# Poisson Process arrivals

This is a homework problem. Between 10 AM and 6 PM visitors arrive at the Tate Modern Gallery in accordance with a Poisson process at the rate of 6 per minute. Determine the probability that 10 visitors arrive between 1:00 PM and 1:02 PM, given that 800 visitors arrive between noon and 2:00 PM.

My attempt:

I know that the 10 arrivals (in permuted order) are identically and independently, $S_1,\ldots, S_{10} \sim \text{Uniform(0,120)}$

so $P(\text{Arrival between 1:00 PM and 1:02 PM}) = \displaystyle\int_{60}^{62} \dfrac{1}{120} dt = \dfrac{2}{120}$

So my guess is that

$$P(\text{10 arrive between 1:00 PM and 1:02 PM}|N(120)=800) = \displaystyle{800 \choose 10}\left(\dfrac{2}{120}\right)^{10}\left(1-\dfrac{2}{120}\right)^{790}$$

Haha, I think I have figured it out as I typed out the question. Do you guys think this looks right?

> dbinom(10, size=800, prob=1/60)
[1] 0.0791953

• 800 per hour is 800/60 per minute and you want the probable number in 2 minutes.
– DWin
Commented Dec 19, 2013 at 2:21

The link Mathias gave is quite useful. Use $P(A|B) = \frac{P(A \cap B)}{P(B)}$ where A is the event that 10 people arrive in the 61st or 62nd minutes and B is the event that 800 people arrive in the first 120 minutes.

$P(A \cap B) = P(A)*P(B|A)$. The marginal $P(A)$ is easy; $P(X=10; \lambda=2*6)$. $P(B|A)$ is the same as the probability that we have 790 people arrive in the remaining 118 minutes since we are given 10 arrive in the 61st or 62nd minutes. This is simply $P(X=790; \lambda = 118*6)$. $P(B)$ is simply $P(X=800; \lambda = 120*6)$

Then you can combine those with the first equation.

To beat a dead horse, here is the explicit solution worked in R:

(A = dpois(x=10, lambda = 2*6))
[1] 0.1048373

(BgivenA = dpois(x=790, lambda = 118*6))
[1] 0.0001462412

(B = dpois(x=800, lambda = 120*6))
[1] 0.0001935913

(AgivenB = A*BgivenA/B)
[1] 0.0791953

• Not quite right, the 800 people come between noon and 2:00pm and we want to know whats prob 10 of them come between 1:00 and 1:02. Commented Dec 19, 2013 at 2:34
• You are correct, I will edit my explanation. The reasoning stays the same. Commented Dec 19, 2013 at 2:45
• Note: this is equivalent to the answer you posted, just an alternate solution. Commented Dec 19, 2013 at 3:00
• Thanks for verifying my answer and providing some insight using a different method! Commented Dec 19, 2013 at 3:04

I don´t quite understand why you worked with an integral when you´re dealing with a discrete variable. Anyways, this link should help:

https://math.stackexchange.com/questions/328230/poisson-process-conditional-probability-question

• I'm using an integral because the arrivals are uniformly (i.e. continous) distributed over the interval noon to 2:00 PM. Your answer doesn't specifically answer my question. Commented Dec 19, 2013 at 2:21