Precalculus (6th Edition) Blitzer

The probability is $\frac{7}{13}$.
We know that there are fifty-two outcomes in a $52$ card deck, so $\text{n}\left( \text{S} \right)=\text{ 52}$. And there are four outcomes when the pulled card is a queen, so $\text{n}{{\left( \text{E} \right)}_{\text{a queen}}}=\text{ 4}$. And the probability of getting a queen is given below, $\text{P}{{\left( \text{E} \right)}_{\text{a queen}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{a queen}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{4}{52}\text{ }$ And there are twenty-six outcomes when the pulled card is a red card, so $\text{n}{{\left( \text{E} \right)}_{\text{a red card}}}=\text{ 26}$. And the probability of getting a red card is given below $\text{P}{{\left( \text{E} \right)}_{\text{a red card}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{a red card}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{26}{52}\text{ }$ There are two outcomes when the pulled card is a red queen card, so $\text{n}{{\left( \text{E} \right)}_{\text{a red queen card}}}=\text{ 2}$ And the probability of getting a red queen card is $\text{P}{{\left( \text{E} \right)}_{\text{a red queen card}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{a red queen card}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{2}{52}$ Thus, the probability of getting a queen or a red card is \begin{align} & {{\text{P}}_{\text{a queen or a red card}}}\text{ = }{{\text{P}}_{\text{a queen}}}+\text{ }{{\text{P}}_{\text{a red card}}}-\text{ }{{\text{P}}_{\text{a red queen card}}} \\ & =\text{ }\frac{4}{52}\text{ }+\text{ }\frac{26}{52}\text{ }-\text{ }\frac{2}{52} \\ & =\text{ }\frac{4+26-2}{52} \\ & =\text{ }\frac{28}{52} \\ & =\text{ }\frac{7}{13} \end{align}