analytic geometrygeometryprobability
Please help me with this one.
Two points (B and C) are selected randomly on a line of length L. Find the probability that the segment BC has a length less than L / 4. It is assumed that the probability of a point falling on the segment is proportional to the length and does not depend on its location on the numerical axis OX.
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 ${1\over16}$.
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.
Best Answer
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.