Double Integral Used for For Finding Probability of two Uniform RVs

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 * ( + )

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

The answer should be 14/16.

Re: Double Integral Used for For Finding Probability of two Uniform RVs

I didnt integrate the entire thing out, but since you're trying to find the probability that the distance between the two points is greater than L/4, then, so the lower limit of your last integral should be x+L/4

Re: Double Integral Used for For Finding Probability of two Uniform RVs

Yeah sorry about that, edited.

But I did it with those limits and I'm still not getting it right.

It's the calculus part I really need help with

Re: Double Integral Used for For Finding Probability of two Uniform RVs

so,

also,

your limits on integration will be:

therefore,

that should give you what you're looking for.