Example 6.9. In an urn, there are 6 black, 5 red, and 4 white balls. Three balls are drawn sequentially. Find the probability that the first ball will be black, the second - red, and the third - white.
Solution. Consider the following events: $A$ - the first ball drawn is black, $B$ - the second ball is red, $C$ - the third ball is white. Let $D$ be the event that the balls are drawn in the sequence: black, red, white. Obviously, $D=A \cdot B \cdot C$.

By formula (4) we have:

$$
P(D)=P(A) \cdot P(B / A) \cdot P(C / A B) .
$$

We will find the probabilities entering the right-hand side of this equality. The probability that the first ball drawn is black, $P(A)=\frac{6}{15}=\frac{2}{5}$. The probability of drawing a red ball from the urn given that a black ball was initially drawn, $P(B / A)=\frac{5}{14}$, since after removing the black ball, 14 balls remain in the urn, 5 of which are red. The probability of drawing a white ball from the urn after a black and a red ball have been drawn, $P(C / A B)=\frac{4}{13}$ (after removing the black and red balls, 13 balls remain in the urn, 4 of which are white).

Thus,

$$
P(D)=\frac{2}{5} \cdot \frac{5}{14} \cdot \frac{4}{13}=\frac{4}{91} \text {. }
$$