Q.If $X \sim B(n,p)$ and $P(X=0) = (1-p)^n = \dfrac{1}{64}$ , then $n+p$ equals

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Q.If X∼B(n,p)X \sim B(n,p)X∼B(n,p) and P(X=0)=(1−p)n=164P(X=0) = (1-p)^n = \dfrac{1}{64}P(X=0)=(1−p)n=641​, then n+pn+pn+p equals