To solve this problem, let's break it down step-by-step for each event.
### Event A: The sum is greater than 8
1. Possible Outcomes When Rolling Two Dice:
When rolling two dice, each die has 6 faces, so the total number of possible outcomes is [tex]\(6 \times 6 = 36\)[/tex].
2. Sums Greater Than 8:
- If we list the possible sums and consider only those greater than 8, the favorable sums are: 9, 10, 11, and 12.
- We need to count the occurrences of each of these sums.
- Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) [tex]\( \rightarrow 4 \)[/tex] outcomes
- Sum = 10: (4, 6), (5, 5), (6, 4) [tex]\( \rightarrow 3 \)[/tex] outcomes
- Sum = 11: (5, 6), (6, 5) [tex]\( \rightarrow 2 \)[/tex] outcomes
- Sum = 12: (6, 6) [tex]\( \rightarrow 1 \)[/tex] outcome
- Total favorable outcomes for sums greater than 8: [tex]\(4 + 3 + 2 + 1 = 10\)[/tex].
3. Probability of Event A:
- Probability [tex]\( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{10}{36} \approx 0.28 \)[/tex].
### Event B: The sum is not divisible by 6
1. Possible Sums:
The possible sums of two dice range from 2 to 12.
2. Sums Divisible by 6:
- The sums that are divisible by 6 within this range are: 6 and 12.
- Sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) [tex]\( \rightarrow 5 \)[/tex] outcomes
- Sum = 12: (6, 6) [tex]\( \rightarrow 1 \)[/tex] outcome
- Total outcomes for sums divisible by 6: [tex]\(5 + 1 = 6\)[/tex].
3. Sums Not Divisible by 6:
- Total outcomes for all sums: 36
- Outcomes where sum is not divisible by 6: [tex]\(36 - 6 = 30\)[/tex].
4. Probability of Event B:
- Probability [tex]\( P(B) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{30}{36} \approx 0.83 \)[/tex].
### Final Answer:
(a) [tex]\( P(A) = 0.28 \)[/tex]
(b) [tex]\( P(B) = 0.83 \)[/tex]
