1. **State the problem:** Mark draws one card from a standard deck of 52 cards. He receives 0.35 if the card is a diamond, 0.55 if the card is an ace, and 0.80 if the card is the ace of diamonds. We want to find the expected value, i.e., how much he should pay for one draw. 2. **Formula for expected value:** $$E(X) = \sum (\text{value} \times \text{probability})$$ 3. **Identify probabilities:** - Number of diamonds = 13 - Number of aces = 4 - Ace of diamonds is counted in both categories, so we must avoid double counting. 4. **Calculate probabilities:** - Probability of diamond = $\frac{13}{52} = \frac{1}{4}$ - Probability of ace = $\frac{4}{52} = \frac{1}{13}$ - Probability of ace of diamonds = $\frac{1}{52}$ 5. **Calculate expected value:** We add the values for diamonds and aces but subtract the ace of diamonds once because it is included in both categories: $$E = 0.35 \times P(\text{diamond}) + 0.55 \times P(\text{ace}) - 0.35 \times P(\text{ace of diamonds}) + 0.80 \times P(\text{ace of diamonds})$$ 6. **Simplify the expression:** $$E = 0.35 \times \frac{1}{4} + 0.55 \times \frac{1}{13} - 0.35 \times \frac{1}{52} + 0.80 \times \frac{1}{52}$$ 7. **Calculate each term:** $$0.35 \times \frac{1}{4} = 0.0875$$ $$0.55 \times \frac{1}{13} \approx 0.0423$$ $$-0.35 \times \frac{1}{52} \approx -0.0067$$ $$0.80 \times \frac{1}{52} \approx 0.0154$$ 8. **Sum all terms:** $$E = 0.0875 + 0.0423 - 0.0067 + 0.0154 = 0.1385$$ **Final answer:** Mark should pay approximately 0.14 for one draw.