Show that if $$NP ⊆ BPP$$ then $$NP = RP$$
Solution: It is enough to show that if $NP ⊆ BPP$ then $3SAT∈ RP$. Let $A$ be a $BPP$ algorithm for $3SAT$. Given a formula $ϕ$ with n variables, we first run $A$ on $ϕ$. If $A$ rejects, we reject. Otherwise, we try to construct a satisfying assignment for $ϕ$ one variable at a time. That is, we try instantiating $x_1$ to $0$, and then use $A$ to decide if the resulting formula is satisfiable: if so, then we permanently set $x_1$ to $0$ and proceed with $x_2$; otherwise we set $x_1$ to $1$ and proceed with $x_2$. If we manage to construct a satisfying assignment we accept, otherwise we reject, and so on. If $ϕ$ is unsatisfiable, then we always reject. If $ϕ$ is satisfiable, then we construct a satisfying assignment and accept provided that the $n + 1$ invocations of $A$ that we make are all correct. We can ensure that this happens with high probability by replacing each invocation of $A$ with $O(\log n)$ independent ones and taking the majority answer
It is fine solution from https://people.eecs.berkeley.edu/~luca/cs278-04/notes/sol.pdf
My only question is: Why $O(\log n)$ ? After all, we can run it polynomially to make error probablity exponetially small.