Example 12. In each of the three boxes, there are 30 parts. In the first box, there are 27, in the second 28, and in the third 25 standard parts. One part is randomly taken from each box. What is the probability that all three parts taken will be standard.
Solution. The probability that a standard part is drawn from the first box (event $A$), $P(A)=27 / 30=9 / 10$. The probability that a standard part is drawn from the second box (event $B$) $P(B)=28 / 30=14 / 15$. The probability that a standard part is drawn from the third box (event $C$) $P(C)=25 / 30=5 / 6$. Since events $A, B, C$ are independent, we obtain by formula (1.8.14)

$$
P(A B C)=P(A) P(B) P(C)=\frac{9}{10} \cdot \frac{14}{15} \cdot \frac{5}{6}=0.7
$$