Given that two people were at a location at the same time, find the probability they were there at a particular time 
Roger and Stacy each go to the county fair on the same day. They each separately show up at a random time between $12:00$ PM and $9:00$ PM. Roger stays for $2$ hours and Stacy stays for $3$ hours.
Given that, at some time, Roger and Stacy were at the fair simultaneously, what is the probability that they were both there at exactly $6:00$ PM?

This is what I have so far:
$0 \leq r \leq 9$ and $0 \leq s \leq 9$ (in arrival times expressed in hours after noon).
If they were both at the fair at the same time, Roger must have arrived sometime between $2$ hours before Stacy and $3$ hours after Stacy arrived. So we have the inequalities $s-2 \leq r \leq s+3$. Now this is where I get stuck. What do I do next, and how do I finish this problem? There is a similar problem in the book, but the solution does not make much sense.

For @NickH:
Here was the statement for the similar problem:
Roger and Stacy each go to the county fair on the same day. They each separately show up at a random time between 12:00 and 6:00. Roger stays for an hour, whereas Stacy stays for 2 hours. If we know that at some time they were both at the fair simultaneously, what is the probability that they were both there at exactly 3:00??
The solution to that problem is below:

 A: Let $R$ be Roger's arrival time and $S$ be Stacy's; both $R$ and $S$ are independent Uniform$(0,9)$ random variables. 
You are asked to find the probability that they are both there at 6 so that means $4\leq R \leq 6$ and $3\leq S\leq 6$. 
However, you are also conditioning on the fact that they are both at the fair at the same time: as you correctly said, this means that $R-3\leq S\leq R+2$. 
Finally, it should be clear that the event that they're there at 6 implies that they're at the fair at the same time, i.e., $\lbrace 4\leq R \leq 6,3\leq S\leq 6 \rbrace \subset \lbrace R-3\leq S\leq R+2\rbrace$. 
To sum up, this is what you are solving for:
\begin{equation*}
\mathbf{P}(4\leq R \leq 6,3\leq S\leq 6\big|R-3\leq S\leq R+2)=\frac{\mathbf{P}(4\leq R \leq 6,3\leq S\leq 6}{\mathbf{P}(R-3\leq S\leq R+2)},
\end{equation*}
 which should be no trouble--the numerator is really just the area of a rectangle (divided by $1/81$) and the denominator is the area of the region bounded by two lines and the 9$\times$9 square (divided by $1/81$).
A: I'm user198152 (yes, I really should register). The solution is not that complicated. If we're agreed that we're solving for:
\begin{equation*}
\frac{\mathbf{P}(4\leq R\leq 6,3\leq S\leq 6)}{\mathbf{P}(R-3\leq S\leq R+2)},
\end{equation*}
we'll proceed as follows, using the following graphs. Remember that $R$ and $S$ are independent uniform r.v.s so their joint density is just $\frac{1}{81}$ on $[0,9]\times[0,9]$. For the numerator,
hence, 
\begin{equation*}
\mathbf{P}(4\leq R\leq 6,3\leq S\leq 6)=6/81.
\end{equation*}
Similarly, for the denominator,
\begin{equation*}
\mathbf{P}(R-3\leq S\leq R+2)=\frac{1}{81}(\text{area of shaded region})=\frac{1}{81}\frac{77}{2}.
\end{equation*} 
The final answer is then $\frac{12}{77}$. Hope that makes sense.
