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The problem involves rolling an ordinary (fair) die twice. The sum of the values from the two rolls is recorded. You are asked to compute the probability of two specific events:

1. **Event A**: The sum is greater than 9.
2. **Event B**: The sum is not divisible by 2 and not divisible by 6.

We'll start by calculating the total number of possible outcomes when rolling two dice.

### Total Possible Outcomes:
Each die has 6 faces (1 through 6), so when you roll two dice, the total number of outcomes is:
\[
6 \times 6 = 36
\]
There are 36 possible outcomes when rolling two dice.

### Event A: The sum is greater than 9.
The possible sums when rolling two dice range from 2 (1+1) to 12 (6+6). We need to find the sums that are greater than 9.

Possible sums greater than 9 are: **10, 11, and 12**.

Now, let's find how many ways these sums can occur:
- Sum = 10: (4, 6), (5, 5), (6, 4) → 3 outcomes.
- Sum = 11: (5, 6), (6, 5) → 2 outcomes.
- Sum = 12: (6, 6) → 1 outcome.

So, there are \(3 + 2 + 1 = 6\) outcomes where the sum is greater than 9.

Now, the probability of **Event A** is:
\[
P(A) = \frac{6}{36} = \frac{1}{6} \approx 0.17
\]

### Event B: The sum is not divisible by 2 and not divisible by 6.
We need to find the sums that are **odd** (since they are not divisible by 2) and **not divisible by 6**.

First, list the odd sums: 3, 5, 7, 9, 11.

Next, exclude those that are divisible by 6. The only odd sum divisible by 6 is 6 itself, but since 6 is even, there are no odd sums divisible by 6.

Now, we count how many outcomes give these odd sums:
- Sum = 3: (1, 2), (2, 1) → 2 outcomes.
- Sum = 5: (1, 4), (2, 3), (3, 2), (4, 1) → 4 outcomes.
- Sum = 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 outcomes.
- Sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes.
- Sum = 11: (5, 6), (6, 5) → 2 outcomes.

So, there are \(2 + 4 + 6 + 4 + 2 = 18\) outcomes where the sum is odd and not divisible by 6.

Now, the probability of **Event B** is:
\[
P(B) = \frac{18}{36} = \frac{1}{2} = 0.50
\]

### Final Results:
- **Probability of Event A** (sum > 9): \(P(A) = 0.17\)
- **Probability of Event B** (sum is not divisible by 2 and not divisible by 6): \(P(B) = 0.50\)

Let me know if you'd like more details or have any questions!

### 5 Related Questions:
1. What is the probability of getting an even sum when rolling two dice?
2. How would the probabilities change if a biased die were used?
3. What is the probability of rolling a sum exactly equal to 7?
4. How can you calculate the probability of a sum divisible by 3?
5. What is the probability that the sum of the two dice equals 12?

### Tip:
When working with probabilities involving dice, it's helpful to first list all possible outcomes to ensure you consider every scenario.

An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a die is rolled twice in succession, and that the face values of the two rolls are added together. This sum is recorded as the outcome of a single trial of a random experiment. Compute the probability of each of the following events:

Event A: The sum is greater than 9.
Event B: The sum is not divisible by 2 and not divisible by 6.

Round your answers to two decimal places.

Calculate the probability of specific events with two dice rolls. Learn how to determine the probability of sums greater than 9 and sums that are neither divisible by 2 nor by 6. This detailed solution explains the concepts using basic probability theory and combinatorics, suitable for Grades 6-9.

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