If Cheryl gets two marbles of the same color, then Claudia and Carol must take all four marbles of the two other colors. The probability of this happening, given that Cheryl has two marbles of a certain color is $\frac{4}{6} \cdot \frac{3}{5} \cdot \frac{2}{4} \cdot \frac{1}{3} = \frac{1}{15}$. Since there are three different colors, our final probability is $3 \cdot \frac{1}{15} = \textbf{ (C)} \frac{1}{5}$.
A box contains 2 red marbles, 2 green marbles, and 2 yellow marbles. Carol takes 2 marbles from the box at random; then Claudia takes 2 of the remaining marbles at random; and then Cheryl takes the last 2 marbles. What is the probability that Cheryl gets 2 marbles of the same color?
$\textbf{(A)}\ \frac{1}{10} \qquad\textbf{(B)}\ \frac{1}{6} \qquad\textbf{(C)}\ \frac{1}{5} \qquad\textbf{(D)}\ \frac{1}{3} \qquad\textbf{(E)}\ \frac{1}{2}$