Task 1. In the production of costumes, the defect rate is $2 \%$. Calculate the probability that when inspecting a batch of 800 costumes, the deviation from the established defect rate will be less than $1 \%$.
Solution. It is necessary to evaluate the probability $p\left(\left|\frac{m}{800}-0.02\right| \leqslant\right.$ $\leqslant 0.01)$, where $m$ is the number of defective suits out of 800, and $\frac{m}{800}$ is the proportion of defective suits. According to Bernoulli's theorem, the required probability is estimated from below by the number

$$
1-\frac{pq}{n \cdot \varepsilon^{2}}=1-\frac{0.02 \cdot 0.98}{0.01^{2} \cdot 800}=1-0.245=0.755
$$

Answer. The required probability is not less than 0.755.

Problem 2. The average yield of a given agricultural crop is 30 centners per hectare. What yield can be expected with a probability of at least 0.75?

Solution. The expected value is known to be $M(X)=30$ (RV $X$ - yield). By the context of the problem, we are dealing with the third Chebyshev's inequality: $p(X \leqslant \varepsilon) \geqslant 0.75$. Therefore, we need to find $\varepsilon$ from the equation $1-30 / \varepsilon=0.75$. We have $\varepsilon-30=0.75 \varepsilon$, from which $\varepsilon=30: 0.25=120$. Thus, we can write the estimate $p(X \leqslant 120) \geqslant$ $\geqslant 0.75$, i.e., the expected yield can be up to 120 centners per hectare.

Problem 3. The amount of water required by an enterprise over a day is a random variable $X$ with an expected value of $200 \mathrm{~m}^{3}$. Find the probability that the water consumption in the next day will exceed $800 \mathrm{~m}^{3}$.

Solution. We will use the first Chebyshev's inequality with $M(X)=200$ and $\varepsilon=800$. The required probability satisfies the inequality $p(X>800) \leqslant \frac{200}{800}=0.25$.

Answer. $p \leqslant 0.25$.