Let $X,Y$ random variables IID with $\mathbb{P}\left(X=x\right)=pq^{x}$. Show that
$$\mathbb{P}\left(\left.X=x\right|X+Y=n\right)=\dfrac{1}{n+1}$$
for $x=0,1,...,n$.
Hint: If $X,Y$ are two random variables independent then
$$\mathbb{P}\left(Z=z\right)=\sum_{x}\mathbb{P}\left(X=x\right)\mathbb{P}\left(Y=z-x\right)$$
where $Z=X+Y$.
My attempt:
$$\mathbb{P}\left(\left.X=x\right|X+Y=n\right)=\dfrac{\mathbb{P}\left(X=x,X+Y=n\right)}{\mathbb{P}\left(X+Y=n\right)}=$$
$$=\dfrac{\mathbb{P}\left(X=x,Y=n-x\right)}{\mathbb{P}\left(X+Y=n\right)}=\dfrac{\mathbb{P}\left(X=x\right)\mathbb{P}\left(Y=n-x\right)}{\mathbb{P}\left(X+Y=n\right)}=...$$
Observe that $$ P(X=x)P(Y=n-x)=pq^x(pq^{n-x})=p^2q^n \quad (0\leq x\leq n) $$ while $$ P(X+Y=n)=\sum_{x=0}^nP(X=x)P(Y=n-x)=p^2q^n(n+1) $$ whence $$ \frac{P(X=x)P(Y=n-x)}{P(X+Y=n)}=\frac{1}{n+1};\quad (0\leq x\leq n). $$