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Let $X \thicksim B(n,p)$ $$f(x) = \begin{pmatrix}n\\x\end{pmatrix}p^x(1-p)^{n-x}, x = 0,1,2, ..., n$$

Then solve the inequality $f(x) \ge f(x-1)$ and show that $f(x)$ become maximized when $x = [(n+1)p]$, which denotes the maximum integer equal to or smaller than $(n+1)p$.


I had solved given inequality and derived below inequality $$x\le (n-x+1)p(1-p)$$ and if well summarized about x,

$$x\le \dfrac{p(1-p)(n+1)}{p(1-p)+1}$$

Anyhow could I derive above conclusion from this inequality?

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Check your calculations. I ended up with $$(n-x+1)p \geq (1-p)x $$ which reduces to $$(n+1)p \geq x $$ Which, in turn, means that $f$ increases for $x$'s up to $x=(n+1)p$.

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