20. (12 marks)\\
(I) Let the daily production quantities of products $A$ and $B$ be $x$ and $y$ respectively, with corresponding profit $z$. Then we have

$$\left\{ \begin{array} { l } 
2 x + 1.5 y \leq W \\
x + 1.5 y \leq 12 \\
2 x - y \geq 0 \\
x \geq 0 , \quad y \geq 0
\end{array} \right.$$

The objective function is $z = 1000x + 1200y$.

\begin{figure}[h]
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\caption{Solution diagram 1 for Question 20}
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\caption{Solution diagram 2 for Question 20}
\end{center}
\end{figure}

\begin{figure}[h]
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\caption{Solution diagram 3 for Question 20}
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When $W = 12$, the planar region represented by (1) is shown in Figure 1, with three vertices $A ( 0,0 ) , B ( 2.4,4.8 ) , C ( 6,0 )$ respectively.\\
Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$,\\
when $x = 2.4 , y = 4.8$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept,\\
maximum profit $Z = z _ { \max } = 2.4 \times 1000 + 4.8 \times 1200 = 8160$.\\
When $W = 15$, the planar region represented by (1) is shown in Figure 2, with three vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 7.5,0 )$ respectively.\\
Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$,\\
when $x = 3 , y = 6$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept,\\
maximum profit $Z = z _ { \max } = 3 \times 1000 + 6 \times 1200 = 10200$.\\
When $W = 18$, the planar region represented by (1) is shown in Figure 3,\\
with four vertices $A ( 0,0 ) , B ( 3,6 ) , C ( 6,4 ) , D ( 9,0 )$ respectively.\\
Rewriting $z = 1000 x + 1200 y$ as $y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$,\\
when $x = 6 , y = 4$, the line $l : y = - \frac { 5 } { 6 } x + \frac { z } { 1200 }$ has maximum $y$-intercept,\\
maximum profit $Z = z _ { \max } = 6 \times 1000 + 4 \times 1200 = 10800$.\\
Thus the distribution of maximum profit $Z$ is

\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$Z$ & 8160 & 10200 & 10800 \\
\hline
$P$ & 0.3 & 0.5 & 0.2 \\
\hline
\end{tabular}
\end{center}

Therefore, $E ( Z ) = 8160 \times 0.3 + 10200 \times 0.5 + 10800 \times 0.2 = 9708$.\\
(II) From (I), the probability that daily maximum profit exceeds 10000 yuan is $p _ { 1 } = P ( Z > 10000 ) = 0.5 + 0.2 = 0.7$. By the binomial distribution, the probability that at least one day out of 3 days has maximum profit exceeding 10000 yuan is\\
$p = 1 - \left( 1 - p _ { 1 } \right) ^ { 3 } = 1 - 0.3 ^ { 3 } = 0.973$.\\