Solution. The first distributor can choose 5 out of 10 mailboxes in $\binom{10}{5}$ ways, and the same applies to the second distributor. Therefore, the total number of cases is $\binom{10}{5}^{2}$.

If we want at least 8 mailboxes to have flyers, then the second distributor must place flyers in at least 3 of the empty mailboxes. However, they can also place flyers in 4 or 5 empty mailboxes, as this would still meet the requirement of the problem.

Let's calculate how many good solutions there are in total. The case with 5 additional empty mailboxes can occur in $\binom{10}{5}\binom{5}{5}$ ways. (The first 5 mailboxes are chosen from 10, and the other 5 are chosen from the empty mailboxes.) The case with 4 additional empty mailboxes has $\binom{10}{5}\binom{5}{4}\binom{5}{1}$ possibilities. Finally, the case with 3 additional empty mailboxes can be chosen in $\binom{10}{5}\binom{5}{3}\binom{5}{2}$ ways.

The number of favorable cases:

$$
\binom{10}{5}\left[\binom{5}{5}+\binom{5}{4}\binom{5}{1}+\binom{5}{3}\binom{5}{2}\right]=\binom{10}{5}(1+25+100)=\binom{10}{5} \cdot 126
$$

The desired probability:

$$
P=\frac{\binom{10}{5} \cdot 126}{\binom{10}{5}^{2}}=\frac{126}{\binom{10}{5}}=\frac{126}{252}=\frac{1}{2}
$$
In a stairwell, there are 10 mailboxes. One distributor drops a flyer into 5 mailboxes. Later, another distributor also drops a flyer into 5 mailboxes. What is the probability that this way, at least 8 mailboxes will contain a flyer?