Consider the square with vertices $(0,0),(1,0),(1,1),(0,1)$.
Choose two independent uniformly random points $P$ and $Q$ inside the square.
Draw a line $l_P$ connecting $(0,0)$ and $P$.
Draw another line $l_Q$ connecting $(1,0)$ and $Q$.
What is the probability that $l_P$ and $l_Q$ intersect inside the square?
I will post my answer. Alternative solutions are welcome.
This question and answer serve to flesh out a comment I made at a similar question.
Best Answer
Let $p$ be the $x$-coordinate of the intersection of $l_P$ and $y=1$.
Let $q$ be the $x$-coordinate of the intersection of $l_Q$ and $y=1$.
If $p<1$ then $p$ is uniformly distributed in $[0,1]$, because the probability that $p<x$ is proportional to the area of the triangle with vertices $(0,0)$, $(0,1)$ and $(x,1)$, which is proportional to $x$.
Similarly, if $q>0$ then $q$ is uniformly distributed in $[0,1]$.
For the lines not to intersect inside the square, first we require that $p<1$, which has probability $\frac12$. Then we require that $q>0$, which has probability $\frac12$. Finally we require that $p<q$, which (given that the first two requirements have been met) has probability $\frac12$ (since $p$ and $q$ are each uniformly distributed in $[0,1]$). So the probability that the lines do not intersect inside the square, is $\left(\frac12\right)^3=\frac18$.
Therefore the probability that the lines intersect inside the square is $\frac78$.