Prove that $(1-p)^n \le 1-p^n$

Question

Prove that $(1-p)^n \le 1-p^n$ for $0\le p \le 1$

I tried using binomial expansion formula. But it is of no use.

Answer 1

It's not so sophisticated. The only result you need is that, if $\;0\le x\le 1$, then $\; x^n\le x$.

So, as $\;0\le 1-p\le 1$, one has $\;(1-p)^n\le 1-p$.

Next $\;p^n\le p$, so $-p\le -p^n$ and $1-p\le1-p^n$.

The result follows by transitivity.

Answer 2

If you use binomial distribution ( we have

$$ 1 = \sum_{k=0}^n{n\choose k}p^k(1-p)^{n-k} \geq (1-p)^n + p^n$$ and thus conclusion.

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Binomial distribution: Flip a coin $n$-times with probability $0\leq p\leq 1$ showing head each time. If $X$ is a number of heads then $P(X=k)={n\choose k}p^k(1-p)^{n-k}$. Clearly $$P(X=0)+P(X=1)+...+P(X=n) =1$$

Answer 3

Let there be two independent experiments that both have probability $p$ to succeed.

Then $p^2+(1-p)^2$ is the probability of the event that both experiments succeed or both fail.

The probability of an event never exceeds $1$ so:$$p^2+(1-p)^2\le1$$