[Combinatorics]
Problem: If $a$ is chosen randomly from the set $\{1,2,3,4,5\}$ and $b$ is chosen randomly from the set $\{6,7,8\}$, what is the probability that $a^{b}$ is an even number?

Solution: Since there are 5 choices for $a$ and 3 choices for $b$, there are fifteen possible ways of choosing $a$ and $b$.

If $a$ is even, $a^{b}$ is even; if $a$ is odd, $a^{b}$ is odd.

So the choices of $a$ and $b$ which give an even value for $a^{b}$ are those where $a$ is even, or 6 of the choices (since there are two even choices for $a$ and three ways of choosing $b$ for each of these). (Notice that in fact the value of $b$ does not affect whether $a^{b}$ is even or odd, so the probability depends only on the choice of $a$.)

Thus, the probability is $\frac{6}{15}=\frac{2}{5}$.

Final answer: $\frac{2}{5}$