A and B throw two dice each. If A gets a sum of 9 on his two dice, then the probability of B getting a higher sum is

Options

A. `1/6`
B. `1/36`
C. `2/9`
D. `11/36`

Answer

To solve the problem, we need to calculate the probability that player B gets a sum greater than 9 on two dice, given that player A has already obtained a sum of 9 on their two dice.

### Step-by-Step Solution:

1. **Identify Possible Sums Greater than 9**:
- The possible sums that are greater than 9 when rolling two dice are 10, 11, and 12.

2. **Determine Favorable Outcomes for Each Sum**:
- **Sum of 10**: The combinations that yield a sum of 10 are:
- (4, 6)
- (5, 5)
- (6, 4)
- Total combinations for sum of 10 = 3
- **Sum of 11**: The combinations that yield a sum of 11 are:
- (5, 6)
- (6, 5)
- Total combinations for sum of 11 = 2
- **Sum of 12**: The only combination that yields a sum of 12 is:
- (6, 6)
- Total combinations for sum of 12 = 1

3. **Calculate Total Favorable Outcomes**:
- Total favorable outcomes for sums greater than 9:
- Sum of 10: 3 outcomes
- Sum of 11: 2 outcomes
- Sum of 12: 1 outcome
- Total favorable outcomes = 3 + 2 + 1 = 6

4. **Calculate Total Possible Outcomes**:
- When rolling two dice, the total number of possible outcomes is:
- Total outcomes = 6 (sides on die 1) × 6 (sides on die 2) = 36

5. **Calculate the Probability**:
- Probability of B getting a sum greater than 9:
\[ P(\text{B's sum} > 9) = \frac{\text{Number of favorable outcomes}}{\text{Total possible outcomes}} = \frac{6}{36} = \frac{1}{6} \]

### Final Answer:
The probability that player B gets a higher sum than player A (who has a sum of 9) is \(\frac{1}{6}\).

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