2011 AMC 10A Problem 20

A chord of length $r$ in a circle of radius $r$ subtends a $60^\circ$ arc. Fix the first chord, with endpoints at angles $0^\circ$ and $60^\circ.$ If the second chord starts at angle $\theta,$ its other endpoint is $60^\circ$ clockwise from there. The endpoints of the two chords alternate exactly when $\theta$ lies in either of the two $60^\circ$ arcs immediately adjacent to the fixed chord’s endpoints. Thus the favorable starting positions occupy $120^\circ$ of the circle. The desired probability is then $$\dfrac{2}{6} = \dfrac{1}{3}.$$ Thus, D is the correct answer.