Probability to choose at least one green ball and no red balls Assume we have $n$ red balls, $n$ green balls, and unknown number of white balls.
We select each ball to a set with probability $p=\frac{1}{n}$ and not choosing it with probability $1-p=1-\frac{1}{n}$ independently with each other.
Show a constant lower bound (does not depend on $n$) on the probability the set will contain at least one green ball and won't contain any red balls.
I've said the probability to choose at least one green ball is $1-(1-p)^n$ , and the probability not to choose any red balls is $(1-p)^n$ .
So the requested probability is $(1-(1-p)^n)(1-p)^n=(1-(1-\frac{1}{n})^n)(1-\frac{1}{n})^n$
How can I find a constant lower bound for this?
Thanks

Let $x=\left(1-\frac1n\right)^n$. Then you've found that the desired probability is $x(1-x)$. This has a maximum at $x=\frac12$ and monotonically increases towards that maximum.  As $n\to\infty$, $x$ monotonoically increases towards $\frac1{\mathrm e}\lt\frac12$. Thus the probability monotonically increases with $n$.
