Probability Help! X and Y are geometric RV's with parameter p. What are ... a) $P\{X + Y = n\} (n = 1,2,...)$? 
b) $P\{X = k | X + Y = n\} (1 ≤ k < n)$?
My work thus far...
a)
$\begin{align}P(X + Y = 1) & = P((X = 0 \cap Y = 1) \cup (X = 1 \cap Y = 0)) \\ & = P(X = 0)P(Y = 1)+P(X = 1)P(Y = 0) \\ ~ & = 2P(0)P(1) \\ ~ & = 2q p^2 \end{align}$
$\begin{align}P(X + Y = 2) & = P((X = 0 \cap Y = 2) \cup (X = 2\cap Y = 0)\cup (X = 1\cap Y = 1)) \\ ~ & = 2P(0)P(2) + P(1)P(1) \\ ~ & = 2(p^2)(q^2) + (q^2)(p^2) \\ ~ & = 3(q^2)(p^2) \end{align}$
Does this seem like I'm on the correct path... Any help is much appreciated!
 A: I am assuming $X$ and $Y$ are independent.  Without this assumption, it is not possible to answer your question without further information.
You have the right idea, but doing the calculation for each $n$ is clearly going to take a while.  So, you have to be a little more clever:  Suppose we let $Z = X+Y$.  We wish to find $\Pr[Z = n]$.  Note that we can write $$\Pr[Z = n] = \sum_{y=0}^n \Pr[(X = n-y) \cap (Y = y)] = \sum_{y=0}^n \Pr[X = n-y]\Pr[Y = y].$$  Now you know what each of these probabilities is, so write them out, and calculate the sum.
A: For the shifted geometric distribution, of $X$ failures before success on trial $X+1$, $X\sim\mathcal{SGeo}(p)\iff \Pr(X=x)=p(1-p)^x$
$\begin{align} \text{Given: } & X\bot Y, X\sim\mathcal{SGeo}(p), Y\sim\mathcal{SGeo}(p) 
\\ \Pr(X+Y=n) & = \Pr\left(\bigcup\limits_{x=0}^{n} (X=x\cap Y= n-x)\right) & \forall n\in\{0..\infty\}
\\ ~ & = \sum\limits_{x=0}^{n} \Pr(X=x)\Pr(Y=n-x)
\\ ~ & = \sum\limits_{x=0}^{n} p(1-p)^x\cdot p(1-p)^{n-x}
\\ ~ & = (n+1)p^2(1-p)^n
\\ ~ & ~
\\ \Pr(X=k\mid X+Y=n) & = \dfrac{\Pr(X=k\cap X+Y=n)}{\Pr(X+Y=n)} & \forall n\in\{0..\infty\}, k\in\{0..n\}
\\ ~ & = \text{et cetera}
\end{align}$

For the geometric distribution of $X-1$ failures before success on trial $X$, $X\sim\mathcal{Geo}(p)\iff \Pr(X=x)=p(1-p)^{x-1}$
$\begin{align} \text{Given: } & X\bot Y, X\sim\mathcal{Geo}(p), Y\sim\mathcal{Geo}(p)
\\ \Pr(X+Y=n) & = \Pr\left(\bigcup\limits_{x=1}^{n-1} (X=x\cap Y= n-x)\right) & \forall n\in\{2..\infty\}
\\ ~ & = \sum\limits_{x=1}^{n-1} \Pr(X=x)\Pr(Y=n-x)
\\ ~ & = \sum\limits_{x=1}^{n-1} p(1-p)^{x-1}\cdot p(1-p)^{n-x-1}
\\ ~ & = (n-1)p^2(1-p)^{n-2}
\\ ~ & ~
\\ \Pr(X=k\mid X+Y=n) & = \dfrac{\Pr(X=k\cap X+Y=n)}{\Pr(X+Y=n)} & \forall n\in\{2..\infty\}, k\in\{1..n-1\}
\\ ~ & = \text{et cetera}
\end{align}$
