To find the probability that the sum is odd given that the sum is 6 or greater, we can apply the formula for conditional probability:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Where:
- \( P(A|B) \) is the probability of event A occurring given that event B has occurred.
- \( P(A \cap B) \) is the probability of both events A and B occurring.
- \( P(B) \) is the probability of event B occurring.
Step 1: Identify the outcomes.
When rolling two six-sided dice, the possible sums can be from 2 to 12. The outcomes are:
- Sums that are odd: 3, 5, 7, 9, and 11 (so, A = {3, 5, 7, 9, 11}).
- Sums that are 6 or greater: 6, 7, 8, 9, 10, 11, and 12 (so, B = {6, 7, 8, 9, 10, 11, 12}).
Step 2: Find \( A \cap B \) (i.e., outcomes that are both odd and 6 or greater).
- The odd sums that are 6 or greater are: 7, 9, and 11. So, \( A \cap B = {7, 9, 11} \).
Step 3: Count the outcomes.
- The number of successful outcomes for \( A \cap B \) is 3 (the sums 7, 9, and 11).
- The total outcomes for event B (sums that are 6 or greater) is 7 (the sums 6, 7, 8, 9, 10, 11, and 12).
Step 4: Calculate the probabilities.
- \( P(A \cap B) = \frac{3}{12} \) since there are a total of 12 possible outcomes when rolling two dice (ranging from 2 to 12).
- \( P(B) = \frac{7}{12} \) since there are 7 outcomes that are 6 or higher.
Step 5: Calculate \( P(A|B) \).
Using the conditional probability formula:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{3}{12}}{\frac{7}{12}} = \frac{3}{7} \]
So the answer is:
\(\frac{3}{7}\)
Thus, the correct response is:
37