19 Given that there are 4 genuine items and 3 defective items among 7 items.
(1) If 3 items are randomly drawn from these 7 items at once, find the probability that the number of genuine items is not less than the number of defective items;
(2) If 5 items are randomly drawn from these 7 items at once, let the number of defective items be $\xi$, find the mathematical expectation of $\xi$.
19 (1) The number of results from randomly drawing 3 items from these 7 products at once is $\mathrm{C}_{7}^{3}$, where the number of results with the number of good items not less than the number of defective items is $\mathrm{C}_{4}^{2} \cdot \mathrm{C}_{3}^{1}+\mathrm{C}_{4}^{3}$. Therefore, the probability of the desired random event is
$$
P=\frac{\mathrm{C}_{4}^{2} \cdot \mathrm{C}_{3}^{1}+\mathrm{C}_{4}^{3}}{\mathrm{C}_{7}^{3}}=\frac{22}{35}
$$

Thus, the probability that the number of good items is not less than the number of defective items is $\frac{22}{35}$.
(2) All possible values of the number of defective items $\xi$ are $1, 2, 3$, and
$$
\begin{array}{l}
P(\xi=1)=\frac{\mathrm{C}_{4}^{4} \cdot \mathrm{C}_{3}^{1}}{\mathrm{C}_{7}^{5}}=\frac{3}{21}=\frac{1}{7}, \\
P(\xi=2)=\frac{\mathrm{C}_{4}^{3} \cdot \mathrm{C}_{3}^{2}}{\mathrm{C}_{7}^{5}}=\frac{12}{21}=\frac{4}{7}, \\
P(\xi=3)=\frac{\mathrm{C}_{4}^{2} \cdot \mathrm{C}_{3}^{3}}{\mathrm{C}_{7}^{5}}=\frac{6}{21}=\frac{2}{7} .
\end{array}
$$

Therefore, the mathematical expectation of the random variable $\xi$ is $E \xi=1 \times \frac{1}{7}+2 \times \frac{4}{7}+3 \times \frac{2}{7}=\frac{15}{7}$.