Conditional pmf of $X \mid X+Y=n$, with $X,Y$ independent geometric r.v.

Question

Suppose, $X, Y$ are independent geometric random variables with the same parameter $p$. We want to find the conditional probability function of $X$ given that $X+ Y =n$, where $n >1$.

Answer 1

I assume that the geometric distribution starts from $k=1$. Then for $1\le k\le n$ we have that \begin{align}P(X=k \mid X+Y=n)&=\frac{P(X+Y=n \mid X=k)P(X=k)}{P(X+Y=n)}=\frac{P(Y=n-k)P(X=k)}{P(X+Y=n)}\\[0.3cm]&=\frac{p(1-p)^{n-k-1}p(1-p)^{k-1}}{P(X+Y=n)}=\frac{p^2(1-p)^{n-2}}{P(X+Y=n)}\tag{1}\end{align} Now for any $n\in \mathbb N$\begin{align}P(X+Y=n)&=\sum_{k=1}^{n-1}P(X=k)P(Y=n-k)=\sum_{k=1}^{n-1}p(1-p)^{k-1}p(1-p)^{n-k-1}\\[0.2cm]&=(n-1)p^2(1-p)^{n-2}\end{align} so, substituting back in $(1)$ gives $$P(X=k \mid X+Y=n)=\frac{p^2(1-p)^{n-2}}{(n-1)p^2(1-p)^{n-2}}=\frac{1}{n-1}$$ So $X \mid X+Y=n$ is uniformly distributed in $\{1,2,\ldots, n-1\}$.