12.1.1. In a box, there are 6 identical pairs of black gloves and 4 identical pairs of beige gloves. Find the probability that two randomly drawn gloves form a pair.
Solution. Consider the event $A$ - two randomly drawn gloves form a pair; and hypotheses: $B_{1}$ - a pair of black gloves is drawn, $B_{2}$ - a pair of beige gloves is drawn, $B_{3}$ - the drawn gloves do not form a pair.

The probability of hypothesis $B_{1}$ by the multiplication theorem is the product of the probabilities that the first glove is black and the second glove is black, i.e.,

$$
P\left(B_{1}\right)=P_{1 \text { black }} \cdot P_{2 \text { black }}=\frac{12}{20} \cdot \frac{11}{19}=\frac{33}{95} .
$$

Similarly, the probability of hypothesis $B_{2}$ is:

$$
P\left(B_{2}\right)=P_{1 \text { beige }} \cdot P_{2 \text { beige }}=\frac{8}{20} \cdot \frac{7}{19}=\frac{14}{95} .
$$

Since hypotheses $B_{1}, B_{2}$, and $B_{3}$ form a complete set of events, the probability of hypothesis $B_{3}$ is:

$$
P\left(B_{3}\right)=1-\left(P\left(B_{1}\right)+P\left(B_{2}\right)\right)=1-\frac{33}{95}-\frac{14}{95}=\frac{48}{95}
$$

By the formula of total probability, we have:

$$
P(A)=P\left(B_{1}\right) \cdot P_{B_{1}}(A)+P\left(B_{2}\right) \cdot P_{B_{2}}(A)+P\left(B_{3}\right) \cdot P_{B_{3}}(A),
$$

where $P_{B_{1}}(A)$ is the probability that a pair of black gloves is formed and $P_{B_{1}}(A)=1$; $P_{B_{2}}(A)$ is the probability that a pair of beige gloves is formed and $P_{B_{2}}(A)=1$; and finally, $P_{B_{3}}(A)$ is the probability that a pair is formed by gloves of different colors and $P_{B_{3}}(A)=0$.

Substituting into the formula of total probability:

$$
P(A)=\frac{33}{95} \cdot 1+\frac{14}{95} \cdot 1+\frac{48}{95} \cdot 0=\frac{47}{95}
$$

Thus, the probability that two randomly drawn gloves form a pair is $\frac{47}{95}$.