18. Amelia tosses a coin, with the probability of landing heads up being $\frac{1}{3}$; Brian also tosses a coin, with the probability of landing heads up being $\frac{2}{5}$. Amelia and Brian take turns tossing the coins, and the first one to get heads wins. All coin tosses are independent. Starting with Amelia, the probability that she wins is $\frac{p}{q}\left((p, q)=1, p 、 q \in \mathbf{Z}_{+}\right)$. Then the value of $q-p$ is $(\quad)$.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
18. D.

Let $P_{0}$ be the probability of Amelia winning.
Notice,
$P_{0}=P($ Amelia wins in the first round $)+$ $P($ both fail to win in the first round $) \cdot P_{0}$, where, if both fail to win in the first round, it still starts with Amelia, and her probability of winning remains $P_{0}$.
In the first round, the probability of Amelia winning is $\frac{1}{3}$.
In the first round, the probability of Amelia not winning is $1 - \frac{1}{3} = \frac{2}{3}$, and the probability of Brian not winning is $1 - \frac{2}{5} = \frac{3}{5}$. Therefore, the probability of both not winning is $\frac{2}{3} \times \frac{3}{5} = \frac{2}{5}$.
Thus, $P_{0} = \frac{1}{3} + \frac{2}{5} P_{0} \Rightarrow P_{0} = \frac{5}{9}$.
Therefore, the answer is $9 - 5 = 4$.