To solve this, we need to determine the probability of rolling a sum of no more than 10 with two dice. The event $D$ represents all sums between 2 and 10 inclusive. Let's break it down step by step.

Total Possible Outcomes

When rolling two dice, each die has 6 faces, so the total number of possible outcomes is: $6 \times 6 = 36$ Thus, there are 36 possible outcomes when rolling two dice.

Favorable Outcomes

The event $D$ represents the sums between 2 and 10. Let's count how many outcomes produce these sums:

• Sum = 2: (1,1) → 1 outcome
• Sum = 3: (1,2), (2,1) → 2 outcomes
• Sum = 4: (1,3), (2,2), (3,1) → 3 outcomes
• Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
• Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes
• Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
• Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
• Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
• Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes

Now, let's add up the number of favorable outcomes: $1 + 2 + 3 + 4 + 5 + 6 + 5 + 4 + 3 = 33$

Probability

The probability of the event $D$ (sum between 2 and 10) is the ratio of favorable outcomes to total outcomes: $P(D) = \frac{33}{36} = \frac{11}{12}$

Thus, the probability of the event $D = \{2, 3, \dots, 10\}$ is $\frac{11}{12}$.

Would you like further details or have any questions?

Here are 5 related questions:

1. What is the probability of rolling a sum greater than 10 with two dice?
2. How many outcomes result in a sum of exactly 7 when rolling two dice?
3. What is the probability of rolling a sum less than or equal to 5?
4. What is the probability of rolling a double (e.g., (1,1), (2,2), etc.)?
5. If a third die is added, how would the probability of getting a sum of less than or equal to 10 change?

Tip: When calculating probabilities with dice, listing all possible outcomes is a helpful strategy to ensure accuracy.