## Precalculus (6th Edition) Blitzer

The required probability is, $\frac{1}{12}$
There are 36 equally likely outcomes; therefore, the number of elements in the sample is 36, that is, $n\left( S \right)=36$ Assume $E$ to be the event of obtaining two numbers which sum to 4; then $E=\left\{ \left( 1,3 \right),\left( 2,2 \right),\left( 3,1 \right) \right\}$ Therefore, $n\left( E \right)=3$ Thus, the probability of obtaining two numbers whose sums is 4 is: \begin{align} & P\left( E \right)=\frac{n\left( E \right)}{n\left( S \right)} \\ & =\frac{3}{36} \\ & =\frac{1}{12} \end{align}