Let $X(t)$ be a Poisson process with rate $\lambda = 6$ describing arrivals per hour of customers at a bank. Let the probability of a customer being male be $2/3$. Suppose 10 males has arrived during the two first business hours. How many females would you expect to have arrived?
Attempt: Let $M$ be male customers in the first two hours and $F$ be the female ones. Then $M$ should be $Po(8)$ and $F$ should be $Po(4)$, right?
What we're looking for is $E[F | M+F = 10]$.
Two problems: 1) I don't think you can just divide $E(F) = 4$ by $P(X+Y=10)$ since X and X+Y aren't independent (right...?).
2) I'm not sure how to calculate $P(X+Y=10)$ for two Poisson processes.
When each point of a homogeneous Poisson point process (HPPP) of intensity (rate) $\lambda$ is marked (selected) with a probability $p$, independently of the other points, the result is two independent HPPPs (consisting of the marked and non-marked points) with intensities $\lambda p$ and $\lambda (1-p)$, respectively.
In this case, we have a HPPP representing the customer arrivals and each customer is randomly "marked" as being male. Therefore, $\mathbb{E}(F|M) = \mathbb{E}(F)$.