Solution
= Solution
The <sample space> outcomes with qualifying sums have totals three, four and five. Their ordered patterns are $(1,1,1)$, the three permutations of $(1,1,2)$, and the three permutations each of $(1,1,3)$ and $(1,2,2)$. Thus there are $1+3+3+3=10$ qualifying outcomes:
$$
\boxed{P(F)=\frac{10}{216}=\frac5{108}.}
$$