When trying to calculate the value of $P(X= 1 | X+Y=9)$, I tried doing:
$P(X= 1 | X+Y=9) = \frac{P(X=1\cap X+Y=9)}{P(X+Y = 9)} = \frac{P(X=1)P(Y=8)}{P(X+Y = 9)} $, but I'm constantly getting an incorrect answer.
Could someone shed some light on this?
The simplification is correct.
You just used $P(A|B) = \frac{P(A \cap B)}{P(B)}$ and $P(AB)=P(A)P(B)$ if $A$ and $B$ are independent.
You have to look for other factors, including potential computational mistake such as when you evaluate $P(X+Y=9)$ or perhaps a mistake in the answer key.
Remark:
$$\frac{P(X=1)P(Y=8)}{P(X+Y=9)}=\frac{u_1\cdot \frac{u_2^8}{8!}}{\frac{(u_1+u_2)^{9}}{9!}}=\binom91\left( \frac{u_1}{u_1+u_2}\right)\left( 1-\frac{u_1}{u_1+u_2}\right)^8$$