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

Your approach is correct.
The numbers $a_n:=\left(1-{1\over n}\right)^n$ converge monotonically increasing to ${1\over e}<{1\over2}$. This implies that the probabilities
$$p_n:=(1-a_n)a_n\qquad(n\geq2)$$
increase monotonically to the limit ${e-1\over e^2}=0.233$, and one has
$$\min_{n\geq2} p_n=p_2={3\over16}=0.1875\ .$$
