## Precalculus (6th Edition) Blitzer

The probability is $\frac{5}{6}$.
We know that there are six outcomes, so $\text{n}\left( \text{S} \right)=\text{ 6}$. And there are three outcomes in the red stopping event, so $\text{n}{{\left( \text{E} \right)}_{\text{red}}}=\text{ 3}$. And the probability of stopping on red is given below, $\text{P}{{\left( \text{E} \right)}_{\text{red}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{red}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{3}{6}\text{ }$ And the event of stopping on a number greater than 3 can be represented by ${{\left( \text{E} \right)}_{\text{greater than 3}}}\text{ }=\text{ }\left\{ 4,5,6 \right\}$ There are three outcomes in this event, so $\text{n}{{\left( \text{E} \right)}_{\text{greater than 3}}}=\text{ 3}$. And the probability of stopping on greater than $3$ is given below, $\text{P}{{\left( \text{E} \right)}_{\text{greater than 3}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{greater than 3}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{3}{6}\text{ }$ There is one outcome in red greater than $3$ event, so $\text{n}{{\left( \text{E} \right)}_{\text{red greater than 3}}}=\text{ 1}$. And the probability of stopping on red greater than $3$ is given below $\text{P}{{\left( \text{E} \right)}_{\text{red greater than 3}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{red greater than 3}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{1}{6}\text{ }$ And the probability of stopping on red or a number greater than 3 is \begin{align} & {{\text{P}}_{\text{red or greater than 3}}}={{\text{P}}_{\text{red}}}+{{\text{P}}_{\text{greater than 3}}}-{{\text{P}}_{\text{red greater than 3}}} \\ & =\frac{3}{6}+\frac{3}{6}-\frac{1}{6} \\ & =\frac{3+3-1}{6} \\ & =\frac{5}{6} \end{align}