I guess the answer is that this unknown.
Maybe this implies some "lowness" result on NP relative to BPP?



Does EXP $\in$ P/poly imply NP=RP?
Note that it would suffice to prove $EXP \subseteq P/poly$ implies $NP = BPP$ (since the latter implies $NP = RP$).


I am pretty sure this is not known. $NP \neq RP$ does not seem to imply any circuit lower bounds for $EXP$.


Note that $EXP \subseteq P/poly$ does imply $P \neq NP$. If $EXP \subseteq P/poly$ then there is some fixed $k$ for which $TIME[2^n]$ has $n^k$-size circuits. Therefore $P \subseteq TIME[2^n]$ has $n^k$-size circuits. But $P = NP$ implies that $\Sigma\_2 P = P$, and we know by Kannan's theorem that $\Sigma\_2 P$ does not have $n^k$-size circuits for any fixed $k$ -- this is a contradiction. (This result is due to Meyer, cited in the famous Karp-Lipton paper.)


I would suspect that one can prove $EXP \subseteq P/poly$ implies $RP \neq NP$ or maybe $ZPP \neq NP$, but the above proof doesn't do the job. (Technically, since I believe $EXP \not\subseteq P/poly$, the negation should imply everything...)


