Example. Find the probability that among six cards randomly taken from a deck of 36 cards, there will be exactly three black figure cards.
Solution. When the number of equally possible elementary outcomes is finite, the probability of an event is equal to the ratio of the number of elementary outcomes favorable to this event to the total number of elementary outcomes.

1. Define the space of elementary outcomes.

An elementary outcome is an arbitrary set of 6 cards from a deck of 36 cards.

2. Check the equiprobability of elementary outcomes.

In this case, the equiprobability of elementary outcomes is determined by the method of selecting cards. We assume that it guarantees the equiprobability of elementary outcomes.

3. Select those elementary outcomes that lead to the occurrence of the event (favorable outcomes).

The criterion for cards to belong to set $A$ is a black figure, i.e., the ace, king, queen, or jack of spades or clubs. $A$ contains only these cards, and $B$ contains all others.

Favorable outcomes are formed by arbitrarily selecting $m_{1}=3$ cards from set $A$ and arbitrarily selecting $m-m_{1}=$ $=6-3=3$ cards from set $B$.

4. Count the total number of elementary outcomes $N$ (the number of elements in the space of elementary outcomes).

Since an elementary outcome is an arbitrary set of $m=6$ cards and the total number of cards is 36, the number of elementary outcomes is equal to the number of subsets of 6 elements in a set of 36 elements, i.e.,

$$
N=C_{36}^{6}=\frac{36!}{6!(36-6)!}=\frac{31 \cdot 32 \cdot 33 \cdot 34 \cdot 35 \cdot 36}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}=1947792
$$

5. Count the number of elementary outcomes $N\left(m_{1}\right)$ favorable to the given event.

For this, we determine the number of elements in sets $A$ and $B$, i.e., the numbers $n_{A}$ and $n_{B}\left(n_{A}+n_{B}=n\right)$. We have $n_{A}=8$ (ace, king, queen, jack of spades and clubs) and $n_{B}=28\left(n_{A}+n_{B}=8+28=36=n\right)$.

Favorable outcomes are formed by arbitrarily selecting $m_{1}=3$ cards from set $A$, consisting of $n_{A}=8$ cards, and arbitrarily selecting $m-m_{1}=6-3=3$ cards from set $B$, consisting of $n_{B}=28$ cards. Therefore, the number of favorable outcomes is

$$
N(3)=C_{8}^{3} C_{28}^{3}=\frac{8!}{3!5!} \cdot \frac{28!}{3!25!}=183456
$$

6. According to the classical definition of probability

$$
P(3)=\frac{N(3)}{N}=\frac{C_{8}^{3} C_{28}^{3}}{C_{36}^{6}}=\frac{183456}{1947792}=\frac{546}{5797} \approx 0.0942
$$

Answer. $\quad P(3)=C_{8}^{3} C_{28}^{3} / C_{36}^{6} \approx 0.0942$.

Conditions of the Problems.

1. Among $N$ lottery tickets, $M$ are winning. Find the probability that out of $n(n \leqslant N)$ purchased tickets, exactly $m(m \leqslant M)$ tickets will be winning.
2. In the "Sportlotto" game, a participant marks 6 out of 49 sports on a card. Find the probability that he will guess at least three out of the six sports obtained as a result of the draw.
3. Find the probability that among six cards randomly drawn from a deck of 36 cards, there will be exactly two aces.
4. Find the probability that among six cards randomly drawn from a deck of 36 cards, there will be five cards of one color and the sixth of another.
5. In a box, there are 10 white and 15 black balls. Five balls are randomly drawn. Find the probability that among them there will be exactly two white balls.
6. From a complete set of dominoes (28 pieces), seven tiles are randomly selected. Find the probability that among them there will be at least one tile with six dots.
7. From a complete set of dominoes (28 pieces), seven tiles are randomly selected. Find the probability that among them there will be at least two doubles.
8. From a complete set of dominoes (28 pieces), seven tiles are randomly selected. Find the probability that the total number of dots on each tile is less than 7.
9. A student can satisfactorily answer 20 out of 25 tickets. The teacher allows the student to replace one ticket they do not like with another. What is the probability that the student will pass the exam?
10. From a student group of 10 boys and 12 girls, 5 people are randomly selected for a survey. Find the probability that among them there will be at least one girl.

Answers. 1. $C_{M}^{m} C_{N-M}^{n-m} / C_{N}^{n}$. 2. $\left(C_{6}^{3} C_{43}^{3}+C_{6}^{4} C_{43}^{2}+C_{6}^{5} C_{43}^{1}+1\right) / C_{49}^{6}$. 3. $C_{4}^{2} C_{32}^{4} / C_{36}^{6}$. 4. $2 C_{18}^{5} C_{18}^{1} / C_{36}^{6}$. 5. $C_{10}^{2} C_{15}^{3} / C_{25}^{5}$. 6. $1-C_{21}^{7} / C_{28}^{7}$. 7. $1-\left(C_{21}^{7}+C_{7}^{1} C_{21}^{6}\right) / C_{28}^{7}$. 8. $C_{16}^{7} / C_{28}^{7}$. 9. $1-C_{5}^{2} / C_{25}^{2}$. 10. $1-C_{10}^{5} / C_{22}^{5}$.

## 6.3. Arrangement of Balls in Boxes

Problem Statement. There are $r$ balls that are randomly distributed into $n$ boxes, with each box capable of holding all the balls at once. It is required to find the probability of some random event $A$ defined by the distribution of the balls in the boxes.

Plan of Solution. When the number of equally possible elementary outcomes is finite, the probability of an event is equal to the ratio of the number of elementary outcomes favorable to this event to the total number of elementary outcomes.

1. Define the space of elementary outcomes.

Number all the boxes and all the balls. Then an elementary outcome of the random experiment is a set of numbers $\left(i_{1}, i_{2}, \ldots, i_{r}\right)$, where $i_{1}$ is the number of the box in which the first ball landed, $i_{2}$ is the number of the box in which the second ball landed, and so on. Clearly, each of the numbers $i_{1}, i_{2}, \ldots, i_{r}$ can take any natural value from 1 to $n$.

2. Check the equiprobability of elementary outcomes.

It is assumed that it is guaranteed by the method of conducting the random experiment, i.e., the distribution of balls into boxes.

3. Select those elementary outcomes that lead to the occurrence of event $A$ (favorable outcomes).

The criterion for selection is the description of event $A$.

4. Count the total number of elementary outcomes $N$ (the number of elements in the space of elementary outcomes).

The number of elementary outcomes is $n^{r}$, since each of the $r$ positions in the set $\left(i_{1}, i_{2}, \ldots, i_{r}\right)$ can be any of the $n$ numbers $1,2, \ldots, n$ and these positions are filled independently of each other.

5. Count the number of elementary outcomes $N_{A}$ favorable to event $A$.

This is done directly or using combinatorics.

6. According to the classical definition of probability

$$
\mathrm{P}(A)=\frac{N_{A}}{N} \text {. }
$$