A company wants to popularize their confectionery by hiding prize vouchers in some of the boxes. The management believes that the campaign will be effective and the cost will be manageable if a customer who buys 10 boxes of the product has a probability of about $50 \%$ of finding at least 1 voucher. How many boxes should contain a voucher for this to be true?
Solution. When selecting 10 boxes, we find 0, 1, 2, ..., or 10 gift vouchers in the boxes. Let's call these elementary events and denote them by $A_{0}, A_{1}, A_{2}, \ldots, A_{10}$. Then $A_{i} \cap A_{j} \neq \emptyset(0 \leq i \neq j \leq 10)$ is the impossible event and $A_{0} \cup A_{1} \cup \ldots \cup A_{10}=1$ is the certain event.

Let $n$ be the number of boxes in which the vouchers must be placed so that the requirement of the problem is met. Then the probability that we will find a gift voucher in at least one of the boxes is $p=\frac{1}{n}$, and that we will not find one is $q=1-\frac{1}{n}=\frac{n-1}{n}$.

According to the known formula for the probability of elementary events:

$$
p\left(A_{i}\right)=\binom{10}{i} p^{i} q^{10-i}, \quad i=0,1, \ldots, 10
$$

According to the problem statement, when selecting 10 boxes, this probability must be at least $50 \%$, that is,

$$
p\left(A_{1}\right)+p\left(A_{2}\right)+\ldots+p\left(A_{10}\right)=\sum_{i=1}^{10}\binom{10}{i} p^{i} q^{10-i} \geq \frac{1}{2}
$$

from which

$$
1=\binom{10}{0} p^{0} q^{10}+\sum_{i=1}^{10}\binom{10}{i} p^{i} q^{10-i} \geq\binom{ 10}{0} p^{0} q^{10}+\frac{1}{2}
$$

Hence $q^{10} \leq \frac{1}{2}$. Substituting the $q=\frac{n-1}{n}$ relationship, we get $\left(\frac{n-1}{n}\right)^{10} \leq \frac{1}{2}$. Solving for $n$:

$$
1-\frac{1}{n}=\frac{n-1}{n} \leq \sqrt[10]{\frac{1}{2}}, \quad \text { and } \quad n \leq \frac{\sqrt[10]{2}}{\sqrt[10]{2}-1} \approx 14.93
$$

Therefore, a gift voucher should be placed in approximately every 15th box.