2014 AMC 10B Problem 19

Fix the first endpoint $A$ on the outer circle. Draw the two chords from $A$ that are tangent to the inner circle, and call their other endpoints $B$ and $C$. A chord from $A$ meets the inner circle exactly when its second endpoint lies on the minor arc $BC$. If $O$ is the common center and $D$ is a tangency point, then $\triangle AOD$ is right, with $OA=2$ and $OD=1$. Hence $\angle OAD=30^\circ$. The two tangent chords therefore make a $60^\circ$ angle at $A$, so the intercepted minor arc $BC$ measures $120^\circ$. Therefore, the probability is $$\dfrac {120^\circ}{360^\circ} = \dfrac 13 .$$ Thus, the correct answer is D .