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I have to show that if $X\sim Poisson(\lambda_1), Y\sim Poisson(\lambda_2)$ are independent then $$ Pr(X=k| X+Y=n)={n \choose x}p^k q^{n-k} $$ where $p=\frac{\lambda_1}{\lambda_1+\lambda_2}$ and $q=\frac{\lambda_2}{\lambda_1+\lambda_2}$.

My attempt: \begin{align*} Pr(X=k| X+Y=n) & = \frac{Pr(X=k,X+Y=n)}{Pr(X+Y=n)}\\ & = \frac{Pr(X+Y=n,X=k)}{Pr(X+Y=n)}\\ & = \frac{Pr(X+Y=n| X=k)\cdot Pr(X=k)}{Pr(X+Y=n)}\\ & = \frac{Pr(Y=n-k)\cdot Pr(X=k)}{Pr(X+Y=n)}\\ & = \frac{\frac{e^{-\lambda_2}\lambda_2^{n-k}}{(n-k)!}\cdot \frac{e^{-\lambda_1}\lambda_1^{n-k}}{k!}}{\frac{e^{-\lambda_1-\lambda_2}{(\lambda_1+\lambda_2)}^{n}}{n!}}\\ &= {n\choose k}p^k q^{n-k} \end{align*}

Please check my solution.

Thanks.

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    Something looks off in the second to the last line. – 2017-02-18
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    Yeah that was typos, so I corrected it. Thanks. – 2017-02-18
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    I still see some typos in the same place. Assuming they are typos, this should be conceptually correct. – 2017-02-18

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Some comments. $X$ and $Y$ are independent? (Independence has been used in the fourth equality). In fifth equality there are typos regarding factorials. Is $X+Y$ random variable a poisson random variable? (Check it's validity if the score for this question in test is set "very" high!)

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    Sorry they are independent. – 2017-02-18
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    No need to be sorry. Glad to be of help. – 2017-02-18