The question asks for the probability of the sum of two dice being less than 9.

To solve this, we need to consider all possible outcomes when two dice are rolled. Each die has 6 faces, so the total number of possible outcomes is:

$6 \times 6 = 36$

Next, let's count how many combinations of the dice give a sum less than 9.

• 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

Summing up the outcomes for sums less than 9:

$1 + 2 + 3 + 4 + 5 + 6 + 5 = 26$

The probability is the number of favorable outcomes (26) divided by the total possible outcomes (36):

$\frac{26}{36} = \frac{13}{18}$

Thus, the correct answer is:

B. $\frac{13}{18}$

Would you like further details or explanations on any part of this? Here are 5 questions that extend this concept:

1. How would the probability change if the condition was "greater than 9" instead of "less than 9"?
2. Can you calculate the probability of getting an even sum with two dice?
3. What is the probability of getting a double (e.g., (1,1), (2,2), etc.) when rolling two dice?
4. How does the sample space change if you roll three dice instead of two?
5. If you roll two dice, what is the probability of getting a sum exactly equal to 7?

Tip: When solving probability questions involving dice, always start by identifying the total possible outcomes and then carefully count the favorable outcomes.