46. On a library shelf, 15 textbooks are randomly arranged, with five of them in hardcover. The librarian randomly takes three textbooks. Find the probability that at least one of the taken textbooks will be in hardcover (event $A$).
Solution. First method. The requirement - at least one of the three taken textbooks is in a hardcover - will be fulfilled if any of the following three mutually exclusive events occur: $B$ - one textbook is in a hardcover, $C$ - two textbooks are in a hardcover, $D$ - three textbooks are in a hardcover.

The event of interest $A$ can be represented as the sum of events: $A=B+C+D$. By the addition theorem,

$$
P(A)=P(B)+P(C)+P(D)
$$

We will find the probabilities of events $B, C$, and $D$ (see the solution to problem 17, chapter 1, §1):

$$
\begin{gathered}
P(B)=C_{5}^{1} \cdot C_{10}^{2} / C_{15}^{3}=45 / 91, P(C)=C_{5}^{2} \cdot C_{10}^{1} / C_{15}^{3}=20 / 91 \\
P(D)=C_{5}^{3} / C_{15}^{3}=2 / 91 .
\end{gathered}
$$

Substituting these probabilities into equation (*), we finally get

$$
P(A)=45 / 91+20 / 91+2 / 91=67 / 91
$$

Second method. The events $A$ (at least one of the taken three textbooks has a hardcover) and $\bar{A}$ (none of the taken textbooks have a hardcover) are complementary, so $P(A)+P(\bar{A})=1$ (the sum of the probabilities of two complementary events is one). Therefore, $P(A)=1-P(\bar{A})$.

The probability of the event $\bar{A}$ (none of the taken textbooks have a hardcover)

$$
P(\bar{A})=C_{10}^{8} / C_{13}^{8}=24 / 91
$$

The required probability

$$
P(A)=1-P(\bar{A})=1-24 / 91=67 / 91
$$