1. **State the problem:** Paul receives 8 milk candies, 6 chocolate candies, and 10 butter candies. We want to find the probability that Paul will eat milk or chocolate candies (a), and the probability that he will eat chocolate or butter candies (b). 2. **Define the sample space (S):** The total number of candies Paul has is $$8 + 6 + 10 = 24$$. So, $$S = 24$$. 3. **Define events:** - Let $$A$$ be the event Paul eats milk candies. - Let $$B$$ be the event Paul eats chocolate candies. For part (a): - $$A = 8$$ (milk candies) - $$B = 6$$ (chocolate candies) 4. **Calculate probability for (a):** We want $$P(A \cup B)$$, the probability Paul eats milk or chocolate candies. Using the formula for union of two events: $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$ Since candies are distinct types, $$A$$ and $$B$$ are mutually exclusive (no overlap), so $$P(A \cap B) = 0$$. Calculate probabilities: $$ P(A) = \frac{8}{24} = \frac{1}{3} $$ $$ P(B) = \frac{6}{24} = \frac{1}{4} $$ So, $$ P(A \cup B) = \frac{1}{3} + \frac{1}{4} = \frac{4}{12} + \frac{3}{12} = \frac{7}{12} $$ 5. **For part (b):** - Let $$A$$ be chocolate candies: $$6$$ - Let $$B$$ be butter candies: $$10$$ Calculate probabilities: $$ P(A) = \frac{6}{24} = \frac{1}{4} $$ $$ P(B) = \frac{10}{24} = \frac{5}{12} $$ Again, mutually exclusive, so $$ P(A \cup B) = P(A) + P(B) = \frac{1}{4} + \frac{5}{12} = \frac{3}{12} + \frac{5}{12} = \frac{8}{12} = \frac{2}{3} $$ **Final answers:** - (a) $$\boxed{\frac{7}{12}}$$ - (b) $$\boxed{\frac{2}{3}}$$