Parrots and crows are visiting your backyard. They arrive according to independent Poisson processes with rates $\lambda$ and $\mu$, respectively.
What is the distribution of the total number of parrots at time $t$ given that the total number of birds seen at time $t$ is $n$?
My idea was to use that the distribution of the total number of birds at time $t$ is given by a Poisson process with parameter $\lambda+\mu$. Then use something like $P(\#birds = n)=P(\#crows = m)P(\#parrots = n-m)$
Which leads to $P(\#parrots)=\frac{m!(\mu+\lambda)^n}{n!\mu^m}e^{-\lambda}$.
But this is not really going anywhere.
I probably need to do something with that it's a stochastic process, but I really just don't see what I have to do.
Let $X_t$ denote the number of parrots at time $t$ and let $Y_t$ denote the number of crows at time $t$.
$$\begin{aligned}P\left(X_{t}=k\mid X_{t}+Y_{t}=n\right) & =\frac{P\left(X_{t}=k\wedge Y_{t}=n-k\right)}{P\left(X_{t}+Y_{t}=n\right)}\\ & =\frac{P\left(X_{t}=k\right)P\left(Y_{t}=n-k\right)}{P\left(X_{t}+Y_{t}=n\right)}\\ & =\frac{e^{-\lambda}\frac{\lambda^{k}}{k!}e^{-\mu}\frac{\mu^{n-k}}{\left(n-k\right)!}}{e^{-\lambda-\mu}\frac{\left(\lambda+\mu\right)^{n}}{n!}}\\ & =\binom{n}{k}\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k} \end{aligned} $$So we are dealing with binomial distribution equipped with parameters $n$ and $p=\frac{\lambda}{\lambda+\mu}$.