cumulative binomial distribution upper bound

Question

This question is about the probability that out of $2n$ births there are more girls than boys, with the probability that a girl is born is $p=0.485$ The number of girls born has obviously a binomial distribution. It is to show that the probability that more girls than boys are born is lower than $$\binom{2n}{n}(1-p)^n\frac{1-p}{1-2p}$$ The actual chance is obviously $$\sum_{k=n+1}^{2n}\binom{2n}{k}p^k(1-p)^{2n-k}$$ But I have no idea how to show that this is less than the one above.

Answer 1

Using that $\binom{2n}{k}$ is maximized when $k=n$ and the geometric sum: \begin{align} & \sum_{k=n+1}^{2n}\binom{2n}{k}p^k(1-p)^{2n-k}\\ % & =(1-p)^{2n} \sum_{k=n+1}^{2n}\binom{2n}{k} \left(\frac{p}{1-p}\right)^k\\ % & \le(1-p)^{2n} \binom{2n}{n} \sum_{k=n+1}^{2n} \left(\frac{p}{1-p}\right)^k \\ % & \le (1-p)^{2n} \binom{2n}{n} \sum_{k=0}^{\infty} \left(\frac{p}{1-p}\right)^k \\ % & = (1-p)^{2n} \binom{2n}{n} \frac{1}{ 1 - \frac{p}{1-p}} \\ & = (1-p)^{2n} \binom{2n}{n} \frac{1-p}{ 1-2p } \\ & \le (1-p)^{n} \binom{2n}{n} \frac{1-p}{ 1-2p }. \end{align}