Poisson Process Conditional Proability I'm completely stumped with this problem:

Let $\{X_t\;|\;t > 0\}$ be a Poisson process with parameter $\lambda$. Let $T_1, T_2, \ldots$ be the interarrival times for the process, and $W_n = T_1 + \ldots + T_n$.
Now, let $r$ and $n$ be integers such that $1 \leq r \leq n$, and fix $t > 0$.
  Find the conditional distribution of $W_r$ given the event $\{X_t = n\}$.

I realize that $\{X_t = n\}$ is equivalent to $\{W_n \leq t, W_{n+1} > t\}$ by definition, but I have no clue where to go from there.
 A: Have you tried computing the joint probability $\mathbb{P}(X_t = n,W_r \leq s)$ ? The law of $W_1$ conditionally on $\{X_t = n\}$ is rather simple. 
A: For $s\in\left(0,t\right)$ we have:
$$\begin{aligned}P\left(W_{r}\leq s\wedge X_{t}=n\right) & =P\left(X_{s}\geq r\wedge X_{t}=n\right)\\
 & =\sum_{k=r}^{n}P\left(X_{s}=k\wedge X_{t}=n\right)\\
 & =\sum_{k=r}^{n}P\left(X_{s}=k\wedge X_{t}-X_{s}=n-k\right)\\
 & =\sum_{k=r}^{n}P\left(X_{s}=k\right)P\left(X_{t}-X_{s}=n-k\right)\\
 & =\sum_{k=r}^{n}e^{-\lambda s}\frac{\left(\lambda s\right)^{k}}{k!}e^{-\lambda\left(t-s\right)\frac{\left(\lambda\left(t-s\right)\right)^{n-k}}{\left(n-k\right)!}}\\
 & =e^{-\lambda t}\frac{\lambda^{n}}{n!}\sum_{k=r}^{n}\binom{n}{k}s^{k}\left(t-s\right)^{n-k}\\
 & =P\left(X_{t}=n\right)\sum_{k=r}^{n}\binom{n}{k}\left(\frac{s}{t}\right)^{k}\left(1-\frac{s}{t}\right)^{n-k}
\end{aligned}
$$
so that: $$P\left(W_{r}\leq s\mid X_{t}=n\right)=\sum_{k=r}^{n}\binom{n}{k}\left(\frac{s}{t}\right)^{k}\left(1-\frac{s}{t}\right)^{n-k}$$
Actually it can be shown that - if $U_1,\dots U_n$ are iid uniform on $(0,t)$ - then the vector of order statistics $(U_{(1)},\dots,U_{(n)})$ has the same distribution as $(W_1,\dots,W_n)$.
