TITLE: Is a non-abelian free group fully residually a finite non-abelian simple group?
QUESTION [5 upvotes]: It is well known that a non-abelian free group is residually a finite simple group.  Katz and Magnus proved, in fact, that non-abelian free groups are residually alternating and residually $PSL_{2}$.  S. J. Pride has some nice results along these lines as well.  The best result that I know of is the theorem of Weigel that can be formulated as follows.  If $\mathfrak{X}$ is a group-theoretic class containing an infinite set of pairwise non-isomorphic finite non-abelian simple groups, then every non-abelian free group is residually an $\mathfrak{X}$-group.

My question is this:

Is a non-abelian free group fully residually a finite non-abelian simple group?

It seems likely that the answer to such an obvious question is known, but I have not been able to find it in the literature.
I should probably add that I suspect we can probably replace "finite non-abelian simple" with "alternating", but I haven't yet given any thought to the other infinite series.  I'd like to learn whether anything is known before spending more time on this.

REPLY [4 votes]: Okay I have found an online version of the paper by Dixon, Pyber, Seress, and Shalev at
http://mathstat.carleton.ca/~jdixon/Residual.pdf so I was able to check things. We need the following:
Theorem 3: Let $S$ be a finite simple group and let $w$ be a non-trivial element of the free group $F_2$ on $X,Y$. Then the probability that two randomly chosen elements $x$ and $y$ of $S$ satisfy both that $x$ and $y$ generate $S$ and $w(x,y) \neq 1$ tends to $1$ as $|S|$ tends $\infty$.
Now, clearly replacing $w$ by any finite set of words, the theorem is still true. Therefore, the fully residually case is true.
Now, the pro-$p$ case is more difficult because while the random generation has probability $1$ the not satisfying an identity has only positive probability. But it may be that the proof itself still works for finite number of words.