# prove change in total probability of success in binomial distribution

A binomial distribution of $$n$$ samples and probability of success $$p$$ is defined as

$$P(k) = \binom{n}{k} \cdot p^kq^{n-k}$$.

For a given value of $$r$$ where $$r \in \mathbb{N} \quad \text{and } r \gt 1$$, if we decrease the probability of success by $$\hat{p} = \frac{p}{r}$$ and increase the samples $$\hat{n} = nr$$, I want to prove

$$P(k\ge 1) - P(\hat{k}\ge 1) \gt 0$$

$$\Rightarrow \left(1-P(k= 0)\right) - \left(1-P(\hat{k}= 0)\right) \gt 0$$

$$\Rightarrow P(\hat{k}= 0) - P(k= 0) \gt 0$$

$$\Rightarrow \binom{nr}{0} \cdot \hat{p}^0\hat{q}^{nr} - \binom{n}{0} \cdot p^0q^{n} \gt 0$$

$$\Rightarrow \left(1-\hat{p}\right)^{nr} - \left(1-p\right)^{n} \gt 0$$

$$\Rightarrow \left(1-\frac{p}{r}\right)^{nr} - \left(1-p\right)^{n} \gt 0$$

I'm able to show it for specific examples but unfortunately unable to prove the last inequality in generic form.

First, by expanding $$(1-p/r)^r$$ to the power series, we have

$$(1-p/r)^r=1-p + [(r-1)p^2/(2! r)-(r-1)(r-2)p^3/(3!r^2)+…+(-p/r)^r]$$

that is,

$$(1-p/r)^r = 1-p + A$$

where $$A > 0$$ is the term in the square brackets. Thus, we have

$$(1-p/r)^r > 1-p$$, (for r > 1). Then we have

$$\left(1-\frac{p}{r}\right)^{nr} \gt \left(1-p\right)^{n}$$