Solution:

Since we only care about the units digit, we can turn the set $$\{11,13,15,17,19\}$$ into $$\{1,3,5,7,9\}.$$ Then we can case on the value of $m.$

$m = 1$

Any value of $n$ works. This occurs with a $\frac{1}{5}$ probability.

$m = 3$

Looking at powers of $3,$ we see that this sequence of units digits repeats: $$3, 9, 7, 1, \ldots$$

This means that $n$ must be a multiple of $4.$ There are $5$ such values. This means that $n$ works $\frac{5}{20} = \frac{1}{4}$ of the time. The total probability is $$\dfrac{1}{5} \cdot \dfrac{1}{4} = \dfrac{1}{20}.$$

$m = 5$

Powers of $5$ always end in $5,$ which means that this case will never work.

$m = 7$

The units digits repeat in this pattern: $$7, 9, 3, 1,\ldots$$ This means that $n$ must be a multiple of $4$ to work. As when $m = 3,$ this case works with a probability of $\frac{1}{20}.$

$m = 9$

The units digit alternates between $1$ and $9.$ This means that $n$ has to be even. This happens with a $\frac{1}{2}$ chance. The total probability is then $$\dfrac{1}{5} \cdot \dfrac{1}{2} = \dfrac{1}{10}.$$

The total probability is therefore $$\dfrac{1}{5} + 2 \cdot \dfrac{1}{20} + \dfrac{1}{10} = \dfrac{2}{5}.$$

Thus, E is the correct answer.