Problem: Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. [i.e., the two points X and Y are independent random variables such that X is uniformly distributed over (0,L/2) and Y is uniformly distributed over (L/2,L).] Find the probability that the distance between the two points is greater than L/4.

So we are looking for P(Y-x > L/4) and f(x,y) = 4/L^2

This probability would be: 4/L^2 * ($\displaystyle \int_{0}^{L/8}dx \int_{L/2}^{L}dy $ + $\displaystyle \int_{L/8}^{L/2}dx \int_{x+L/4}^{L}dy $ )

But whenever I do this integration i keep getting (4/L^2)(L^2/4) = 1 and

The answer should be 14/16.