{5} \cdot 0.5^{5} \cdot (0.5)^{3}$

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.

• The combinatorial approach is perfectly fine if one uses the property of Poisson processes that given that the number of arrivals in $(0, T]$ is $n$, then the conditional distribution of the $n$ arrival times is the same as that of the order statistics of $n$ $U(0,T]$ random variables. Thus, the probability that exactly $k$ of the arrivals are in $(0, \hat{T}]$ is just the binomial probability $\binom{n}{k}p^k(1-p)^{n-k}$ where $p = \hat{T}/T$ which has value $0.5$ in this instance. – Dilip Sarwate Dec 8 '13 at 1:57