Example 14. In a batch of 12 parts, 8 are standard. Find the probability that among 5 randomly selected parts, 3 will be standard.
Solution. We apply formula (3.6.3). Since in this case $N=12, M=8, n=5, m=3$, then

$$
\begin{gathered}
P(X=3)=\frac{C_{M}^{m} C_{N-M}^{n-m}}{C_{N}^{n}}=\frac{C_{8}^{3} \cdot C_{4}^{2}}{C_{12}^{5}}=\frac{8!}{3!5!} \cdot \frac{4!}{2!2!}: \frac{12!}{5!7!}= \\
=\frac{8!4!5!7!}{3!5!2!2!12!}=\frac{8!4!7!}{3!2!2!12!}=\frac{7!}{9 \cdot 10 \cdot 11 \cdot 12}=\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}{9 \cdot 10 \cdot 11 \cdot 12}=\frac{14}{33}
\end{gathered}
$$

## Problems

1. In a box, there are 10 parts, 7 of which are standard. What is the probability that among 6 parts taken at random, 4 will be standard?
2. According to which law is the random variable $X$ distributed if the probability density function of this variable is defined by the function $p(x)=3 e^{-3 x}$ for $x \geq 0, p(x)=0$ for $x<0$?

For $\delta > 0$, the inequality

$$
P(X \leq \delta) \geq 1-\frac{a}{\delta}
$$

holds. In this case, the inequality

$$
P(X>\delta) < \frac{a}{\delta}
$$

also holds.

For $\varepsilon > 0$, the inequalities

$$
P(|X-\alpha|>\varepsilon)<\frac{D(X)}{\varepsilon^{2}}
$$

$$
P(|X-a| \leq \varepsilon) \geq 1-\frac{D(X)}{\varepsilon^{2}}
$$

are satisfied. The inequality (4.1.3) is called the first Chebyshev's inequality, and the inequality (4.1.4) is called the second Chebyshev's inequality.