## Precalculus (6th Edition) Blitzer

The probability is $\frac{2}{13}$.
We know that there are fifty-two outcomes in a $52$ card deck, so $\text{n}\left( \text{S} \right)=\text{ 52}$. There are four outcomes when the pulled card is an ace, so $\text{n}{{\left( \text{E} \right)}_{\text{an ace}}}=\text{ 4}$. The probability of getting an ace is given below $\text{P}{{\left( \text{E} \right)}_{\text{an ace}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{an ace}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{4}{52}\text{ = }\frac{1}{13}$. And there are four outcomes when the pulled card is a king, so $\text{n}{{\left( \text{E} \right)}_{\text{a king}}}=\text{ 4}$. The probability of getting a king is given below $\text{P}{{\left( \text{E} \right)}_{\text{a king}}}\text{ = }\frac{\text{n}{{\left( \text{E} \right)}_{\text{a king}}}}{\text{n}\left( \text{S} \right)}\text{ = }\frac{4}{52}\text{ = }\frac{1}{13}$. Thus, the probability of getting an ace or a king is \begin{align} & {{\text{P}}_{\text{an ace or a king}}}\text{ = }{{\text{P}}_{\text{an ace}}}+\text{ }{{\text{P}}_{\text{a king}}} \\ & =\text{ }\frac{1}{13}\text{ }+\text{ }\frac{1}{13} \\ & =\text{ }\frac{2}{13} \end{align}