1. **Stating the problem:** We roll two six-sided balanced dice. We want to find the probabilities of: i. Sum equals 8 ii. Sum equals 6 or 9 iii. Sum equals 3, 8, or 12 iv. Sum is not an even number 2. **Total number of outcomes:** Each die has 6 faces, so total outcomes when rolling two dice are $$6 \times 6 = 36$$. 3. **Sum equals 8:** Possible pairs that sum to 8 are: (2,6), (3,5), (4,4), (5,3), (6,2). There are 5 such pairs. Probability $$P(8) = \frac{5}{36}$$. 4. **Sum equals 6 or 9:** Pairs summing to 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 pairs. Pairs summing to 9: (3,6), (4,5), (5,4), (6,3) → 4 pairs. Total pairs = 5 + 4 = 9. Probability $$P(6 \text{ or } 9) = \frac{9}{36} = \frac{1}{4}$$. 5. **Sum equals 3, 8, or 12:** Sum 3: (1,2), (2,1) → 2 pairs. Sum 8: 5 pairs (from step 3). Sum 12: (6,6) → 1 pair. Total pairs = 2 + 5 + 1 = 8. Probability $$P(3,8,\text{or}\ 12) = \frac{8}{36} = \frac{2}{9}$$. 6. **Sum is not an even number:** Check sums that are odd: 3, 5, 7, 9, 11. Count pairs for each: - 3: 2 pairs - 5: (1,4), (2,3), (3,2), (4,1) → 4 pairs - 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 pairs - 9: 4 pairs (from step 4) - 11: (5,6), (6,5) → 2 pairs Total odd sum pairs = 2 + 4 + 6 + 4 + 2 = 18. Probability $$P(\text{not even}) = \frac{18}{36} = \frac{1}{2}$$. **Final answers:** i. $$\frac{5}{36}$$ ii. $$\frac{1}{4}$$ iii. $$\frac{2}{9}$$ iv. $$\frac{1}{2}$$