Example 3. For what number of independent trials will the probability of the inequality $\left|\frac{m}{n}-p\right|<0.2$ being satisfied exceed 0.96, if the probability of the event occurring in a single trial $p=0.7$?
Solution. According to the problem, we have: $\varepsilon=0.2, p=0.7$, so $q=0.3$; it is required to determine $n$ using inequality (4.2.5). The condition $P>0.96$ is equivalent to the inequality

$$
\frac{p q}{n \varepsilon^{2}} < 0.04
$$

or

$$
n > \frac{p q}{\varepsilon^{2} \cdot 0.04}
$$

Substituting the values $p=0.7, q=0.3$, and $\varepsilon=0.2$ into the last inequality, we find that

$$
n > \frac{0.7 \cdot 0.3}{(0.2)^{2} \cdot 0.04} = \frac{0.21}{0.0016} = 131.25
$$

Therefore, the required inequality is satisfied for the number of independent trials starting from 132.

Example 4. To determine the average yield of a field with an area of 1800 hectares, a sample of $1 \mathrm{~m}^{2}$ was taken from each hectare. It is known that the variance on each hectare of the field does not exceed 6. Estimate the probability that the deviation of the average sample yield from the average yield of the entire field does not exceed 0.25 centners.

Solution. In the right-hand side of inequality (4.2.1.), which defines the required probability, the values are given: $\varepsilon=0.25, C=6$, and $n=1800$. Therefore,

$$
P > 1 - \frac{C}{n \varepsilon^{2}} = 1 - \frac{6}{1800 \cdot 0.0625} = 1 - 0.053 = 0.947
$$

Example 5. The variance of each of the 1000 independent random variables $X_{\kappa} (\kappa=1,2, \ldots, 1000)$ is 4. Estimate the probability that the deviation of the arithmetic mean of these variables from the arithmetic mean of their mathematical expectations in absolute value does not exceed 0.1.

Solution. According to inequality (4.2.1.) with $C=4, \varepsilon=0.1$, we get

$$
P\left(\left|\frac{1}{1000} \sum_{k=1}^{1000} X_{k} - \frac{1}{1000} \sum_{k=1}^{1000} M\left(X_{\kappa}\right)\right| \leq 0.1\right) \geq 1 - \frac{4}{1000 \cdot 0.1^{2}} = 0.6
$$

Thus, $P \geq 0.6$.