## Precalculus (6th Edition) Blitzer

The required probability is, $\frac{1}{2}$
We know that the sample space of equally likely outcomes is $\left\{ MMM,MMF,MFM,MFF,FMM,FMF,FFM,FFF \right\}$ Thus $n\left( S \right)=8$ Assume $E$ to be the event of obtaining at least two female children; then, we get $E=\left\{ MFF,FMF,FFM,FFF \right\}$ Therefore, $n\left( E \right)=4$ So the probability of getting at least two female children is: \begin{align} & P\left( E \right)=\frac{n\left( E \right)}{n\left( S \right)} \\ & =\frac{4}{8} \\ & =\frac{1}{2} \end{align}