170. In a batch of 10 parts, there are 8 standard parts. Two parts are randomly selected. Formulate the distribution law of the number of standard parts among the selected ones *).
S o l u t i o n. The random variable $X$ - the number of standard parts among the selected parts - has the following possible values: $x_{1}=0 ; x_{2}=1 ; x_{3}=2$. We will find the probabilities of the possible values of $X$ using the formula (see problem 17, chapter $1, \S 1$ )

$$
P(X=k)=C_{n}^{k} \cdot C_{N-n}^{m-k} / C_{N}^{m}
$$

( $N$ - the number of parts in the batch, $n$ - the number of standard parts in the batch, $m$ - the number of selected parts, $k$ - the number of standard parts

*) The law under consideration is called hypergeometric. See: G m u r m a n V. E. Probability Theory and Mathematical Statistics. M., 1977, chapter VI, § 8 .
among the selected), we find:

$$
\begin{aligned}
& P(X=0)=\frac{C_{8}^{0} \cdot C_{2}^{2}}{C_{10}^{2}}=\frac{1}{10 \cdot 9 /(1 \cdot 2)}=\frac{1}{45} ; \\
& P(X=1)=\frac{C_{8}^{1} \cdot C_{2}^{1}}{C_{10}^{2}}=\frac{8 \cdot 2}{45}=\frac{16}{45} \\
& P(X=2)=\frac{C_{8}^{2} \cdot C_{2}^{0}}{C_{10}^{2}}=\frac{8 \cdot 7 /(1 \cdot 2)}{45}=\frac{28}{45}
\end{aligned}
$$

We will form the required distribution law:

$$
\begin{array}{cccc}
X & 0 & 1 & 2 \\
p & 1 / 45 & 16 / 45 & 28 / 45
\end{array}
$$

Check: $1 / 45+16 / 45+28 / 45=1$.