Example 11. Among 25 students in a group, where 10 are girls, 5 tickets are being raffled. Find the probability that 2 of the ticket winners will be girls.
Solution. The number of all equally possible cases of distributing 5 tickets among 25 students is equal to the number of combinations of 25 elements taken 5 at a time, i.e., $C_{25}^{5}$. The number of groups of three boys out of 15 who can receive tickets is $C_{15}^{3}$. Each such trio can be combined with any pair of ten girls, and the number of such pairs is $C_{10}^{2}$. Therefore, the number of groups of 5 students formed from a group of 25 students, each of which includes three boys and two girls, is the product $C_{15}^{3} \cdot C_{10}^{2}$. This product is equal to the number of favorable cases for distributing five tickets among the students of the group so that three tickets go to boys and two tickets to girls.

According to formula (1.2.1), we find the desired probability

$$
\begin{aligned}
& P(A)=\frac{C_{15}^{3} \cdot C_{10}^{2}}{C_{25}^{5}}=\frac{15!}{3! \cdot 12!} \cdot \frac{10!}{2! \cdot 8!} : \frac{25!}{5! \cdot 20!}=\frac{20! \cdot 15! \cdot 10! \cdot 5!}{25! \cdot 12! \cdot 8! \cdot 3! \cdot 2!}= \\
& =\frac{13 \cdot 14 \cdot 15 \cdot 9 \cdot 10 \cdot 4 \cdot 5}{25 \cdot 24 \cdot 23 \cdot 22 \cdot 21 \cdot 2}=\frac{13 \cdot 5 \cdot 3}{23 \cdot 22}=\frac{195}{506} \approx 0.385
\end{aligned}
$$

Remark. The last formula is a special case of formula (1.3.10): $N=25, M=15; n=5, m=3$.