15. On the table, there are two stacks of notebooks. In the first stack, there are 5 notebooks with blue covers, and in the second stack, there are 5 notebooks with red covers. The notebooks in each of these stacks are numbered with the digits $1,2,3,4,5$ and are arranged in a random order of numbers. A student takes one notebook from each stack. Find the probability that the student will take notebooks numbered 5 from both stacks under the mandatory conditions:

a) a notebook №5 with a blue cover will be drawn;

b) at least one notebook №5 will be drawn;

c) two notebooks with the same number will be drawn.
Solution. The experiment consists of extracting two notebooks from stacks. We will construct a fifth-order square matrix characterizing all 25 equally possible outcomes of this experiment, representing a complete group of mutually exclusive events:

$$
D=\left(\begin{array}{lllll}
11 & 12 & 13 & 14 & 15 \\
21 & 22 & 23 & 24 & 25 \\
31 & 32 & 33 & 34 & 35 \\
41 & 42 & 43 & 44 & 45 \\
51 & 52 & 53 & 54 & 55
\end{array}\right)
$$

The elements of matrix $D$ are two-digit numbers. The first digit in each of these numbers is the number of the notebook with a blue cover, and the second is the number of the notebook with a red cover. Consider the events:

$A$ - extracting notebooks numbered 5 from both stacks, given that a notebook numbered 5 with a blue cover will definitely be extracted;

$B$ - extracting notebooks numbered 5 from both stacks, given that at least one notebook numbered 5 will definitely be extracted;

$C$ - extracting notebooks numbered 5 from both stacks, given that two notebooks with the same numbers will definitely be extracted.

The probabilities of events $A, B$, and $C$ will be found using formula (1). The favorable outcome for event $A$ (as well as for events $B$ and $C$) is the outcome represented in matrix $D$ by the element 55; thus, $m=1$.

The total number of possible outcomes for events $A, B$, and $C$ will be different.

a) The possible outcomes of the experiment for event $A$ are represented in matrix $D$ by the elements of its last row: $51,52,53,54,55$. Thus, the total number of all equally possible outcomes forming a complete group of mutually exclusive events is $n=5$.

The probability of event $A$, according to formula (1), is $P(A)=1 / 5$.

b) The possible outcomes of the experiment for event $B$ are represented in matrix $D$ by numbers containing the digit 5: $15,25,35,45,51,52,53,54,55$. The total number of all equally possible mutually exclusive outcomes of the experiment forming a complete group is $n=9$.

The probability of event $B$, according to formula (1), is $P(B)=1 / 9$.

c) The possible outcomes of the experiment for event $C$ are represented in matrix $D$ by numbers consisting of two identical digits: $11,22,33,44,55$. The total number of all equally possible mutually exclusive outcomes of the experiment forming a complete group is $n=5$.

According to formula (1), we get $P(C)=1 / 5$.

## 1.2. Relative Frequency and Statistical Probability

The relative frequency (frequency) of event $A$ is defined by the equation

$$
W(A)=\frac{m}{n}
$$

where $\boldsymbol{n}$ is the total number of trials conducted; $\boldsymbol{m}$ is the number of trials in which event $A$ occurred (or the frequency of event $A$).

In the statistical definition, the probability of an event is taken as its relative frequency, found based on the results of a large number of trials.