## Precalculus (6th Edition) Blitzer

The probability of getting a number greater than 4 and less than 3 is $\frac{2}{3}$.
We know that there are six outcomes in a dice i.e.$\left( \text{S} \right)\text{ = }\left\{ 1,2,3,4,5,6 \right\}$, so $\text{n}\left( \text{S} \right)=\text{ 6}$. And the event of getting a number less than $3$ can be represented by ${{\left( \text{E} \right)}_{\text{less than 3}}}\text{ }=\text{ }\left\{ 1,2 \right\}$ Since there are two outcomes in this event, $\text{n}{{\left( \text{E} \right)}_{\text{less than 3}}}=\text{ 2}$. And the probability of getting a number less than $3$ is $\text{P}{{\left( \text{E} \right)}_{\text{less than 3}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{less than 3}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{2}{6}\text{ }$. And the event of getting a number greater than $4$ can be represented by ${{\left( \text{E} \right)}_{\text{greater than 4}}}\text{ }=\text{ }\left\{ 5,6 \right\}$ There are two outcomes in this event, so $\text{n}{{\left( \text{E} \right)}_{\text{greater than 4}}}=\text{ 2}$. And the probability of getting a number greater than 4 is given below, $\text{P}{{\left( \text{E} \right)}_{\text{greater than 4}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{greater than 4}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{2}{6}\text{ }$. Therefore, the probability of getting a number less than 3 or greater than 5 is \begin{align} & {{\text{P}}_{\text{less than 3 or greater than 4}}}\text{ = }{{\text{P}}_{\text{less than 3}}}+\text{ }{{\text{P}}_{\text{greater than 4}}} \\ & =\text{ }\frac{2}{6}\text{ }+\text{ }\frac{2}{6} \\ & =\text{ }\frac{4}{6} \\ & =\text{ }\frac{2}{3} \end{align}