Problem 1. In an urn, there are two red, five blue, and three white balls. Find the probability that two balls drawn simultaneously will be blue (event $A$).
Solution. There are a total of 10 balls in the urn. The total number of all elementary outcomes is equal to the number of ways to draw 2 balls out of 10: $n=C_{10}^{2}=\frac{10 \cdot 9}{1 \cdot 2}=45$. The number of outcomes favorable to event $A$ is equal to the number of ways to draw 2 balls out of 5, i.e., $m=C_{5}^{2}=10$. Thus, according to formula (1), we have $p(A)=\frac{10}{45}=\frac{2}{9}$.

Answer. $p(A) \approx 0.222$.

Problem 2. There are 5 balls numbered from 1 to 5 in the urn. Find the probability that the number of the ball drawn at random does not exceed 5 (event $A$).

Solution. The following outcomes are possible: $E_{1}$ - a ball with number 1 is drawn, $E_{2}$ - a ball with number 2, and so on. The number of all possible outcomes is equal to the number of balls, i.e., five. Since the number of any ball does not exceed 5, the number of outcomes favorable to event $A$ is equal to the number of all outcomes, i.e., $m=n=5$ and the desired probability is $p(A)=1$ (thus, event $A$ is certain).

Problem 3. A part taken at random can be of the first grade (event $A$), second grade (event $B$), or third grade (event $C$). What do the following events represent: $A+B$, $\overline{A+C}$, $A C$, $A B+C$?

Solution (see points $2^{\circ}, 3^{\circ}$). $A+B$ is the event that will occur if at least one of the events $A$ or $B$ occurs; in our case, it is a part of the first or second grade.

$\overline{A+C}$ - this is the event opposite to the event $A+C$ (a part of the first or third grade), meaning a part of the second grade is drawn.

$A C$ - this is an impossible event, as a part cannot be both of the first and third grade at the same time. $A B+C$ - the sum of an impossible event and event $C$, equal to $C$, i.e., this is a part of the third grade.