18. A factory produces 2 large products daily, with a production cost of 2000 yuan per item. The probability of a product being of first-grade quality is 0.5; the probability of it being of second-grade quality is 0.4. The factory price of each first-grade product is 10000 yuan, and that of each second-grade product is 8000 yuan. If the product quality does not meet the standards of first or second grade, in addition to the unrecovered cost, the production of each item will also result in a loss of 1000 yuan.
(1) Find the probability that in 3 consecutive days of production, exactly one day will have both products of first-grade quality;
(2) Given that one of the 2 large products produced by the factory on a certain day is of first-grade quality, find the probability that the other one is also of first-grade quality;
(3) Find the distribution and expectation of the profit $\xi($ yuan $)$ the factory earns daily from producing this product.
(1) Let $A=$ “exactly one day out of three consecutive production days has both products as first-class”, $B=$ “both products produced in a day are first-class”, then $P(B)=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$, so $P(A)=C_{3}^{1} \cdot \frac{1}{4} \cdot\left(\frac{3}{4}\right)^{2}=\frac{27}{64}$.
(2) Let $C=$ “one of the two products produced on a certain day is first-class”, then the required event is $B \mid C$. Since $P(C)=1-\left(\frac{1}{2}\right)^{2}=\frac{3}{4}, P(B C)=P(B)=\frac{1}{4}$, so $P(B \mid C)=\frac{P(B C)}{P(C)}=\frac{1}{3}$.
(3) $\xi=-6000,3000,5000,12000,14000,16000$, and $P(\xi=-6000)=0.1^{2}=0.01$,
$$
\begin{array}{l}
P(\xi=3000)=2 \cdot 0.1 \cdot 0.4=0.08, P(\xi=5000)=2 \cdot 0.1 \cdot 0.5=0.1, \\
P(\xi=12000)=0.4^{2}=0.16, P(\xi=14000)=2 \cdot 0.4 \cdot 0.5=0.4, \\
P(\xi=16000)=0.5^{2}=0.25, \quad \xi \text { has the following distribution: }
\end{array}
$$
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$\xi$ & -6000 & 3000 & 5000 & 12000 & 14000 & 16000 \\
\hline$P$ & 0.01 & 0.08 & 0.1 & 0.16 & 0.4 & 0.25 \\
\hline
\end{tabular}
The mathematical expectation of $\xi$ is $E \xi=-60+240+500+1920+5600+4000=12200$ (yuan).