9. 6 identical red balls, 3 identical white balls, 3 identical yellow balls are arranged in a straight line, then the probability that balls of the same color are not adjacent is $\qquad$ .
Obviously, the 5 gaps between 6 red balls need to be filled with 5 white or yellow balls, and there are 2 ways to choose the balls, so the number of filling methods is $\frac{2 \cdot 5!}{3!\cdot 2!}$. For the remaining ball, since it cannot be adjacent to a ball of the same color, there are 5 ways to place it.
Thus, the number of arrangements where balls of the same color are not adjacent is $\frac{5 \cdot 2 \cdot 5!}{3!\cdot 2!}$. The total number of arrangements of 12 balls is $\frac{12!}{6!\cdot 3!\cdot 3!}$, so the probability that balls of the same color are not adjacent is $P=\frac{5 \cdot 6!\cdot 6!}{12!}=\frac{5}{924}$.