2005 AMC 12A Problem 14

The die has $21$ dots, so a dot is removed from the face with $n$ dots with probability $\dfrac{n}{21}.$ If a dot is removed from an odd face, that face becomes even, leaving two odd faces and hence probability $\dfrac{2}{6}=\dfrac{1}{3}$ of an odd top. If a dot is removed from an even face, that face becomes odd, leaving four odd faces and hence probability $\dfrac{4}{6}=\dfrac{2}{3}.$ The removed dot lies on an odd face with probability $\dfrac{1 + 3 + 5}{21}$ and on an even face with probability $\dfrac{2 + 4 + 6}{21}.$ Hence the answer is $$\dfrac{1}{3} \cdot \dfrac{9}{21} + \dfrac{2}{3} \cdot \dfrac{12}{21} = \dfrac{33}{63} = \dfrac{11}{21}.$$ Thus, the correct answer is D.