1. **Problem Statement:** Find the probability of picking certain cards from a standard deck and answer related probability questions. 2. **Background:** A standard deck has 52 cards: 26 black cards (13 spades + 13 clubs), 13 clubs, 12 face cards (4 kings + 4 queens + 4 jacks), and 4 aces. 3. **Formulas:** Probability of an event $E$ is given by: $$P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}$$ 4. **Calculations:** **a. Probability of black card:** Number of black cards = 26 Total cards = 52 $$P(\text{black}) = \frac{26}{52} = \frac{1}{2} = 0.5 = 50\%$$ **b. Probability of club:** Number of clubs = 13 $$P(\text{club}) = \frac{13}{52} = \frac{1}{4} = 0.25 = 25\%$$ **c. Expected number of face cards in 100 draws with replacement:** Number of face cards = 12 $$P(\text{face card}) = \frac{12}{52} = \frac{3}{13} \approx 0.2308$$ Expected number in 100 draws: $$100 \times 0.2308 = 23.08 \approx 23 \text{ times}$$ 5. **Complement probabilities:** **a. Probability of not getting a club:** $$P(\text{not club}) = 1 - P(\text{club}) = 1 - \frac{13}{52} = \frac{39}{52} = \frac{3}{4} = 0.75 = 75\%$$ **b. Probability of not getting a face card:** $$P(\text{not face card}) = 1 - P(\text{face card}) = 1 - \frac{12}{52} = \frac{40}{52} = \frac{10}{13} \approx 0.7692 = 76.92\%$$ **c. Probability of getting an ace on second draw if first ace drawn and not replaced:** Initially, 4 aces in 52 cards. After drawing one ace and not replacing, remaining aces = 3, remaining cards = 51. $$P(\text{ace on second draw}) = \frac{3}{51} = \frac{1}{17} \approx 0.0588 = 5.88\%$$ This probability decreases because the deck now has fewer aces and fewer total cards. **Final answers:** - $P(\text{black}) = \frac{1}{2} = 0.5 = 50\%$ - $P(\text{club}) = \frac{1}{4} = 0.25 = 25\%$ - Expected face cards in 100 draws = 23 - $P(\text{not club}) = \frac{3}{4} = 0.75 = 75\%$ - $P(\text{not face card}) = \frac{10}{13} \approx 0.7692 = 76.92\%$ - $P(\text{ace on second draw after first ace drawn without replacement}) = \frac{1}{17} \approx 0.0588 = 5.88\%$