19. (12 points) A company decides whether to invest in a project through voting by three people: A, B, and C. Each of them has one "agree," one "neutral," and one "disagree" vote. When voting, each person must and can only cast one vote, and the probability of each person casting any of the three types of votes is $\frac{1}{3}$, with no influence between the three people's votes. The rule is: if the voting result includes at least two "agree" votes, the company will decide to invest in the project; otherwise, the project will be abandoned. Find:
(1) The probability that the company decides to invest in the project;
(2) The probability that the company abandons the project and the voting result includes at most one "neutral" vote.
19. (1) The probability that the company decides to invest in this project is
$$
P=\mathrm{C}_{3}^{2}\left(\frac{1}{3}\right)^{2}\left(\frac{2}{3}\right)+\mathrm{C}_{3}^{3}\left(\frac{1}{3}\right)^{3}=\frac{7}{27} .
$$
(2) The company decides to abandon the investment in this project and the voting results have at most one neutral vote, as shown in Table 1.
Table 1
\begin{tabular}{|c|c|c|c|}
\hline Event & Number of Agree Votes & Number of Neutral Votes & Number of Disagree Votes \\
\hline$A$ & 0 & 0 & 3 \\
\hline$B$ & 1 & 0 & 2 \\
\hline$C$ & 1 & 1 & 1 \\
\hline$D$ & 0 & 1 & 2 \\
\hline
\end{tabular}

Then $P(\Lambda)=\mathrm{C}_{3}^{3}\left(\frac{1}{3}\right)^{3}=\frac{1}{27}$,
$P(B)=\mathrm{C}_{3}^{1}\left(\frac{1}{3}\right)^{3}=\frac{3}{27}$,
$P(C)=\mathrm{C}_{3}^{1} \mathrm{C}_{2}^{1}\left(\frac{1}{3}\right)^{3}=\frac{6}{27}$,
$P(D)=\mathrm{C}_{3}^{1}\left(\frac{1}{3}\right)^{3}=\frac{3}{27}$.
Since events $A, B, C, D$ are mutually exclusive,
$$
\begin{array}{l}
P(A+B+C+D) \\
=P(A)+P(B)+P(C)+P(D) \\
=\frac{13}{27} .
\end{array}
$$