Example 12. In a box, there are 15 red, 9 blue, and 6 green balls. Six balls are drawn at random. What is the probability that 1 green, 2 blue, and 3 red balls are drawn (event $A$)?
Solution. There are a total of 30 balls in the box. In this experiment, the number of all equally possible elementary outcomes will be $C_{30}^{6}$. Let's calculate the number of elementary outcomes favorable to event $A$. Three red balls out of 15 can be chosen in $C_{15}^{3}$ ways, two blue balls out of 9 can be chosen in $C_{9}^{2}$ ways, and one green ball out of 6 can be chosen in $C_{6}^{1}$ ways. Therefore (by the multiplication principle in combinatorics), the number of outcomes favorable to event $A$ will be $m=C_{15}^{3} \cdot C_{9}^{2} \cdot C_{6}^{1}$. Using formula (1.2.1), we find the desired probability

$$
P(A)=\frac{C_{15}^{3} \cdot C_{9}^{2} \cdot C_{6}^{1}}{C_{30}^{6}}=\frac{24!\cdot 15!\cdot 9!\cdot 6!\cdot 6!}{30!\cdot 12!\cdot 7!\cdot 5!\cdot 3!\cdot 2!}=\frac{24}{145} \approx 0.17
$$