 Solution

The side of the inscribed equilateral triangle has length $\sqrt{3}r$. If $d$ is the distance of a chord midpoint from the center, a right triangle gives chord length $2\sqrt{r^2-d^2}$. Thus a chord is sufficiently long exactly when $d<r/2$.

(a) Uniform midpoint area puts the midpoint in the radius-$r/2$ disk with probability(b) Fix the first endpoint by rotational symmetry. Independence and uniformity make the angle $\theta$ of the second endpoint uniform on $[0,2\pi )$. Its chord length is $2r\sin (\theta /2)$, so the required event is $2\pi /3<\theta <4\pi /3$. Hence(c) If the midpoint distance itself has uniform distribution on $[0,r]$, the event $d<r/2$ has probabilityThese are three different probability measures on chords. Their different answers are Bertrand's paradox: specifying what is uniform is indispensable. Uniform area, uniform endpoints and uniform radial distance are not interchangeable sampling rules. Equality at the length threshold has probability zero in all three models.