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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 : The sum is greater than 7.
Event : The sum is divisible by 2.
Round your answers to two decimal places.

To solve the problem, let's first analyze the events based on the scenario described: rolling a fair die twice and summing the face values.

### Step 1: Total Possible Outcomes
When rolling a die twice, each roll has 6 possible outcomes, so the total number of outcomes is:
\[
6 \times 6 = 36
\]

### Step 2: Event (a) - The sum is greater than 7
We need to find the probability that the sum of the two rolls is greater than 7. The possible sums greater than 7 are: 8, 9, 10, 11, and 12.

Let's list the possible combinations for these sums:
- **Sum = 8:** (2,6), (3,5), (4,4), (5,3), (6,2)
- **Sum = 9:** (3,6), (4,5), (5,4), (6,3)
- **Sum = 10:** (4,6), (5,5), (6,4)
- **Sum = 11:** (5,6), (6,5)
- **Sum = 12:** (6,6)

Counting these, we get:
- 5 outcomes for 8
- 4 outcomes for 9
- 3 outcomes for 10
- 2 outcomes for 11
- 1 outcome for 12

Total favorable outcomes for event (a) = \(5 + 4 + 3 + 2 + 1 = 15\).

So, the probability of event (a) is:
\[
P(A) = \frac{15}{36} = 0.4167
\]
Rounded to two decimal places: 
\[
P(A) \approx 0.42
\]

### Step 3: Event (b) - The sum is divisible by 2
The sums that are divisible by 2 are: 2, 4, 6, 8, 10, 12.

Let's list the possible combinations for these sums:
- **Sum = 2:** (1,1)
- **Sum = 4:** (1,3), (2,2), (3,1)
- **Sum = 6:** (1,5), (2,4), (3,3), (4,2), (5,1)
- **Sum = 8:** (2,6), (3,5), (4,4), (5,3), (6,2)
- **Sum = 10:** (4,6), (5,5), (6,4)
- **Sum = 12:** (6,6)

Counting these, we get:
- 1 outcome for 2
- 3 outcomes for 4
- 5 outcomes for 6
- 5 outcomes for 8
- 3 outcomes for 10
- 1 outcome for 12

Total favorable outcomes for event (b) = \(1 + 3 + 5 + 5 + 3 + 1 = 18\).

So, the probability of event (b) is:
\[
P(B) = \frac{18}{36} = 0.5
\]
Rounded to two decimal places:
\[
P(B) = 0.50
\]

### Final Answers:
- \( P(A) \approx 0.42 \)
- \( P(B) = 0.50 \)

Would you like any additional details or have further questions?

### Related Questions:
1. What is the probability of the sum being less than 7?
2. What is the probability of rolling a sum of exactly 7?
3. How would the probabilities change if the die was not fair?
4. What is the probability of rolling the same sum twice in two consecutive trials?
5. How does the probability change if one of the dice is loaded?
6. What is the probability of rolling a prime number as the sum?
7. What is the expected value of the sum when rolling two dice?
8. What is the variance and standard deviation of the sum of the two dice rolls?

### Tip:
When dealing with dice probabilities, always list out all possible outcomes, especially for small numbers of dice, to ensure accuracy in your calculations.

Explore the probabilities of rolling dice twice and calculating sums, focusing on events such as sums greater than 7 and sums divisible by 2. Understand the concept of random variables and apply it to dice rolling scenarios. Suitable for high school students learning about probability and random experiments.

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