## Precalculus (6th Edition) Blitzer

The required solution of probability is, $\frac{1}{2}$
If the die is rolled, then the sample space of equally likely outcomes is $S=\left\{ 1,2,3,4,5,6 \right\}$ As there are six outcomes in the sample space, thus, $n\left( S \right)=6$ Also, the event of getting an odd number can be represented by $E=\left\{ 1,3,5 \right\}$ Since there is only one outcome in this event, therefore, $n\left( E \right)=3$ Hence, the probability of obtaining an odd number is: \begin{align} & P\left( E \right)=\frac{n\left( E \right)}{n\left( S \right)} \\ & =\frac{3}{6} \\ & =\frac{1}{2} \end{align}