## Precalculus (6th Edition) Blitzer

The probability is $\frac{5}{6}$.
We know that there are six outcomes, so $\text{n}\left( \text{S} \right)=\text{ 6}$. The event of stopping on yellow can be represented by ${{\left( \text{E} \right)}_{\text{yellow}}}\text{ }=\text{ }\left\{ 1 \right\}$ There is one outcome in this event, so $\text{n}{{\left( \text{E} \right)}_{\text{yellow}}}=\text{ 1}$. The probability of stopping on yellow is given below, $\text{P}{{\left( \text{E} \right)}_{\text{yellow}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{yellow}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{1}{6}\text{ }$ And the probability of not stopping on yellow is given below, \begin{align} & {{\text{P}}_{\text{not yellow}}}\text{ = 1}-\text{ }{{\text{P}}_{\text{yellow}}} \\ & \text{= 1}-\text{ }\frac{1}{6}\text{ } \\ & \text{= }\frac{5}{6} \end{align}