Problem 814

Solution. In this trial, there are 36 equally possible elementary outcomes (see Table 1.1). The random variable $X$ can take integer values from 2 to 12, where the values 2 and 12 occur once, 3 and 11 - twice, 4 and 10 - three times, 5 and 9 - four times, 6 and 8 - five times, and the value 7 - six times.

Let's calculate the probabilities of these values:

$$\begin{gathered} p_{1}=P(X=2)=\frac{1}{36}, p_{2}=P(X=3)=\frac{2}{36}=\frac{1}{18}, p_{3}=P(X=4)=\frac{3}{36}=\frac{1}{12} \\ p_{4}=P(X=5)=\frac{4}{36}=\frac{1}{9}, p_{5}=P(X=6)=\frac{5}{36}, p_{6}=P(X=7)=\frac{6}{36}=\frac{1}{6} \\ p_{7}=P(X=8)=\frac{5}{36}, p_{8}=P(X=9)=\frac{4}{36}=\frac{1}{9}, p_{9}=P(X=10)=\frac{3}{36}=\frac{1}{12} \\ p_{10}=P(X=11)=\frac{2}{36}=\frac{1}{18}, p_{11}=P(X=12)=\frac{1}{36} \end{gathered}$$

Therefore, the distribution law of the random variable $X$ can be given by the table

| $X$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | $\frac{1}{36}$ | $\frac{1}{18}$ | $\frac{1}{12}$ | $\frac{1}{9}$ | $\frac{5}{36}$ | $\frac{1}{6}$ | $\frac{5}{36}$ | $\frac{1}{9}$ | $\frac{1}{12}$ | $\frac{1}{18}$ | $\frac{1}{36}$ |

Note that

$$\frac{1}{36}+\frac{1}{18}+\frac{1}{12}+\frac{1}{9}+\frac{5}{36}+\frac{1}{6}+\frac{5}{36}+\frac{1}{9}+\frac{1}{12}+\frac{1}{18}+\frac{1}{36}=1$$

i.e., the equality (2.1.2) is satisfied.

Example 7: In a box, there are 7 pencils, 4 of which are red. Three pencils are randomly drawn from this box. Find the distribution law of the random variable $X$, which is equal to the number of red pencils in the sample.

Solution. In a sample of three pencils, there may be no red pencils, one, two, or three pencils. Therefore, the random variable $X$ can take only four values: $x_{1}=0, x_{2}=1, x_{3}=2, x_{4}=3$.

Let's find the probabilities of these values:

$$\begin{array}{ll} p_{1}=P(X=0)=\frac{C_{4}^{0} \cdot C_{3}^{3}}{C_{7}^{3}}=\frac{1}{35} ; \quad p_{2}=P(X=1)=\frac{C_{4}^{1} \cdot C_{3}^{2}}{C_{7}^{3}}=\frac{12}{35} \\ p_{3}=P(X=2)=\frac{C_{4}^{2} \cdot C_{3}^{1}}{C_{7}^{3}}=\frac{18}{35} ; \quad p_{4}=P(X=3)=\frac{C_{4}^{3} \cdot C_{3}^{0}}{C_{7}^{3}}=\frac{4}{35} \end{array}$$

Therefore, the random variable $X$ has the following distribution law:

| $X$ | 0 | 1 | 2 | 3 |
| :---: | :---: | :---: | :---: | :---: |
| $P^{\cdot}$ | $\frac{1}{35}$ | $\frac{12}{35}$ | $\frac{18}{35}$ | $\frac{4}{35}$ |

Note that $1 / 35+12 / 35+18 / 35+4 / 35=1$, i.e., the equality $(2.1 .2)$ is satisfied.