= Solution
The side of the inscribed <equilateral triangle> has length $\sqrt3\,r$. 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 $d<r/2$.
\b[(a)] Uniform midpoint area puts the midpoint in the radius-$r/2$ <disk> with <probability>
$$
\boxed{\frac{\pi(r/2)^2}{\pi r^2}=\frac14.}
$$
\b[(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
$$
\boxed{\frac{4\pi/3-2\pi/3}{2\pi}=\frac13.}
$$
\b[(c)] If the midpoint distance itself has <uniform distribution> on $[0,r]$, the event $d<r/2$ has <probability>
$$
\boxed{\frac12.}
$$
These 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.
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