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

$$\boxed{{\displaystyle \frac{4}{7}}}.$$