1. **Stating the problem:** A cube with 3 black faces and 3 white faces is thrown. If a black face appears, a card is drawn from box A containing cards labeled P, M, R. If a white face appears, a card is drawn from box B containing cards labeled K, U, 7, 9. (a) Complete the list of possible outcomes. 2. **Possible outcomes:** - Black face with cards from box A: (Black, P), (Black, M), (Black, R) - White face with cards from box B: (White, K), (White, U), (White, 7), (White, 9) So the complete set is: $$\{(Black, P), (Black, M), (Black, R), (White, K), (White, U), (White, 7), (White, 9)\}$$ 3. **Calculating probabilities:** - Probability of black face = $\frac{3}{6} = \frac{1}{2}$ - Probability of white face = $\frac{3}{6} = \frac{1}{2}$ - Probability of each card in box A = $\frac{1}{3}$ - Probability of each card in box B = $\frac{1}{4}$ (b)(i) Probability of white face and card labeled with a letter: - Cards labeled with letters in box B: K, U - Probability = Probability(white face) $\times$ Probability(card labeled letter) = $\frac{1}{2} \times \frac{2}{4} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ (b)(ii) Probability of black face or card labeled with number: - Cards labeled with numbers: 7, 9 (in box B) - Probability(black face) = $\frac{1}{2}$ - Probability(card labeled number) = Probability(white face) $\times$ Probability(card number in box B) = $\frac{1}{2} \times \frac{2}{4} = \frac{1}{4}$ Using the formula for union: $$P(Black \cup Number) = P(Black) + P(Number) - P(Black \cap Number)$$ Since black face and card number cannot happen together (card number only if white face), $$P(Black \cap Number) = 0$$ Therefore, $$P = \frac{1}{2} + \frac{1}{4} - 0 = \frac{3}{4}$$ --- **Slug:** cube cards **Subject:** probability **Desmos:** {"latex":"y=0","features":{"intercepts":true,"extrema":true}} **q_count:** 2