Answer
The probability is $\frac{5}{6}$.
Work Step by Step
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
$\begin{align}
& {{\text{P}}_{\text{red or greater than 3}}}={{\text{P}}_{\text{red}}}+{{\text{P}}_{\text{greater than 3}}}-{{\text{P}}_{\text{red greater than 3}}} \\
& =\frac{3}{6}+\frac{3}{6}-\frac{1}{6} \\
& =\frac{3+3-1}{6} \\
& =\frac{5}{6}
\end{align}$
