Example 8. An urn contains 8 red and 6 blue balls. Three balls are drawn sequentially without replacement from the urn. Find the probability that all 3 balls are blue.
Solution. Let's introduce the notations: $A_{1}$ - "the first ball is blue", $A_{2}$ - "the second ball is blue", $A_{3}$ - "the third ball is blue", $A$ - "all 3 balls are blue", then $A=A_{1} A_{2} A_{3}$. We will use formula (1.8.11), which for $n=3$ takes the form

$$
P\left(A_{1} A_{2} A_{3}\right)=P\left(A_{1}\right) P\left(A_{2} / A_{1}\right) P\left(A_{3} / A_{1} A_{2}\right)
$$

(see formula (1.8.12) in other notations).

Since

$$
P\left(A_{1}\right)=\frac{6}{14}, P\left(A_{2} / A_{1}\right)=\frac{5}{13}, P\left(A_{3} / A_{1} A_{2}\right)=\frac{4}{12}
$$

Then

$$
P(A)=P\left(A_{1} A_{2} A_{3}\right)=\frac{6}{14} \cdot \frac{5}{13} \cdot \frac{4}{12}=\frac{5}{91} \approx 0.055
$$

Remark. This probability can also be found by direct counting using formula (1.2.1).

Since in this case

$$
n=C_{14}^{3}=\frac{14!}{3!\cdot 11!}, m=C_{6}^{3}=\frac{6!}{3!\cdot 3!}, P(A)=\frac{C_{6}^{3}}{C_{14}^{3}}
$$

then

$$
P(A)=\frac{6!}{3!\cdot 3!}: \frac{14!}{3!\cdot 11!}=\frac{6!\cdot 3!\cdot 11!}{14!\cdot 3!\cdot 3!}=\frac{6!\cdot 11!}{14!\cdot 3!}=\frac{4 \cdot 5 \cdot 6}{12 \cdot 13 \cdot 14}=\frac{5}{91}
$$