# If $NP \subseteq BPP$ then $NP = RP$. Confusion on the probability that M gives at least one wrong answer in BPP in n invocations

I was looking at the proof of if $$NP \subseteq BPP$$ then $$NP = RP$$ here.

At the end of the proof the author states: "Note that if $$M$$ always gives correct answers on calls to $$M$$, then when $$\phi$$ is satisfiable, $$N$$ (the newly constructed TM) constructs a satisfying assignment to $$\phi$$ and hence accepts. The probability that this happens is at least $$1 − n2^{− \Omega(n)}$$, which is at most 1/2 for large enough n, since the probability that $$M$$ gives at least one wrong answer is at most $$n2^{− \Omega(n)}$$ by the union bound."

I am really confused, why is the probability of getting no wrong answer in n invocations $$1 − n2^{− \Omega(n)}$$? Shouldn't it be $$(2^{− \Omega(n)})^n$$ since the invocations are independent? I mean, we have $$P(X=0)$$ where X = {number of wrong answers} and this is basically a binomial distribution, right? What am I not seeing?