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We can look at a more general case that I believe will help solve this problem.

\begin{eqnarray*} P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\ &=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\ &=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_2}{\lambda_1+\lambda_2} \right)^{n-x} \end{eqnarray*}

Which is a binomial pmf with $p = \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$ and expected value $E(X|X+Y=n) = np = n \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$

However,I think this only works in the case where $\lambda_1=\lambda_2$. Whcih is not the case in this problem.

Can anyone help to expand on this?

We can look at a more general case that I believe will help solve this problem.

\begin{eqnarray*} P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\ &=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\ &=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_2}{\lambda_1+\lambda_2} \right)^{n-x} \end{eqnarray*}

Which is a binomial pmf with $p = \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$ and expected value $E(X|X+Y=n) = np = n \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$

We can look at a more general case that I believe will help solve this problem.

\begin{eqnarray*} P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\ &=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\ &=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^{n-x} \end{eqnarray*}

Which is a binomial pmf with $p = \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$ and expected value $E(X|X+Y=n) = np = n \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$

We can look at a more general case that I believe will help solve this problem.

\begin{eqnarray*} P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\ &=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\ &=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_2}{\lambda_1+\lambda_2} \right)^{n-x} \end{eqnarray*}

Which is a binomial pmf with $p = \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$ and expected value $E(X|X+Y=n) = np = n \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$

However,I think this only works in the case where $\lambda_1=\lambda_2$. Whcih is not the case in this problem.

Can anyone help to expand on this?

We can look at a more general case that I believe will help solve this problem.

\begin{eqnarray*} P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\ &=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\ &=& \frac{\frac{e^{-(\lambda_1)}\lambda^x}{x!} \frac{e^{-(\lambda_2)}\lambda_2^{(n-x)}}{n-x!}}{\frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^n}{n!} } \\ &=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^{n-x} \end{eqnarray*}

Which is a binomial pmf with $p = \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$ and expected value $E(X|X+Y=n) = np = n \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$

We can look at a more general case that I believe will help solve this problem.

\begin{eqnarray*} P(X|X+Y=n) &=& \frac{P(X=x, Y=n-x)}{P(Z=n)}\\ &=& \frac{P(X=x)P(Y=n-x)}{P(Z=n)} \\ &=& {n \choose x} \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^x \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)^{n-x} \end{eqnarray*}

Which is a binomial pmf with $p = \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$ and expected value $E(X|X+Y=n) = np = n \left( \frac{\lambda_1}{\lambda_1+\lambda_2} \right)$