If two numbers are randomly chosen without replacement from $\{1, 2, 3, 4, 5\}$, what is the probability their sum is greater than their product? Express your answer as a common fraction.
Let $a$ and $b$ be positive integers.  Observe that $ab-(a+b)=(a-1)(b-1)-1$.  This quantity is non-negative unless either $a$ or $b$ equals 1.  Therefore, the only pairs of positive integers for which the sum exceeds the product are those of the form $\{1,n\}$ where $n$ is a positive integer.  In this case, there are 4 such pairs: $\{1,2\}$, $\{1,3\}$, $\{1,4\}$, and $\{1,5\}$.  There are a total of $\binom{5}{2}=10$ pairs of integers, so the probability that the sum exceeds the product is $\frac{4}{10}=\boxed{\frac{2}{5}}$.