Suppose $Y_1, Y_2, \dots, Y_n$ are independent Poisson random variables with means $\lambda_1, \lambda_2, \dots, \lambda_n$.
How do you find the conditional probability function of $Y_1$ given that $\displaystyle\sum_{i=1}^n Y_i=m$?
I know that $\displaystyle\sum_{i=1}^n Y_i \sim Poisson(\displaystyle\sum_{i=1}^n \lambda_i)$.
I've never worked with conditional probabilities, so forgive me if the solution here is trivial...
On
Ah this is quite nice since the answer that comes out is quite intuitive. Anyway you just apply Bayes' theorem, to get (for $0\leq k \leq m$):
$$ \begin{aligned} \Pr\left[Y_1 = k \bigg\vert \sum_{i=1}^n Y_i = m\right] &= \frac{\Pr\left[Y_1 = k \cap \sum_{i=1}^n Y_i = m\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \\ &= \frac{\Pr\left[Y_1 = k \cap \sum_{i=2}^n Y_i = m-k\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \\ &= \frac{\Pr\left[Y_1 = k \right]\Pr\left[\sum_{i=2}^n Y_i = m-k\right]}{\Pr\left[\sum_{i=1}^n Y_i = m\right]} \end{aligned} $$
In the final equality we used the independence of $Y_1$ and $\sum_{i=2}^k Y_i $.
Now you know all of the above expressions, since
$Y_1 \sim \mathrm{Pois}(\lambda_1)$, $\sum_{i=2}^n Y_i \sim \mathrm{Pois}(\sum_{i=2}^n \lambda_i)$ and $\sum_{i=1}^n Y_i \sim \mathrm{Pois}(\sum_{i=1}^n \lambda_i)$.
Plugging all of this in will finally yield:
$$ Y_1 \bigg\vert \sum_{i=1}^n Y_i = m \sim Bin\left(m, \frac{\lambda_1}{\sum_{i=1}^n \lambda_i}\right) $$
You can assume that $Y_1 = n$ for whatever $n$ you want, and then compute the entire probability that the rest of the variables sum to $m - n$, by exploiting the fact that the sum of the rest of the variables is Poisson with mean given by the sum of means of the individual Poissons.