2. A box contains 3 red balls and 3 white balls, all of the same size and shape. Now, a fair die is rolled, and the number of balls taken from the box is equal to the number rolled. What is the probability that the number of red balls taken is greater than the number of white balls taken? $\qquad$ .
Let the number rolled on the die be $X$. When the number of red balls and white balls drawn is equal, we have
$$
P(X=2)=\frac{1}{6} \cdot \frac{C_{3}^{1} \cdot C_{3}^{1}}{C_{6}^{2}}=\frac{1}{10}, P(X=4)=\frac{1}{6} \cdot \frac{C_{3}^{2} \cdot C_{3}^{2}}{C_{6}^{4}}=\frac{1}{10}, P(X=6)=\frac{1}{6}.
$$

When the number of red balls and white balls drawn is not equal, it is clear that the probability of the number of red balls being greater than the number of white balls is equal to the probability of the number of white balls being greater than the number of red balls. Therefore, the probability of the desired event is $\frac{1}{2}\left(1-\frac{1}{5}-\frac{1}{6}\right)=\frac{19}{60}$.