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

 A: 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

A: 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
