Find distribution of $X_1$ by solving $P(X_1 = k |X_1 + X_2 = n)$ where $X_1, X_2$ are independent Poisson variables with parameters $\lambda_1, \lambda_2$ and given $X_1 + X_2 = n$
Solution:
$$P(X_1 = k |X_1 + X_2 = n) = \frac{P(X_1 = k, X_1 + X_2 = n)}{P(X_1 + X_2 = n )}$$
$$= \frac{P(X_1 = k)P(X_2 = n - k)}{P(X_1 + X_2 = n)}$$
$$= \frac{\frac{\lambda_1^ke^{-\lambda_1}}{k!}\frac{\lambda_2^ke^{-\lambda_2}}{(n-k)!}}{e^{-(\lambda_1+\lambda_2)}(\lambda_1 + \lambda_2)^n/n!}$$
$$=\frac{n!}{k!(n-k!)}\left(\frac{\lambda_1}{\lambda_1+\lambda_2} \right)^k\left(\frac{\lambda_2}{\lambda_1 + \lambda_2} \right)^{n-k}$$
This is binomial theorem.
I don't understand the last step.
This part
$$\left(\frac{\lambda_1}{\lambda_1+\lambda_2} \right)^k\left(\frac{\lambda_2}{\lambda_1 + \lambda_2} \right)^{n-k}$$
What I get while trying this was
$$\left(\frac{\lambda_1^k}{(\lambda_1+\lambda_2)^n} \right)\left(\frac{\lambda_2^{n-k}}{(\lambda_1 + \lambda_2)^n} \right)$$
Which is what it equals to I'm not sure how the $n$ was replaced with $k$ and $(n-k)$.
Also on the second step
$$\frac{P(X_1 = k, X_1 + X_2 = n)}{P(X_1 + X_2 = n )} = \frac{P(X_1 = k)P(X_2 = n - k)}{P(X_1 + X_2 = n)}$$
Couldn't they cancell the denominator in the same way ? What property lets them split the AND part of P like $P(A, B) = P(A)P(B)$
$(X_1=k,X_1+X_2=n)=(X_1=k, X_2=n-k)$ and the events $(X_1=k),( X_2=n-k)$ are independent. His justifies the second step.
In line three you have mis-typed $\lambda_2^{k}$ for $\lambda_2^{n-k}$.
In the denominator they have split $(\lambda_1+\lambda_2)^{n}$ as $ (\lambda_1+\lambda_2)^{k}(\lambda_1+\lambda_2)^{n-k}$.