Probability for Uniform distributions Two people decide to meet at a cafeteria sometime between 12:00 and 13:00. If, each and one of them arrive at random chosen times during the hour, and wait 45 minutes on each other (or until the clock strikes 13:00). What is the probability that they will meet?
My teacher told me that this can be solved drawing a quadrate, with I think 0 and 1 at each corner, respectively. Drawing a graph, and finding a cross-section would then yield the answer. The problem is I don't know how to do the graphs. He alse told me that the time each person waits may be regarded as a uniform distribution. So, my guess is that he means that
$$
t_1 \in U(0,60), t_2 \in U(0,60)
$$ 
But from here i don't know where to go. Anyone?
 A: Here is a plot of the starting time in minutes for $x$ (horizontal) and $y$ (vertical) and the cases where they will meet.  Formally:  $(x,y)\ s.t. |x-y| \leq 45 \wedge 0 \leq x \leq 60 \wedge 0 \leq y \leq 60$.

The area of this region divided by the total possible region is the probability they will meet:  $15/16$.
A: For simplicity, imagine they wait the full time regardless of whether they meet.
Anyone arriving after 12:15 will wait less than 45 minutes.
There's a $\frac14$ chance for each of them that they'll actually wait 45 minutes. 
Consider the cases where neither of them wait the full 45 minutes.
Trick question: What's the probability they meet?
Consider the cases where both of them wait the full 45 minutes.
Trick question: What's the probability they meet? 
In the remaining cases, one of their arrivals is uniformly distributed from 12:00 to 12:15; consider that as a distribution of time earlier than 12:15 that they arrive.
The other arrival is uniformly distributed from 12:15 to 13:00; consider that as a distribution of time after 12:15 that they arrive.
They'll meet if the sum of those two times is less than 45 minutes.
I don't know what a quadrate is in this context, but it sounds like you have the tools you need to figure out the probability of that sum being less than 45 minutes, and to combine all the cases into a net probability. 
