MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 added 1 characters in body

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 to for finite number of words.

1

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 to finite number of words.