The Venn diagram represents the results of the experiment of rolling two fair, six-sided number cubes and adding the sum of the two rolls. Consider event A as “sum is odd” and event B as “sum is 6 or greater.” Find the probability that the sum is odd, given that the sum is greater than 6.

Answer 1

To find the probability that the sum is odd, given that the sum is greater than \(6\), we can denote this probability as \(P(A | B)\), where \(A\) is the event that the sum is odd and \(B\) is the event that the sum is \(6\) or greater.

First, we need to find the relevant sums when rolling two six-sided number cubes. The possible sums range from \(2\) (1 + 1) to \(12\) (6 + 6). We will analyze the sums that fulfill both conditions.

Sum Greater than 6

We begin by finding the sums that are greater than \(6\). The possible sums greater than \(6\) are \(7\), \(8\), \(9\), \(10\), \(11\), and \(12\). We need to check how many combinations produce each of these sums:

• Sum = 7: The combinations are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → 6 combinations.
• Sum = 8: The combinations are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → 5 combinations.
• Sum = 9: The combinations are (3, 6), (4, 5), (5, 4), (6, 3) → 4 combinations.
• Sum = 10: The combinations are (4, 6), (5, 5), (6, 4) → 3 combinations.
• Sum = 11: The combinations are (5, 6), (6, 5) → 2 combinations.
• Sum = 12: The combination is (6, 6) → 1 combination.

Now, we can tally the total combinations for sums greater than \(6\):

\[ 6 + 5 + 4 + 3 + 2 + 1 = 21 \text{ total combinations.} \]

Sum is Odd

Next, we'll count how many of these sums are odd:

• Odd sums greater than 6: These are \(7\), \(9\), and \(11\).

For each odd sum:

• Sum = 7: 6 combinations (from above).
• Sum = 9: 4 combinations (from above).
• Sum = 11: 2 combinations (from above).

Total combinations for odd sums greater than \(6\):

\[ 6 + 4 + 2 = 12. \]

Calculate Probability

Now that we have identified the total number of favorable outcomes for \(A\) (the sum is odd and greater than \(6\)) and the total number of outcomes for \(B\) (the sum is greater than \(6\)), we can calculate \(P(A | B)\):

\[ P(A | B) = \frac{P(A \cap B)}{P(B)} = \frac{\text{Number of favorable outcomes for A (odd sums > 6)}}{\text{Number of outcomes for B (sums > 6)}} \]

Thus:

\[ P(A | B) = \frac{12}{21} = \frac{4}{7}. \]

So, the probability that the sum is odd, given that the sum is greater than \(6\), is

\[ \boxed{\frac{4}{7}}. \]