Bertini's theorem in char p for base point free linear system I always believed the following statement: if $X$ is a smooth variety over an algebraically closed field of positive characteristic, assuming we know that the general member of a base point free linear system $|L|$ is reduced, then indeed a general member is smooth.
However, I realize this is not obvious, though all the examples I know which fail this Bertini theorem has non-reduced fibers.
So I was wondering whether indeed this statement is true. Or counterexamples are known. 
[REEDIT:] After Laurent Moret-Bailly's nice counterexample. I was then wondering whether a counterexample exists when the linear system induces a birational morphism to its image. The assumption now is closer to the one in Bertini's theorem which says the statement is true when we assume the morphism is an embedding.
This sort of counterexamples will always help birational geometers to think what's going on in characteristic $p$.
 A: [Added at the end: when I started this response I thought I had a complete answer, but as I started typing I realized that I don't. Yet, I think that this might still be a useful observation, so I will post it.]
The only place where $\mathrm{char}=0$ is used is to conclude that the family $f:W\to G$ of hyperplane sections of $X$ mapping to $G=\mathrm{PGL}_n$ is generically smooth. If you know that the general fiber is reduced, then it implies that the general fiber is generically smooth. 
In order to get Bertini, we need that $f$ is generically smooth on $G$. 
In characteristic $0$ one can prove this by applying the above observation, but not for that family!
Define $W_0=\{\text{closed points }x\in W \ |\  T_{W,x}\to T_{G,f(x)}\text{ is not surjective} \}$, let $W_1=\overline W_0$ with the reduced induced scheme structure and consider the induced $f_1:W_1\to G$. 
If $f_1$ is not surjective, we are done. If it is, replace $W_1$ with an irreducible component on which $f_1$ is surjective. Now it follows that this $f_1$ cannot be generically smooth, because its tangent map is not surjective anywhere. 
In other words, in order to get this more general Bertini you would have to know a sort of second order reducedness: You need that if you take the family of singular loci of all the hyperplane sections of $X$, take the reduced induced scheme structure on the family, then the general member of that family will be reduced. 
I suppose the next obvious question is if someone can show an example that this does not hold automatically if your assumption holds or prove that such an example does not exist, so your desired statement is true. I am leaning towards the first possibility, but I don't know such an example off the top of my head. 
A: In characteristic 3, consider the surface $V\subset \mathbb{P}^2\times\mathbb{A}^1$ with equation $y^2z=x^3-tz^3$ (where $t$ is the coordinate on $\mathbb{A}^1$). It is easily seen to be smooth. The general fiber of the projection on $\mathbb{A}^1$ is a plane cubic which is reduced but not smooth. Taking a suitable projective completion $V\hookrightarrow X$ you get a morphism $X\to\mathbb{P}^1$ which is a counterexample.
A: Since there's already an example posted, I'll point out that if your linear system gives a map $X \to \mathbb{P^n}$ with separable residue field extensions (at ALL points), then you are ok 
EDIT: (of course, this is a very rare condition as pointed out in the comments below...):
Also see the question: of mine.  That question has some links to some positive statements.
