Show that if $NP\subseteq BPP$ then also $NP=RP$ - considerations about solution 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.
It's true that we can run $A$ even more times, but $O(\log n)$ times suffices for us to obtain a certainty of $1 - 1/(100n)$ (say) for each $x_i$. Applying the union bound, we deduce that the algorithm finds a valid truth assignment with probability at least $1 - 1/100$.

If you run $A$ polynomially many times then you get an exponentially small error, which is not needed here.