81. A farmer has 15 cows, 5 of which produce more than 4500 liters of milk per year. Three cows are randomly selected from those owned by the farmer. Find the distribution law of the random variable $X$ representing the number of cows with the specified high milk yield among the selected ones.
Solution. The random variable $X$ is the number of cows among the selected ones that produce more than 4500 liters of milk per year. This variable can take one of the following possible values as a result of the trial: $x_{1}=0 ; x_{2}=1 ; x_{3}=2 ; x_{4}=3$. The probabilities of the possible values of the variable $X$ will be found using the formula

$$
P(X=r)=\frac{C_{k}^{r} \cdot C_{N-k}^{l-r}}{C_{N}^{l}}
$$

(see problem 9).

In this problem, $N=15$ is the total number of cows the farmer has, $k=5$ is the total number of cows with the specified high milk yield, $l=3$ is the number of selected cows, and $r$ is the number of cows among the three selected that have the specified high milk yield. The sought probabilities of the possible values of $X$ are:

$$
\begin{aligned}
& p_{1}=P(X=0)=\frac{C_{5}^{0} \cdot C_{10}^{3}}{C_{15}^{3}}=\frac{24}{91} \\
& p_{2}=P(X=1)=\frac{C_{5}^{1} \cdot C_{10}^{2}}{C_{15}^{3}}=\frac{45}{91} \\
& p_{3}=P(X=2)=\frac{C_{5}^{2} \cdot C_{10}^{1}}{C_{15}^{3}}=\frac{20}{91} \\
& p_{4}=P(X=3)=\frac{C_{5}^{3} \cdot C_{10}^{0}}{C_{15}^{3}}=\frac{2}{91}
\end{aligned}
$$

This distribution law is called hypergeometric.

The random variable $X$ under consideration has the following distribution:

$$
X: \begin{array}{ccccc}
x_{i} & 0 & 1 & 2 & 3 \\
p_{i} & 24 / 91 & 45 / 91 & 20 / 91 & 2 / 91^{\circ}
\end{array}
$$

Verification: $24 / 91 + 45 / 91 + 20 / 91 + 2 / 91 = 1$.