## Precalculus (6th Edition) Blitzer

The required probability is, $\frac{1}{2}$
The sample space of equally likely outcomes is $\left\{ HH,HT,TH,TT \right\}$ and $n\left( S \right)=4$ Assume $E$ to be the event of getting two heads; then $E=\left\{ HH,TT \right\}$; thus, $n\left( E \right)=2$ Hence, the probability of two heads is: \begin{align} & P\left( E \right)=\frac{n\left( E \right)}{n\left( S \right)} \\ & =\frac{2}{4} \\ & =\frac{1}{2} \end{align}