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.
Probability – Random Selection of Two Points on a Line
probabilityprobability distributions
Related Solutions
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.
Wlog we can assume $L=1$ and $x,y$ are coordinates of $B$ and $C$ then $\Omega$ is $[0,1]^2$ and the event is set all of $(x,y)$ such that $|x-y|\leq {1\over 4}$. Drawing this set of points in the coordinate system you can see it has area $$1-\Big({3\over 4}\Big)^2= {7\over 16}$$ which is also the answer to your question.
Best Answer
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 ${1\over16}$.