Compute $\ P(X_i | \sum X_i = 45) $ where $\ X_i \sim \mathrm{Pois}$ Let $\ X_i \sim \mathrm{Pois}(10) $ be independent random variables $\ i = 1, 2,3,4,5 $ .
What is the probability of $\ X_1 = 9 $ given $\ \sum X_i = 45 $
now $\ \sum X_i \sim \mathrm{Pois}(50) $ and so I thought 
$$\ P(X_1 = 9 \ | \ \sum_{i=1}^5 X_i = 45 ) = P(X_1 =9  \ | \ \sum_{i=2}^5X_i = 36) = \frac{P(X_1 =9 , \sum_{i=2}^5 X_i = 36)}{\sum_{i=1}^5 X_i = 45 } = \frac{e^{-5}\cdot \frac{5^9}{9!}\cdot e^{-40}\cdot\frac{40^{36}}{36!}}{e^{-50}\cdot\frac{50^{45}}{45!}} = \frac{0.03626 \cdot 0.053939}{0.0458} =  0.04269 $$ 
But apparently that is wrong and the solution is using binomial distribution with parameters $\ n = 45, p = 1/5 $ and final answer is $\ 0.14724 $ 
Is there anyway I can solve it with the Poisson distribution?
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Let $\ds{\mc{P}\pars{\lambda,n} = {\expo{-\lambda}\lambda^{n} \over n!}}$.

\begin{align}
&\bbox[10px,#ffd]{\sum_{x_{1} = 0}^{\infty}\mc{P}\pars{\lambda,x_{1}}
\!\!\sum_{x_{2} = 0}^{\infty}\mc{P}\pars{\lambda,x_{2}}
\!\!\sum_{x_{3} = 0}^{\infty}\mc{P}\pars{\lambda,x_{3}}
\!\!\sum_{x_{4} = 0}^{\infty}\mc{P}\pars{\lambda,x_{4}}
\!\!\sum_{x_{5} = 0}^{\infty}\mc{P}\pars{\lambda,x_{5}}}
\times
\\[2mm] &\ \bracks{x_{1} = 9}
\bracks{z^{45}}\! z^{\sum_{i = 1}^{5}x_{i}}
\\[5mm] = &\
\mc{P}\pars{\lambda,9}\bracks{z^{45}}z^{9}
\bracks{\sum_{x = 0}^{\infty}\mc{P}\pars{x}z^{x}}^{4} =
\mc{P}\pars{\lambda,9}\bracks{z^{36}}
\bracks{\sum_{x = 0}^{\infty}{\expo{-\lambda}\lambda^{x} \over x!}z^{x}}^{4}
\\[5mm] = &\
\mc{P}\pars{\lambda,9}\expo{-4\lambda}\bracks{z^{36}}
\bracks{\sum_{x = 0}^{\infty}{\pars{\lambda z}^{x} \over x!}}^{4} =
\mc{P}\pars{\lambda,9}\expo{-4\lambda}\bracks{z^{36}}
\expo{4\lambda z}
\\[5mm] = &\
\mc{P}\pars{\lambda,9}\expo{-4\lambda}{\pars{4\lambda}^{36} \over 36!} =
{\expo{-\lambda}\lambda^{9} \over 9!}\expo{-4\lambda}{4^{36}\lambda^{36} \over 36!}
\\[5mm] = &\
\bbx{{4^{36} \over 9!\, 36!}
\,\expo{-5\lambda}\lambda^{45}}
\end{align}

$\ds{\lambda = 10 \implies \approx 6.7474 \times 10^{-3}}$.

A: It is because the sum of independent Poisson random variables is a Poisson random variables whose mean is the sum of their means, and all $X_i$ are independent Poisson random variables whose mean is $10$, that we have the following.
$$X_1\sim\mathcal{Pois}(10)\\\sum_{i=2}^5X_i\sim\mathcal{Pois}(40)\\\sum_{i=1}^5X_i\sim\mathcal{Pois}(50)$$
(Emphasis because the fact of independence is crucial to this solution.)
Also, if we define $Y_\lambda\sim\mathcal {Pois}(\lambda)$ then we have $~\mathsf P(Y_\lambda{=}y)~=~\dfrac{\lambda^y\mathsf e^\lambda}{y!}\mathbf 1_{y\in\Bbb N}$, and we can use this to ensure we are substituting the correct values.
So, therefore, we get the following.
$$\begin{align}\mathsf P(X_1{=}9\mid \sum_{i=1}^5X_i{=}45)&=\dfrac{\mathsf P(X_1{=}9)\cdot\mathsf P(\sum_{i=2}^5 X_i{=}36)}{\mathsf P(\sum_{i=1}^5 X_i{=}45)}\\[1ex]&=\dfrac{\mathsf P(Y_{10}{=}9)\cdot\mathsf P(Y_{40}{=}36)}{\mathsf P(Y_{50}{=}45)}\\[1ex]&=\dfrac{\dfrac{10^9\mathsf e^{10}}{9!}\cdot\dfrac{40^{36}\mathsf e^{40}}{36!}}{\dfrac{50^{45}\mathsf e^{50}}{45!}}\\[1ex]&=\dbinom{45}9 \dfrac{10^9\cdot 40^{36}}{50^{45}}\end{align}$$
Which is where the binomial distribution comes from.
[The count of point events which occur in one among several independent Poisson processes has a conditionally binomial distribution when given the total count of events that occur among all of them, with success rate equal to the ratio of the sums of the Poisson rate parameters (favored versus total).]
