relating to this question, I'd like to ask a further one.
Again we have $$f(x)={k-1 \choose x-1} p^x (1-p)^{k-x}$$
We know that this term is maximal for $x=kp$, before increasing, afterwards decreasing.
But what is, if we know that $kp \ll 1$ (for $n\rightarrow \infty$, where $n$ is not very relevant here) and $x$ can only take on constant values from $X$ (Independent from $n$). Then this would imply that $x \gg kp$ for all $x\in X$ and thus the function is decreasing in $X$, is this correct?
Stated differently, is it correct for $kp \ll 1$ and $x=O(1)$ for all $x\in X$, that this term above is monotonically decreasing in the range $X$?
EDIT: Maybe I should ask my question differently, as I am not interested in the normal case for a fixed $x$ and fixed $k$, $p$. I know what's going on there. My Problem is, when I know that only asymptotically $x < kp$ (and not necessarily for small values of $n$), is it safe to assume that asymptotically $f$ is decreasing?
We know $f(x)>f(x+1)$ is equivalent to $x>kp$ or $x\ge \lceil kp\rceil.$ So $f$ is decreasing if $x\gt kp.$
If $kp \lt 1$ then $f$ is decreasing on $X=\{1,2,3,...\}.$ It is not necessary that $ kp<< 1.$