jmc

We consider all the two-element subsets of the six-element set $\{1,2,3,4,5,6\}$. There are $\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$ such subsets. And of these, only the subsets $\{2,4\},\{2,6\},\{3,6\},\{4,6\}$ are not relatively prime. So the probability of the two-element subset's elements having greatest common factor one is $1-\frac{4}{15}=\boxed{\frac{11}{15}}$.