Example 13. In a box, there are 15 balls, of which 5 are blue and 10 are red. Six balls are randomly selected. Find the probability that among the selected balls, 2 are blue.
Solution. The total number of elementary outcomes of the given experiment is equal to the number of combinations of 15 taken 6 at a time, i.e.

$$
C_{15}^{6}=\frac{15!}{6!\cdot 9!}=\frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}=13 \cdot 11 \cdot 7 \cdot 5=5005
$$

The number of favorable outcomes is equal to the product

$$
C_{5}^{2} \cdot C_{10}^{4}=\frac{5!}{2!\cdot 3!} \cdot \frac{10!}{4!\cdot 6!}=\frac{5!\cdot 10!}{2!\cdot 3!\cdot 4!\cdot 6!}=\frac{5 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{1 \cdot 2 \cdot 1 \cdot 2 \cdot 3}=2100
$$

The desired probability is determined by formula (1.3.10):

$$
P=\frac{C_{5}^{2} \cdot C_{10}^{4}}{C_{15}^{6}}=\frac{2100}{5005} \approx 0.4196
$$