19. Given $m \in\{11,13,15,17,19\}$, $n \in\{1999,2000, \cdots, 2018\}$.
Then the probability that the unit digit of $m^{n}$ is 1 is ( ).
(A) $\frac{1}{5}$
(B) $\frac{1}{4}$
(C) $\frac{3}{10}$
(D) $\frac{7}{20}$
(E) $\frac{2}{5}$
19. E.

Consider different cases.
When $m=11$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 20 ways to choose;

When $m=13$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose;

When $m=15$, $n \in\{1999,2000, \cdots, 2018\}$, the unit digit of $m^{n}$ is 5;

When $m=17$, $n \in\{2000,2004, \cdots, 2016\}$, the unit digit of $m^{n}$ is 1, there are 5 ways to choose;

When $m=19$, $n \in\{2000,2002, \cdots, 2018\}$, the unit digit of $m^{n}$ is 1, there are 10 ways to choose.
In summary, the required result is $\frac{20+5+5+10}{5 \times 20}=\frac{2}{5}$.