Example 17 A worker takes care of three machine tools. In one hour, the probabilities that machine tools A, B, and C need attention are $0.9$, $0.8$, and $0.85$, respectively. In one hour, find:
(1) the probability that no machine tool needs attention;
(2) the probability that at least one machine tool needs attention.
Let $A$, $B$, and $C$ represent the need for attention for machines A, B, and C, respectively. Then, $\bar{A}$, $\bar{B}$, and $\bar{C}$ represent the lack of need for attention for machines A, B, and C, respectively. According to the problem, $P(A)=0.9$, $P(B)=0.8$, $P(C)=0.85$, and $P(\bar{A})=0.1$, $P(\bar{B})=0.2$, $P(\bar{C})=0.15$.
(1) None of the machines A, B, and C need attention, which means $\bar{A} \cdot \bar{B} \cdot \bar{C}$. Since $\bar{A}$, $\bar{B}$, and $\bar{C}$ are independent, we have $P(\bar{A} \cdot \bar{B} \cdot \bar{C})=P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})=0.1 \times 0.2 \times 0.15=0.003$.
(2) Let $D$ represent the event that at least one of the machines A, B, and C needs attention. Then, $D=(A \cdot \bar{B} \cdot \bar{C}) \cup (\bar{A} \cdot B \cdot \bar{C}) \cup (\bar{A} \cdot \bar{B} \cdot C) \cup (A \cdot B \cdot \bar{C}) \cup (A \cdot \bar{B} \cdot C) \cup (\bar{A} \cdot B \cdot C) \cup (A \cdot B \cdot C)$. Since $(A \cdot \bar{B} \cdot \bar{C})$, $(\bar{A} \cdot B \cdot \bar{C})$, $(\bar{A} \cdot \bar{B} \cdot C)$, $(A \cdot B \cdot \bar{C})$, $(A \cdot \bar{B} \cdot C)$, $(\bar{A} \cdot B \cdot C)$, and $(A \cdot B \cdot C)$ are mutually exclusive, and $A$, $B$, $C$, $\bar{A}$, $\bar{B}$, and $\bar{C}$ are independent, the probability that at least one machine needs attention is:
$$
\begin{array}{l}
P(D)=P(A \cdot \bar{B} \cdot \bar{C})+P(\bar{A} \cdot B \cdot \bar{C})+P(\bar{A} \cdot \bar{B} \cdot C)+P(A \cdot B \cdot \bar{C})+P(A \cdot \\
\quad \bar{B} \cdot C)+P(\bar{A} \cdot B \cdot C)+P(A \cdot B \cdot C) \\
\quad=P(A) \cdot P(\bar{B}) \cdot P(\bar{C})+P(\bar{A}) \cdot P(B) \cdot P(\bar{C})+P(\bar{A}) \cdot P(\bar{B}) \cdot P(C)+P(A) \cdot \\
\quad P(B) \cdot P(\bar{C})+P(A) \cdot P(\bar{B}) \cdot P(C)+P(\bar{A}) \cdot P(B) \cdot P(C)+P(A) \cdot P(B) \cdot P(C) \\
\quad=0.9 \times 0.2 \times 0.15+0.1 \times 0.8 \times 0.15+0.1 \times 0.2 \times 0.85+0.9 \times 0.8 \times \\
\quad 0.15+0.9 \times 0.2 \times 0.85+0.1 \times 0.8 \times 0.85+0.9 \times 0.8 \times 0.85 \\
\quad=0.997
\end{array}
$$