14. Given a four-person challenge group composed of two male students, Jia and Yi, and two female students, Bing and Ding, participating in a knowledge-based question challenge activity organized by a TV station. The activity consists of four rounds. The probabilities of male students passing the first to fourth rounds are $\frac{5}{6}, \frac{4}{5}, \frac{3}{4}, \frac{2}{3}$, respectively; the probabilities of female students passing the first to fourth rounds are $\frac{4}{5}, \frac{3}{4}, \frac{2}{3}, \frac{1}{2}$, respectively.
(1) Find the probability of male students passing all four rounds;
(2) Let $\xi$ represent the number of people in the four-person challenge group who pass all four rounds. Find the distribution and expectation of the random variable $\xi$.
(1) Let the event that the male student passes all four rounds be denoted as event $A$, then $P(A)=\frac{5}{6} \times \frac{4}{5} \times \frac{3}{4} \times \frac{2}{3}=\frac{1}{3}$;
(2) Let the event that the female student passes all four rounds be denoted as event $B$, then $P(B)=\frac{4}{5} \times \frac{3}{4} \times \frac{2}{3} \times \frac{1}{2}=\frac{1}{5}$, because
$$
\begin{array}{l}
P(\varepsilon=0)=\left(\frac{2}{3}\right)^{2}\left(\frac{4}{5}\right)^{2}=\frac{64}{225} \\
P(\varepsilon=1)=C_{2}^{1} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot\left(\frac{4}{5}\right)^{2}+C_{2}^{1} \cdot \frac{1}{5} \cdot \frac{4}{5} \cdot\left(\frac{2}{3}\right)^{2}=\frac{96}{225} \\
P(\varepsilon=2)=C_{2}^{2} \cdot\left(\frac{1}{3}\right)^{2}\left(\frac{4}{5}\right)^{2}+C_{2}^{2} \cdot\left(\frac{1}{5}\right)^{2}\left(\frac{2}{3}\right)^{2}+C_{2}^{1} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot C_{2}^{1} \cdot \frac{1}{5} \cdot \frac{4}{5}=\frac{52}{225} \\
P(\varepsilon=3)=C_{2}^{1} \cdot \frac{1}{3} \cdot \frac{2}{3}\left(\frac{1}{5}\right)^{2}+C_{2}^{1} \cdot \frac{1}{5} \cdot \frac{4}{5} \cdot\left(\frac{1}{3}\right)^{2}=\frac{12}{225} \\
P(\varepsilon=4)=\left(\frac{1}{3}\right)^{2}\left(\frac{1}{5}\right)^{2}=\frac{1}{225}
\end{array}
$$

Therefore, the distribution is as follows:
$$
E(\varepsilon)=0 \times \frac{64}{225}+1 \times \frac{96}{225}+2 \times \frac{52}{225}+3 \times \frac{12}{225}+4 \times \frac{1}{225}=\frac{240}{225}=\frac{16}{15}
$$