There is a possibility the the polynomial is not separable, and hence has no Galois group.
The standard example is $F=\mathbb{F}_p(t)$, $f(x)=x^p-t$. Here $f(x)$ is solvable by radicals, as $z=t^{1/p}$ is its only root, of multiplicity $p$. But the splitting field of $f(x)$ is $F(z)$, and that is not a Galois extension of $F$.

If $f(x) \in F[x]$ is a polynomial solvable by radicals, does this imply that its Galois group is solvable when $F$ has characteristic $p > 0$?

  We have a tree with 100 apples. There are 10 red apples and the rest
  are green. Lisa is picking apples at random. When she pick the 3rd red
  apple she stop. What is the probability that Lisa has exactly 7 green
  apples (in addition to the 3 red apples she picked)?

