Probability of A waits B at least 10 minutes Suppose a friend and you are to meet at the town sometimes between
17:00-18:00. Let X be the time you arrive and let Y be the time your friend
arrives.
Find the probability that you have to wait at least 10 minutes for your friend.
Could it be solved by using areas: the area between $Y-X\ge10$ and $60 X 60$ over the area of $60X60$ which equals to $35\over36$? Or is there another solution?
 A: It depends "how" the people arrive. Suppose that it is equally likely that a person arrive any time between 17:00 and 18:00. That is, the arrival time is uniformly distributed and your method of taking areas is valid. 
We want the area of a "unit" square where the $X$ dimension is $\frac{1}{6}$ (10 minutes) larger than the $Y$ dimension. This is the area of a triangle formed by drawing a line with slope 1 from $(0,\frac{1}{6})$ up to $(\frac{5}{6},1)$ (see James' image in the comments. Notice this is a triangle with base and height equal to $\frac{5}{6}$. It's area is half of the base times the height:
$$\frac{1}{2}\left(\frac{5}{6}\right)^2=\frac{25}{72}$$
If the arrival times weren't uniform but instead distributed $f(x)$ and $f(y)$ we could use the following integral: 
$$\int_{0}^{\frac{5}{6}}\int_{x+\frac{1}{6}}^{1}f(x)f(y)\,dy\,dx$$
Notice when $f(x)=f(y)=1$ (uniform case), we get:
$$\int_{0}^{\frac{5}{6}}\int_{x+\frac{1}{6}}^{1}\,dy\,dx=\frac{25}{72}$$
This is exactly the integral representation of the triangle area we took previously.  
A: I get $\frac{1}{2} \times \frac{5}{6} \times \frac{5}{6} $
If you draw X and Y arrival time as a square.  The set of points with X more than 10 minutes later for Y is a triangle.  Then $ A = \frac{1}{2} \times b \times h$.
This is sometimes called geometric probability https://math.stackexchange.com/questions/tagged/geometric-probability
