18. (12 points) A factory produces two large products daily, with an input cost of 2000 yuan per product. The probability of a product being of first-class quality is 0.5, and the probability of it being of second-class quality is 0.4. The factory price of each first-class product is 10000 yuan, and the factory price of each second-class product is 8000 yuan. If the product quality does not meet the standards of first-class or second-class, in addition to the unrecovered cost, the production of each product will also result in a loss of 1000 yuan.
(1) Find the probability that in three consecutive days of production, exactly one day has both products of first-class quality;
(2) Given that one of the two products produced on a certain day is of first-class quality, find the probability that the other product is also of first-class quality;
(3) Find the distribution and expectation of the daily profit $\xi$ (yuan) from producing this product.
18. (1) The probability that both items in a day are first-class is $\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$.

Let the event that exactly one day out of three consecutive days produces two first-class items be event $A$.
Then $P(A)=\mathrm{C}_{3}^{1} \times \frac{1}{4}\left(\frac{3}{4}\right)^{2}=\frac{27}{64}$.
(2) The probability that one of the two items is a first-class item (event $B$) is $\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}+\frac{1}{2} \times \frac{1}{2}=\frac{3}{4}$,
the probability that both items are first-class is $\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$.
Given that one item is a first-class item, the probability that the other is also a first-class item is
$$
P(B \mid A)=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3} \text {. }
$$
(3) The possible values of $\xi$ are $16000$, $14000$, $5000$, $12000$, $3000$, and $-6000$.
Then $P(\xi=16000)=0.5^{2}=0.25$,
$$
\begin{array}{l}
P(\xi=14000)=\mathrm{C}_{2}^{1} \times 0.5 \times 0.4=0.4, \\
P(\xi=5000)=\mathrm{C}_{2}^{1} \times 0.5 \times 0.1=0.1, \\
P(\xi=12000)=0.4^{2}=0.16 \\
P(\xi=3000)=\mathrm{C}_{2}^{1} \times 0.1 \times 0.4=0.08, \\
P(\xi=-6000)=0.1^{2}=0.01 .
\end{array}
$$

Therefore, the distribution of $\xi$ is as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$\xi$ & 16000 & 14000 & 5000 & 12000 & 3000 & -6000 \\
\hline$P$ & 0.25 & 0.4 & 0.1 & 0.16 & 0.08 & 0.01 \\
\hline
\end{tabular}
$$
\begin{array}{l} 
\text { Hence } \mathrm{E} \xi= 16000 \times 0.25+14000 \times 0.4+ \\
5000 \times 0.1+12000 \times 0.16+ \\
3000 \times 0.08+(-6000) \times 0.01 \\
=12200 .
\end{array}
$$