For $n\rightarrow \infty$ we consider $$f(p)=\sum_{j=c}^n {n\choose j} p^j (1-p)^{n-j}.$$ We are interested in $\hat{p}:=\arg \max_p f(p)$.
Can we say something about $\hat{p}$ dependent on $n$ and $c$ without further assumptions? Can we say something about $\hat{p}$ if we assume that $\hat{p}:=\arg \max_{p \ : \ np \gg 1} f(p)$.
The binomium of Newton tells us that $$ (x+y)^n=\sum_{i=0}^n \binom n i x^i y^{n-i} $$ setting $x=p$ and $y=1-p$ yields $$ 1 = \sum_{i=0}^n \binom n i p^i(1-p)^{n-i} $$ Because all terms on the right are non-negative, we see that this is maximal (we then have equality) when we start summing at $c=0$. When $n\geq c>0$, we can set $p=1$, to obtain $$ \sum_{i=c}^n \binom n i p^i(1-p)^{n-i}=0+0+\cdots +0+\binom nn p^n 0^0 = 1 $$ (We assume here that $0^0$ is $1$, because there is one way to take zero out of zero objects.) Thus, we achieve the maximum at $p=1$, independent from $c$. When $c=0$, i doesn't matter what $p$ is.
A plot with $n=10$ and $c=0$ to $10$:
