Two points are randomly selected on a line of length $1$ Two points are randomly selected on a line of length $1$. What is the probability that one of the segments is greater than $\frac{1}{2}$? Points can be placed anywhere between [0, 1], for example. Thanks! 
 A: Probability is $3/4$ if points $x$ and $y$ are chosen with uniform probability. That corresponds to the area in color in the picture below.

A: With probability $\frac 12$ both points are on the same side of the midpoint, so we are guaranteed success.
If the points are on opposite sides of the midpoint(a probability $\frac 12$ event, with $P<\frac 12< Q$ say, then again with probability $\frac 12$ we have $Q$ is nearer $1$ than $P$ is near $\frac 12$,so the segment between them has length greater than $\frac 12$.
Thus the total probability Is $$\frac 12+\frac 12\times \frac 12=\frac 34$$
Note:  this is equivalent to asking how probable it is that the three segments formed by the two points can form a triangle (the above shows that the answer is $\frac 14$).  Many proofs for that can be found e.g. here
A: Ok suppose you take some point $x\in[0,1/2]$. Now taking a second point $y\in[0,1]$, there are two situations where you obtain a segment of length at least $1/2$. Firstly if $y\le 1/2$. Secondly if $y\ge x+1/2$. So for taking a first point $x$, the chance that you have a segment of desired length is $1/2+(1-1/2-x)=1 - x$. Now integrating this over $[0,1/2]$ you get $3/8$. Yu can do the same for $x\in[1/2,1]$. So this gives you a chance of $6/8$.
