I was hoping to get verification that I am on the right track and doing/ thinking about the following problem correctly:

Customers arrive at a service facility according to a Poisson process of rate $\lambda$ = 5 per hour. Given that 12 customers arrived during the first two hours of service, what is the conditional probability that 5 customers arrived during the first hour?

This is what I have so far:

Let X(t) = the number of customers arrived up to time t. Fix the times such that 0 < s < t. Then I wanted to use the conditional probability formula Pr{X(t)=n+k|X(s)=n}. So based on my understanding of this question, I have done Pr{X(2)=12|X(1)=5}, so s=1, t=2, n=5 and k=7. Solving this, I get Pr{X(1)=7}, so for a Poisson process with rate $\lambda$=5, this is $\frac{{e^{-35}}5^7}{7!}$ = 15.5$e^{-35}$

Is this correct?


We can mechanically use the defining formula for conditional probability. As an exercise, we will go through the process. But the ultimate result is so simple that it invites further thinking.

Let $B$ be the event there were $12$ in the first two hours, and let $A$ be the event there were $5$ in the first hour. We want $\Pr(A|B)$. This is $\frac{\Pr(A\cap B)}{\Pr(B)}$.

Calculate. The number of arrivals in $2$ hours is Poisson with parameter $10$, so $\Pr(B)=e^{-10}\frac{10^{12}}{12!}$.

For the event $A\cap B$, note this happens if there are $5$ in the first and $7$ in the second. This has probability $\left(e^{-5}\frac{5^{5}}{5!}\right)\left( e^{-5}\frac{5^{7}}{7!}\right)$.

Divide. The powers of $e$ stuff cancels, and we end up with something that may be very familiar to you from your experience with the binomial distribution. With some manipulation, we arrive at $$\binom{12}{5}\left(\frac{1}{2}\right)^{12}.$$

Added: Such a simple answer deserves a simple explanation. We had $12$ occurrences of a customer arriving. Label these $12$ customers in an arbitrary random way. Call a customer prompt if she arrives in the first hour. The probability that a customer randomly chosen from the $12$ is prompt is $\frac{1}{2}$, since she is equally likely to have arrived in the first hour as in the second. So the probability that exactly $5$ of the $12$ customers are prompt is $\binom{12}{5}\left(\frac{1}{2}\right)^{12}$.

Remarks: $1.$ You should track down the general case. We have a Poisson with rate $\lambda$ (per hour). Given that there were $n$ arrivals in the first $a+b$ hours, what is the probability that there are $k$ arivals in the first $a$ hours? Go through the same conditional probability calculation. You will get a familiar "binomial-type" expression. Now explain why the result is "obvious," and the conditional probability machinery was not necessary.

$2.$ Given that there were $12$ successes in $2$ hours, the event that there were $5$ in the first hour sounds not unlikely. So the number obtained in the post is too small by several orders of magnitude.

  • $\begingroup$ Wow, thank you for such a detailed response! I see that I had my conditional probability set up wrong originally (I had them flipped) and that I had calculated Pr(B) wrong. $\endgroup$ – Abi Mar 12 '13 at 18:26
  • $\begingroup$ You are welcome. I have added explanation in the post for "why" the final answer is so simple. $\endgroup$ – André Nicolas Mar 12 '13 at 18:38
  • 2
    $\begingroup$ Why is it the case that $$Pr(A\cap B) = \left(e^{-5}\frac{5^{5}}{5!}\right)\left( e^{-5}\frac{5^{7}}{7!}\right)$$ $\endgroup$ – Sam Apr 23 '17 at 7:58
  • $\begingroup$ yep @sam good question. I also don't see that. $\endgroup$ – Alex Monras Feb 12 '18 at 11:47
  • $\begingroup$ @sam because they are independent $\endgroup$ – superman Feb 26 '18 at 22:48

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