Probability of chossing two points from a segment of length L so that one is 2L/3 greater than another. Two points are selected randomly on a line of length $L$ so as to be on the opposite sides of the midpoint of the line. In other words, two points X and Y are independent random variables such that X is uniformly distributed over $(0,L/2)$ and Y is so over $(L/2,0)$. Find the probability that the distance between these two points is greater than $2L/3$.
Here's what I have done.

Can I get some help to finish this.
 A: Does this make sense to you? I think this aligns with what Ethan has commented.

A: A picture, as provided by J. Wang, is very useful for intuition (and has my +1!), but it does need to be made into a rigorous proof. My suggestion is to try conditioning on the first point, $\ell_1$. Once you know this, you can ask "Which options for $\ell_2$ give the required condition?"---this clearly depends on $\ell_1$.


*

*If $\ell_1 > L/3$, then clearly no $\ell_2$ will suffice.

*If $\ell_1 \in [0,L/3]$, then any $\ell_2 \in [\ell_1+2L/3,L]$ will suffice.


Given $\ell_1$, and that we land in the second scenario, there is a probability $L - (\ell_1 + 2L/3) = L/3 - \ell_1$ that we have the desired property.
Averaging over $\ell_1$, we find that the desired probability is
$$ \textstyle \int_0^{L/3} (L/3 - \ell) f_1(\ell) d\ell
\quad\text{where}\quad
f_1(\ell) = 1/(L/2) = 2/L \text{ is the pdf of $\ell_1$}. $$
Note that this integral is equal to
$$ \textstyle (2/L) \int_0^{L/3} \ell' d \ell' = (2/L) \cdot \tfrac12 (L/3)^2 = L/9. $$
