26. (2004 National College Entrance Examination Hunan Paper) Three machines, A, B, and C, independently process the same type of part. It is known that the part processed by machine A is a first-class part, while the probability that the part processed by machine B is not a first-class part is $\frac{1}{4}$, the probability that the part processed by machine B is a first-class part and the part processed by machine C is not a first-class part is $\frac{1}{12}$, and the probability that the parts processed by machines A and C are first-class parts is $\frac{2}{9}$.
（）Find the probability that the parts processed by machines A, B, and C are first-class parts, respectively;
（II) From the parts processed by machines A, B, and C, one part is taken for inspection. Find the probability that at least one of the parts is a first-class part.
26. ( I ) Let $A$, $B$, and $C$ be the events that the parts processed by machines A, B, and C, respectively, are of first quality. From the given conditions, we have
$$
\left\{\begin{array}{l}
P(A \cdot \overline{B}) = \frac{1}{4}, \\
P(B \cdot \overline{C}) = \frac{1}{12}, \\
P(A \cdot C) = \frac{2}{9}
\end{array} \quad \text{which is equivalent to} \left\{\begin{array}{l}
P(A) \cdot [1 - P(B)] = \frac{1}{4}, \\
P(B) \cdot [1 - P(C)] = \frac{1}{12}, \\
P(A) \cdot P(C) = \frac{2}{9}.
\end{array}\right.\right.
$$

From (1) and (3), we get $P(B) = 1 - \frac{9}{8} P(C)$. Substituting this into (2) yields
$$
27[P(C)]^2 - 51 P(C) + 22 = 0.
$$

Solving this, we get $P(C) = \frac{2}{3}$ or $\frac{11}{9}$ (the latter is discarded). Substituting $P(C) = \frac{2}{3}$ into (3) and (2) respectively, we get $P(A) = \frac{1}{3}$ and $P(B) = \frac{1}{4}$. Therefore, the required results are $\frac{1}{3}, \frac{1}{4}, \frac{2}{3}$.
(II) Let $D$ be the event that at least one first-quality part is found when one part is inspected from each of the parts processed by machines A, B, and C. Then $P(D) = 1 - P(\bar{D}) = 1 - [1 - P(A)][1 - P(B)][1 - P(C)] = 1 - \frac{2}{3} \cdot \frac{3}{4} \cdot \frac{1}{3} = \frac{5}{6}$, or $P(D) = P(A \cdot \bar{B} \cdot \bar{C}) + P(\bar{A} \cdot B \cdot C) + P(\bar{A} \cdot \bar{B} \cdot C) + P(A \cdot B \cdot \bar{C}) + P(A \cdot \bar{B} \cdot C) + P(\bar{A} \cdot B \cdot C) + P(A \cdot B \cdot C) = \frac{5}{6}$.