1. **Problem statement:** Find the probability of drawing two kings from a standard deck of 52 cards. 2. **Formula:** Probability of two dependent events both occurring is the product of the probability of the first event and the conditional probability of the second event given the first. 3. **Step 1:** Probability of drawing the first king: $$\frac{4}{52}$$ 4. **Step 2:** After drawing one king, there are 3 kings left and 51 cards remaining, so probability of drawing the second king: $$\frac{3}{51}$$ 5. **Step 3:** Multiply these probabilities: $$\frac{4}{52} \times \frac{3}{51} = \frac{12}{2652}$$ 6. **Step 4:** Simplify the fraction by canceling common factors: $$\frac{\cancel{12}}{\cancel{2652}} = \frac{1}{221}$$ 7. **Final answer:** The probability of drawing two kings is $$P(\text{Two Kings}) = \frac{1}{221}$$ 1. **Problem statement:** Find the probability of drawing one ace and one queen from a standard deck of 52 cards. 2. **Step 1:** Probability of drawing an ace first and then a queen: $$\frac{4}{52} \times \frac{4}{51} = \frac{16}{2652}$$ 3. **Step 2:** Probability of drawing a queen first and then an ace: $$\frac{4}{52} \times \frac{4}{51} = \frac{16}{2652}$$ 4. **Step 3:** Total probability is sum of these two mutually exclusive events: $$\frac{16}{2652} + \frac{16}{2652} = \frac{32}{2652}$$ 5. **Step 4:** Simplify the fraction by canceling common factors: $$\frac{\cancel{32}}{\cancel{2652}} = \frac{8}{663}$$ 6. **Final answer:** The probability of drawing one ace and one queen is $$P(\text{One Ace and One Queen}) = \frac{8}{663}$$