To compute the probability of these events, let's analyze each possible outcome when a die is rolled twice. The total possible outcomes when rolling a die twice are $6 \times 6 = 36$.

Event A: The sum is greater than 7.

The possible sums range from 2 to 12. We are interested in the sums that are greater than 7, which are 8, 9, 10, 11, and 12. Let's count the outcomes that give each of these sums.

• Sum of 8: $(2,6), (3,5), (4,4), (5,3), (6,2)$ → 5 outcomes
• Sum of 9: $(3,6), (4,5), (5,4), (6,3)$ → 4 outcomes
• Sum of 10: $(4,6), (5,5), (6,4)$ → 3 outcomes
• Sum of 11: $(5,6), (6,5)$ → 2 outcomes
• Sum of 12: $(6,6)$ → 1 outcome

Adding these up, there are $5 + 4 + 3 + 2 + 1 = 15$ outcomes where the sum is greater than 7.

The probability of Event A is therefore: $\frac{15}{36} = \frac{5}{12}$

Event B: The sum is an even number.

The sums that result in even numbers are 2, 4, 6, 8, 10, and 12. We can count the outcomes for each of these sums.

• Sum of 2: $(1,1)$ → 1 outcome
• Sum of 4: $(1,3), (2,2), (3,1)$ → 3 outcomes
• Sum of 6: $(1,5), (2,4), (3,3), (4,2), (5,1)$ → 5 outcomes
• Sum of 8: $(2,6), (3,5), (4,4), (5,3), (6,2)$ → 5 outcomes
• Sum of 10: $(4,6), (5,5), (6,4)$ → 3 outcomes
• Sum of 12: $(6,6)$ → 1 outcome

Adding these up, there are $1 + 3 + 5 + 5 + 3 + 1 = 18$ outcomes where the sum is an even number.

The probability of Event B is therefore: $\frac{18}{36} = \frac{1}{2}$

Final Answers

• Probability of Event A (sum greater than 7): $\frac{5}{12}$
• Probability of Event B (sum is an even number): $\frac{1}{2}$

Would you like further explanations or details?

Here are five related questions you might consider:

1. What is the probability of obtaining a sum less than 5?
2. What is the probability of rolling two dice and getting an odd sum?
3. What is the probability of obtaining a sum exactly equal to 7?
4. How would the probabilities change if the die were biased?
5. What is the probability of getting doubles (the same number on both dice)?

Tip: When working with probabilities in dice, remember that each face combination is equally likely, which makes calculating probabilities straightforward by counting outcomes.