Choosing 2 points on a line 
Two points are selected randomly on a line of length $L$ so as to be on opposite sides of the midpoint of the line. Find the probability that the distance between them is greater than $L/3$.

I was able to figure out that one point will be uniformly distributed on $(0,L/2)$ and the other on $(L/2,L)$, but I don’t know how to proceed further.
 A: WLOG we can assume $L=6$. Let $x$ be coordinate of the left point and $y$ of the right.
Then the sample space is $$\Omega = \{(x,y);\;x\in [0,3];\;y\in [3,6]\}$$
and the event is $$A = \{(x,y)\in\Omega;\; y\geq x+2\}$$
so we have a picture:
Now $$P(A) = {Area (A)\over Area(\Omega)} = {7\over 9}$$
Edit: Yes I missed 3 left squares. 
A: Call the positions of the points on the line $X_1$ and $X_2$. You have three cases:


*

*Case 1: $X_1\in [0,L/6]$, in which case the condition is respected. Add the probability of that to your total.

*Case 2: $X_2\in [5L/6,L]$, in which case the condition is also respected. Add the probability of that to your total (taking care to subtract the intersection of "Case 1" with "Case 2" so you don't double count it).

*Case 3: All other cases. Here I'd think of $X_1$ having a specific value, then find the probability that $X_2$ is placed so that the condition is respected as a function of $X_1$. To make this easy (so you don't have to normalize your probability later), think of the your line being placed in a coordinate system so that $L/6$ is on the point $-1$, $L/2$ is on $0$, and $5L/6$ is on $1$. Finally, integrate over all the values $X_1$ can take. Multiply by the probability that we are in "Case 3", and add to your total.


I find the probability of success to be 7/9 with this method. Do you?
A: Analytic solution. Let $x$ and $y$ be the points in $[0,L]$ such that $0\le x\le \frac L2\le y\le L$. The PDF is $f(x,y)=\frac{4}{L^2}$. We need to find $P\left(y-x\ge \frac L3\right)$.
Note that $y-x\ge \frac L3 \Rightarrow y\ge x+\frac L3\ge \frac L2 \Rightarrow x\ge \frac L6$. Hence, we can divide $x$ into two intervals:
$$\begin{cases}0\le x\le \frac L6\\ \frac L2\le y\le L\end{cases} \ \ \ 
 \ \text{or} \ \ \ \ \ \begin{cases}\frac L6\le x\le \frac L2\\ x+\frac L3\le y\le L\end{cases}$$
The required probability is:
$$\begin{align}P\left(y-x\ge \frac L3\right)&=\int_0^{L/6} \int_{L/2}^{L} \frac 4{L^2} dydx+\int_{L/6}^{L/2} \int_{x+L/3}^{L} \frac 4{L^2} dydx=\\
&=\frac13+\frac49=\\
&=\frac79.\end{align}$$
