$y_1 \sim \operatorname{Po}(e^\lambda), y_2 \sim \operatorname{Po}(e^{\lambda + \psi})$ (they are independent) I want to show that the distribution $y_2| y_1 + y_2 = u$ has distribution $\operatorname{Bin}( u,e^\psi/1+e^\psi)$. I am not getting there. Here is my attempt
$$ \begin{split} f( y_2 = v | y_1 + y_2 = u ) &= \frac{f(y_2 =v , y_1 = u-v)}{f(y_1 + y_2 = u)}\\ &= \frac{e^{-e^{(\lambda + \psi)}} \frac{(e^{\lambda + \psi})^v}{v!} e^{-e^{\lambda}} \frac{(e^\lambda)^{u-v}}{(u-v)!}}{ e^{-(e^{\lambda +\psi} + e^\lambda)} \dfrac{(e^{\lambda + \psi} + e^\lambda)^u}{u!} }\\ &= \frac{u!}{v!(u-v!)} \frac{ (e^{\lambda + \psi })^v }{(e^{\psi} + 1)^u (e^{\lambda})^v} \end{split} $$