Let $X$ and$Y$ be the two points such that $X$ ~ $U(0,\frac{L}{2})$ and $Y$ ~ $U(\frac{L}{2},L)$ What is the probability that the distance between $X$ and $Y$ is greater than $\frac{L}{3}$? I know that it is easier to calculate $1-P(Y-X <\frac{L}{3})$, from where I can set up the limits of my integrals: I need to integrate $X$ across $Y$ from $\frac{L}{2}$ to $X+\frac{L}{3}$, and then integrate across $Y$ from $\frac{1}{6}$ to $\frac{L}{2}$. But here is the part that confuses me: $1-\int _{1/6}^{L/2}\int _{L / 2}^{x+L/3}dydx$ does not give me the right answer, some how there is supposed to be a term $\frac{4}{L^2}$ in front of the double integral $1-\frac{4}{L^2} \int _{1/6}^{L/2}\int _{L / 2}^{x+L/3}dydx$.
Could someone please help me figure out where that $\frac{4}{L^2}$ term comes from?
Thanks!!
EDIT: This answer was completely wrong, so I am redoing it.
Consider the problem geometrically; $(X,Y)$ take values in the subset of $\mathbb R^2$ $$\mathcal D = \left\{(x,y)\in\mathbb R^2: 0<x<\frac L2, \frac L2 <y< L\right\}. $$ Because $(X,Y)$ are uniformly distributed over $\mathcal D$, this problem reduces to finding the ratio of area of the subset $$\mathcal S = \left\{(x,y)\in \mathcal D : y-x > \frac L3\right\}. $$ Now, it is easier to compute the area of $\mathcal D\setminus S$, because this is simply a triangle. Plotting $\mathcal S$ and the line $y-x<\frac L3$, you'll see that the three points of this triangle are $$\left(\frac L6, \frac L2\right), \left(\frac L2, \frac L2\right), \left(\frac L2, \frac{5L}6\right). $$ The area of this triangle is then simply $$\frac12\left(\frac L2-\frac L6\right)\left(\frac{5L}6-\frac L2\right) = \frac1{18}L^2. $$ Hence, the desired probability is $$\frac{A(\mathcal S) - \mathcal A(\mathcal D\setminus \mathcal S)}{A(\mathcal S)} = \frac{\frac{L^2}4 - \frac{L^2}{18}}{\frac{L^2}4} = 7/9.$$