I can perform two counting experiments: in the first experiment I measure the Poisson-distributed $K$ "background" events, while in the second experiment I have additional Poisson-distributed $M$ "signal" events, and I can only measure the total number of events $N=M+K$. The two population are assumed completely independent, so it is easy to derive:
$$ K\sim Poiss(\beta) ; \quad M\sim Poiss(\mu) ; \quad N\sim Poiss(\beta + \mu) ; $$ $$ \mathcal{P}(M|N) \sim Binom(N,p) , \quad p= \mu / (\beta+\mu) , \qquad etc... $$
Now I developed an analysis selection that is suppose to cut a large fraction of background events and keep most of the signal events. If $\chi$ and $\varepsilon$ are the background retention probability and the signal selection efficiency, respectively, and $k$ and $m$ are the events surviving my selection, I can write and easily derive:
$$ k \sim Binom(K,\chi) ; \quad m \sim Binom(M,\varepsilon) ; \quad \mathcal{P}(m|N) \sim Binom(N,\varepsilon p) ; \quad etc... $$
In the first experiment I can estimate $\chi$ and its asymmetric errors using the Bayesian approach suggested in https://arxiv.org/pdf/physics/0701199v1.pdf
However to assess $\varepsilon$ - possibly using the same approach for the error calculation - I have troubles. In the second experiment I do not have directly $m$ and $M$, but rather:
$$ \mathcal{P}(n=k+m|N=K+M) = \sum_{j=0}^n \binom{N}{j}(\varepsilon p)^j (1-\varepsilon p)^{N-j} \binom{N}{n-j}[\chi (1-p)]^{n-j} [1-\chi (1-p)]^{N-(n-j)} $$
I am not sure if there is a sum expression. The terms resemble a multinomial distribution because all the probabilities sum to $1$ and all the exponents sum to $2N$, but this does not help me.
Of course I cannot exclude that I have done mistakes in the definition of the probability and/or in the calculations, or even there is a easier way to assess the efficiency.
Any suggestion is very welcome