35. In an urn, there are 4 white, 6 black, and 5 red balls. Two balls are drawn at random one after the other. Find the probability that both balls are of the same color.
Solution. Consider the events:

$A_{1}$ - the first ball drawn is white;

$B_{1}$ - the second ball drawn is white;

$A_{2}$ - the first ball drawn is black;

$B_{2}$ - the second ball drawn is black;

$A_{3}$ - the first ball drawn is red;

$B_{3}$ - the second ball drawn is red;

$C$ - two balls of the same color are drawn.

Event $C$ represents the sum of the following mutually exclusive events:

$C_{1}$ - two white balls are drawn;

$C_{2}$ - two black balls are drawn;

$C_{3}$ - two red balls are drawn.

Thus, $C=C_{1}+C_{2}+C_{3}$.

Event $C_{1}$ consists of both the first and second balls drawn from the urn being white. This means that event $C_{1}$ is the product of events $A_{1}$ and $B_{1}: C_{1}=A_{1} \cdot B_{1}$. Similarly, we get that $C_{2}=A_{2} \cdot B_{2}$ and $C_{3}=A_{3} \cdot B_{3}$.

The probabilities of events $C_{1}, C_{2}$, and $C_{3}$ will be found using the multiplication rule of probabilities.

Event $B_{1}$ is dependent on event $A_{1}$, as its probability changes when event $A_{1}$ occurs. Using the classical definition of probability, we get that the probability of event $A_{1}$ is $P\left(A_{1}\right)=4 / 15$. The conditional probability of event $B_{1}$, calculated given that event $A_{1}$ has occurred, is $P_{A_{1}}\left(B_{1}\right)=3 / 14$. According to formula (12), we get:

$$
P\left(C_{1}\right)=P\left(A_{1} \cdot B_{1}\right)=P\left(A_{1}\right) \cdot P_{A_{1}}\left(B_{1}\right)=\frac{4}{15} \cdot \frac{3}{14}=\frac{2}{35}
$$

Reasoning similarly, we find

$$
\begin{aligned}
& P\left(C_{2}\right)=P\left(A_{2} \cdot B_{2}\right)=P\left(A_{2}\right) \cdot P_{A_{2}}\left(B_{2}\right)=\frac{6}{15} \cdot \frac{5}{14}=\frac{1}{7} \\
& P\left(C_{3}\right)=P\left(A_{3} \cdot B_{3}\right)=P\left(A_{3}\right) \cdot P_{A_{3}}\left(B_{3}\right)=\frac{5}{15} \cdot \frac{4}{14}=\frac{2}{21} .
\end{aligned}
$$

Having calculated $P\left(C_{1}\right), P\left(C_{2}\right)$, and $P\left(C_{3}\right)$, we find the desired probability $P(C)$ using the addition rule of probabilities for mutually exclusive events:

$$
P(C)=P\left(C_{1}\right)+P\left(C_{2}\right)+P\left(C_{3}\right)=\frac{2}{35}+\frac{1}{7}+\frac{2}{21}=\frac{31}{105} \approx 0.2952
$$