Example 10. The probability of producing a non-standard part under certain technological conditions is 0.1. Estimate the probability that the number of non-standard parts among 10000 will be within the range of 950 to 1030 inclusive.
Solution. The number of non-standard parts under the given conditions is a random variable $X$ distributed according to the binomial law. According to formulas (3.2.5.) and (3.2.6.), for $n=10000, p=0.1$, $q=0.9$, we get $M(X)=10000 \cdot 0.1=1000, D(X)=10000 \cdot 0.1 \cdot 0.9=900$. From this, it is clear that the boundaries of the permissible values of the random variable are not symmetric relative to the mathematical expectation (the left boundary is less than it by $1000-950=50$, and the right boundary is greater by $1030-1000=30$). Therefore, it is impossible to apply Chebyshev's inequality to estimate the probability of the specified event. To make the application of Chebyshev's inequality possible, the right boundary should be greater than the mathematical expectation by 50, i.e., it should be equal to 1050. Considering that the double inequality $950 \leq X \leq 1050$ is equivalent to the inequality $|X-1000| \leq 50$, we apply inequality (4.1.4.) for $\varepsilon=50$ and $D(X)=900$:

$$
P(950 \leq X \leq 1050)=P(|X-1000| \leq 50) \geq 1-\frac{900}{50^{2}}=0.64
$$

Remark. Inequality (4.1.4.) provides a non-trivial estimate of the probability of the event $|X-a| \leq \varepsilon$ only in the case when the variance of the random variable $X$ is sufficiently small (less than $\varepsilon^{2}$). Indeed, the left side of inequality (4.1.4.), expressing the probability of the event, is non-negative. This inequality provides an estimate of the probability of the event $|X-a| \leq \varepsilon$ if the right side of it is positive: $1-D(X) / \varepsilon^{2}>0$, from which $D(X) / \varepsilon^{2}<1$. If $D(X) / \varepsilon^{2} \geq 1$, then inequality (4.1.4.) will be written as: $m(|X-a| \leq \varepsilon)>-\alpha$, where $\alpha>0$, which is obvious (the probability of any event is non-negative).

## Problems

1. The probability of event $A$ occurring in each of 100 independent trials is 0.8. Find the probability that the number of occurrences of event $A$ in these 1000 trials deviates from its mathematical expectation by less than 50 in absolute value.
2. The average number of rainy days in a year at a given location is 90. What is the probability that there will be more than 150 rainy days in a year at this location?
3. A random variable is given by the table:

| $X$ | 1 | 2 | 3 | 5 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $P$ | 0.1 | 0.3 | 0.2 | 0.1 | 0.2 | 0.1 |

Using Markov's inequality, estimate the probability that the random variable $X$ will take a value no greater than 7.

4. The probability of event $A$ occurring in each trial is $p=1/2$. Using Chebyshev's inequality, estimate the probability that the number $X$ of occurrences of event $A$ is within the range from 40 to 60, if 100 independent trials are conducted.
5. Using Chebyshev's inequality, estimate the probability that $|X-M(X)| \leq \frac{3}{7}$. 4. $p \geq 0.75$. 5. $P \geq 0.9$.

## Questions

1. What are Markov's inequalities called?
2. What form does the first Chebyshev's inequality take?
3. What form does the second Chebyshev's inequality take?
4. In what case does inequality (4.1.4.) provide a non-trivial estimate of the probability of the event $|X-a| \leq \varepsilon$?

## § 4.2. Chebyshev's Theorem. Bernoulli's Theorem

## Chebyshev's Theorem

If random variables $X_{1}, X_{2}, \ldots, X_{n}$ are independent, have mathematical expectations $M(X)$ and variances $D\left(X_{i}\right)$, bounded by the same number $C$, then for any number $\varepsilon>0$ the following inequality holds:

$$
P\left(\left|\frac{1}{n} \sum_{i=1}^{n} X_{i}-\frac{1}{n} \sum_{i=1}^{n} M\left(X_{i}\right)\right| \leq \varepsilon\right)>1-\frac{C}{n \varepsilon^{2}}
$$

From this, it follows that

$$
\lim _{n \rightarrow \infty} P\left(\left|\frac{1}{n} \sum_{i=1}^{n} X_{i}-\frac{1}{n} \sum_{i=1}^{n} M\left(X_{i}\right)\right| \leq \varepsilon\right)=1
$$

If all random variables $X_{i}(i=1,2,3, \ldots, n)$ have the same mathematical expectation $M\left(X_{i}\right)=a(i=1,2, \ldots, n)$, then inequality $(4.2 .1$.$) takes the form$

$$
P\left(\left|\frac{1}{n} \sum_{i=1}^{n} X_{i}-a\right| \leq \varepsilon\right)>1-\frac{C}{n \varepsilon^{2}}
$$

By taking the limit as $n \rightarrow \infty$, we obtain from this

$$
\lim _{n \rightarrow \infty} P\left(\left|\frac{1}{n} \sum_{i=1}^{n} X_{i}-a\right| \leq \varepsilon\right)=1
$$

## Bernoulli's Theorem

If in each of $n$ independent trials the probability $p$ of event $A$ occurring is constant, then the probability that the deviation of the frequency $\mathrm{m} / \mathrm{n}$ from the probability $p$ in absolute value does not exceed the number $\varepsilon>0$ is greater than the difference $1-p q / n \varepsilon^{2}$, i.e.,

$$
P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right)>1-\frac{p q}{n \varepsilon^{2}}
$$

From this, it follows that

$$
\lim _{n \rightarrow \infty} P\left(\left|\frac{m}{n}-p\right| \leq \varepsilon\right)=1
$$

Remark 1. Bernoulli's theorem confirms the validity of the statistical definition of probability.

Remark 2. This theorem was published in 1713, it laid the foundation for probability theory as a science. The proof by J. Bernoulli was very complex. A simpler proof was proposed by P.L. Chebyshev in 1846.