10. A bag contains two red balls, three white balls, and four yellow balls, from which four balls are drawn at random. The probability that all three colors are included is $\qquad$
10. $\frac{4}{7}$.

When four balls are drawn from nine balls in a bag, there are $\mathrm{C}_{9}^{4}=126$ different ways to do so, among which the number of ways to draw balls of all three colors is $\mathrm{C}_{2}^{2} \mathrm{C}_{3}^{1} \mathrm{C}_{4}^{1}+\mathrm{C}_{2}^{1} \mathrm{C}_{3}^{2} \mathrm{C}_{4}^{1}+\mathrm{C}_{2}^{1} \mathrm{C}_{3}^{1} \mathrm{C}_{4}^{2}=72$. Therefore, the required probability $P=\frac{72}{126}=\frac{4}{7}$.