Let's analyze the problem step-by-step. We are dealing with an event where a fair die is rolled twice and the sum of the face values is recorded. We need to compute the probabilities of two events and round our answers to two decimal places.
First, let's lay out the basics:
1. Total number of possible outcomes:
An ordinary die has 6 sides, so when rolled twice, there are \( 6 \times 6 = 36 \) possible outcomes.
### Event A: The sum is greater than 9
2. Possible sums greater than 9:
The possible sums greater than 9 are 10, 11, and 12.
3. Ways to achieve each sum:
- Sum of 10: (4, 6), (5, 5), (6, 4) ─ 3 ways
- Sum of 11: (5, 6), (6, 5) ─ 2 ways
- Sum of 12: (6, 6) ─ 1 way
Therefore, there are a total of \( 3 + 2 + 1 = 6 \) ways to achieve a sum greater than 9.
4. Probability of event A:
The probability, \( P(A) \), is the number of favorable outcomes divided by the total number of possible outcomes:
[tex]\[
P(A) = \frac{6}{36} = \frac{1}{6} \approx 0.17
\][/tex]
### Event B: The sum is divisible by 4
5. Possible sums divisible by 4:
The possible sums divisible by 4 are 4, 8, and 12.
6. Ways to achieve each sum:
- Sum of 4: (1, 3), (2, 2), (3, 1) ─ 3 ways
- Sum of 8: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) ─ 5 ways
- Sum of 12: (6, 6) ─ 1 way
Therefore, there are a total of \( 3 + 5 + 1 = 9 \) ways to achieve a sum that is divisible by 4.
7. Probability of event B:
The probability, \( P(B) \), is the number of favorable outcomes divided by the total number of possible outcomes:
[tex]\[
P(B) = \frac{9}{36} = \frac{1}{4} = 0.25
\][/tex]
### Conclusion:
(a) The probability \( P(A) \) that the sum is greater than 9 is:
[tex]\[
P(A) = 0.17
\][/tex]
(b) The probability \( P(B) \) that the sum is divisible by 4 is:
[tex]\[
P(B) = 0.25
\][/tex]
