4. (7 points) From the set $1,2, \ldots, 10$, three numbers (possibly the same) are chosen with equal probability. What is the probability that the sum of these numbers is 10?
Solution: We need to find how many ways there are to solve the equation

$$
x_{1}+x_{2}+x_{3}=10
$$

where $x_{1}, x_{2}, x_{3} \in\{1,2, \ldots, 10\}$. Let's write down a row of ten ones and place two partitions (in different places) between them. Then $x_{1}$ is the number of ones before the left partition, $x_{2}$ is between the left and right partitions, and $x_{3}$ is after the right partition. Since there are a total of 10 ones, $x_{1}+x_{2}+x_{3}=10$. Note that there are 9 places to place the partitions, and we need to choose only 2. Therefore, the number of solutions to the equation is $C_{9}^{2}=36$. Thus, the final probability is $\frac{36}{10^{3}}=0.036$.

(7 points) Four numbers are chosen equi-probably from the set $\{0,1,2, \ldots, 9\}$ (possibly the same). What is the probability that the sum of these numbers is 9?

Solution: It is solved similarly, but there are complications because $x_{i}$ can be 0. We will associate each solution $x_{1}+x_{2}+x_{3}+x_{4}=9$ with a solution $y_{1}+y_{2}+y_{3}+y_{4}=13$, where $y_{i}=x_{i}+1$ and $y_{1}, y_{2}, y_{3}, y_{4} \in\{1,2, \ldots, 10\}$. The number of solutions to the second equation is exactly $C_{12}^{3}=220$. Therefore, the final probability is $\frac{220}{10^{4}}=0.022$.

## Answers:

| Variant | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | 0.036 | 0.036 | 0.036 | 0.036 | 0.036 | 0.036 | 0.036 | 0.036 | 0.036 | 0.036 | 0.022 |