5. A bag contains 5 white balls, 4 red balls, and 3 yellow balls. If 4 balls are drawn at random, the probability that balls of all colors are included is
$\qquad$
Let $A=$ "all colors of balls are present", then
$$
P(A)=\frac{C_{5}^{2} \cdot 4 \cdot 3+5 \cdot C_{4}^{2} \cdot 3+5 \cdot 4 \cdot C_{3}^{2}}{C_{12}^{4}}=\frac{24(120+90+60)}{12 \cdot 11 \cdot 10 \cdot 9}=\frac{6}{11} .
$$