10. A city adopts a lottery system for car license plates, with 300,000 people participating in the lottery, and 30,000 slots available each month.
(1) If those who win the lottery each month exit the lottery, and those who do not win continue to the next month's lottery with an additional 30,000 people joining each month, how long on average will it take for each person to win a slot?
(2) Under the conditions of (1), if the traffic management department can control the proportion of people who win, such that the probability of winning in the first month of each quarter is $\frac{1}{11}$, the second month is $\frac{1}{10}$, and the third month is $\frac{1}{9}$, how long on average will it take for each person to win a slot?
(1) Let $p_{i}$ be the probability that a person gets a number in the $i$-th month, then $p_{i}=\left(1-\frac{1}{10}\right)^{i-1} \cdot \frac{1}{10}=\frac{1}{10} \cdot\left(\frac{9}{10}\right)^{i-1}$.

Thus, $E(X)=\sum_{i=1}^{\infty} i p_{i}=\frac{1}{10} \sum_{i=1}^{\infty} i\left(\frac{9}{10}\right)^{i-1}$,
$$
\frac{9}{10} E(X)=\frac{1}{10} \sum_{i=1}^{\infty} i\left(\frac{9}{10}\right)^{i}=\frac{1}{10} \sum_{i=2}^{\infty}(i-1)\left(\frac{9}{10}\right)^{i-1}=\frac{1}{10} \sum_{i=1}^{\infty}(i-1)\left(\frac{9}{10}\right)^{i-1},
$$

Therefore, $\frac{1}{10} E(X)=\frac{1}{10} \sum_{i=1}^{\infty}\left(\frac{9}{10}\right)^{i-1}=\frac{1}{10} \times \frac{1}{1-\frac{9}{10}}=1 \Rightarrow E(X)=10$.
(2) According to the problem, $P(X=1)=\frac{1}{11}, P(X=2)=\frac{10}{11} \cdot \frac{1}{10}=\frac{1}{11}$,
$$
\begin{array}{l}
P(X=3)=\frac{10}{11} \cdot \frac{9}{10} \cdot \frac{1}{9}=\frac{1}{11}, P(X=4)=\frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \cdot \frac{1}{11}=\frac{8}{11} \cdot \frac{1}{11}, \\
P(X=5)=\frac{8}{11} \cdot \frac{1}{11}, P(X=6)=\frac{8}{11} \cdot \frac{1}{11}, P(X=7)=\left(\frac{8}{11}\right)^{2} \cdot \frac{1}{11}, \cdots,
\end{array}
$$

Thus, $E(X)=\frac{1+2+3}{11}+\frac{4+5+6}{11} \cdot \frac{8}{11}+\frac{7+8+9}{11} \cdot\left(\frac{8}{11}\right)^{2}+\cdots$
$$
\begin{array}{l}
=\frac{1}{11} \sum_{i=1}^{\infty}(9 i-3)\left(\frac{8}{11}\right)^{i-1} \Rightarrow \frac{8}{11} E(X)=\frac{1}{11} \sum_{i=1}^{\infty}(9 i-3)\left(\frac{8}{11}\right)^{i} \\
=\frac{1}{11} \sum_{i=2}^{\infty}(9 i-12)\left(\frac{8}{11}\right)^{i-1}=\frac{1}{11} \sum_{i=1}^{\infty}(9 i-12)\left(\frac{8}{11}\right)^{i-1}+\frac{3}{11},
\end{array}
$$

Therefore, $\frac{3}{11} E(X)=\frac{9}{11} \sum_{i=1}^{\infty}\left(\frac{8}{11}\right)^{i-1}-\frac{3}{11}=\frac{9}{11} \times \frac{1}{1-\frac{8}{11}}-\frac{3}{11}=\frac{30}{11} \Rightarrow E(X)=10$.