Two points are selected on a straight line of length 'a' units at random If two points are selected on a straight line of length 'a' units at random, then what is the probability that none of the three line segments formed by the two random points has length less than a/4.
 A: Choosing two independently and uniformly distributed points $x$, $y\in[0,1]$ is the same as choosing an uniformly distributed point $(x,y)\in Q:=[0,1]^2$. In other words: The joint probability measure at stake is just the area measure on $Q$. Consider a point $(x,y)\in Q$. When $y\leq x$ the requirement that none of the three parts of the stick so generated has length  $<{1\over4}$ is fulfilled iff $y\geq{1\over 4}$, $\>x\leq{3\over4}$, and $y\leq x-{1\over4}$. The points fulfilling these conditions are the points in the lower right red triangle in the following figure. The assumption $y\geq x$ gives rise to a congruent such triangle, colored red in the figure as well. All in all, the set of points where  none of the three pieces of the stick has length $<{1\over4}$ has area ${1\over16}$. It follows that the probability in question is ${1\over16}$.

A: Hint: Calculate $\mathsf P(\frac a 4\leq \min(X,Y-X,a-Y) \mid X\leq Y)$ where $X,Y\sim\mathcal U[0;a]$
Note: without loss of generality, consider $a=1$
