Example: Worker A produces 10 parts, 9 of which are qualified and 1 is defective. Worker B produces 10 parts, 8 of which are qualified and 2 are defective. Worker C produces 10 parts, 7 of which are qualified and 3 are defective. One part is randomly selected from the parts produced by A, B, and C. Find: (1) the probability that all are qualified; (2) the probability that exactly 1 is qualified; (3) the probability that exactly 2 are qualified; (4) the probability that at least 1 is qualified; (5) the probability that at most 1 is qualified.
Analysis: (1) Let events A, B, C be set out. The events discussed in the above five questions are respectively: $\mathrm{A} \cdot \mathrm{B} \cdot \mathrm{C}$; $\mathrm{A} \cdot \overline{\mathrm{B}} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}} \cdot \overline{\mathrm{B}} \cdot \mathrm{C}$; $\mathrm{A} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}+\mathrm{A} \cdot \overline{\mathrm{B}} \cdot \mathrm{C}+\overline{\mathrm{A}} \cdot \mathrm{B} \cdot \mathrm{C}$; the opposite event of "at least one qualified product" is $\overline{\mathrm{A}} \cdot \bar{B} \cdot \bar{C}$; the event of "at most 1 qualified product" is $\mathrm{A} \cdot \overline{\mathrm{B}} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}} \cdot \mathrm{B} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}} \cdot \overline{\mathrm{B}} \cdot \mathrm{C}+\overline{\mathrm{A}} \cdot \overline{\mathrm{B}} \cdot \overline{\mathrm{C}}$.
(3) For question (4), the second method of analysis is simpler, and for question (5), the complexity of the two methods of analysis is about the same.
(3) To find $P$ (A), $P$ (B), $P$ (C), use the formula $P$ (A) $=\frac{m}{n}, P(A)=0.9$, $P(B)=0.8, P(C)=0.7$.
(4) The answers to this problem are $0.494 ; 0.092$, $0.398 ; 0.994 ; 0.698$.