You only need to consider the situations where 10 or 11 of the babies are girls, then subtract those probabilities from 1. This will give probability that any other number up to 9 of the babies are girls.
Use binomial theorem.
[tex]P(x=k) = (nCk) p^k (1-p)^{n-k}[/tex]
n = 11
k = 10,11
p = 1/2
[tex]P(x=10) = 11 (\frac{1}{2})^{11} = \frac{11}{2048} \\ \\ P(x=11) = 1(\frac{1}{2})^{11} = \frac{1}{2048} \\ \\ P(x \leq 9) = 1 - \frac{11}{2048} - \frac{1}{2048} \\ \\ P(x \leq 9)=\frac{2036}{2048} = \frac{509}{512}[/tex]
