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 * ($\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.

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,$\displaystyle P(Y-X > L/4) \implies P(Y>X+L/4)$ 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

$\displaystyle Y-X>\frac{L}{4}$

so, $\displaystyle Y>X+\frac{L}{4}$

also, $\displaystyle X < y-\frac{L}{4}$

your limits on integration will be:

therefore,

$\displaystyle 0 < X < \frac{L}{2}-\frac{L}{4}=\frac{L}{4} \implies \boxed{0<X<\frac{L}{4}}$

$\displaystyle \dfrac{4}{L^2}\bigg[ \int_{L/2}^{L} \int_{0}^{L/4}dx dy\;+\;\int_{L/4}^{L/2} \int_{x+L/4}^{L}dydx\bigg]$

that should give you what you're looking for.