11. Given $m \in\{11,13,15,17,19\}, n \in\{2000,2001, \cdots, 2019\}$, the probability that the unit digit of $m^{n}$ is 1 is
Solve:
$11^{n} \equiv 1, 13^{n} \equiv 3,9,7,1, 15^{n} \equiv 5, 17^{n} \equiv 7,9,3,1, 19^{n} \equiv 9,1(\bmod 10)$, that is, the number of $11^{n}$ with a units digit of 1 is 20, the number of $13^{n}, 17^{n}$ with a units digit of 1 is 5 each, the number of $15^{n}$ with a units digit of 1 is 0, and the number of $19^{n}$ with a units digit of 1 is 10, making a total of 40 numbers that satisfy the condition. Let $A=$ “the units digit of $m^{n}$ is 1”, then $P(A)=\frac{40}{100}=\frac{2}{5}$.