B.1/11
P(sum = 12 | one number is a 6)
= P(sum = 12 and one number is a 6) / P(one of the numbers is a 6)
= P(both numbers are 6) / (1 − P(neither number is a 6))
= (1/36) / (1 − (5/6)²)
= (1/36) / (1 − 25/36)
= (1/36) / (11/36)
= 1/11
Alternate method:
List all possible combinations when rolling 2 dice:
1−1 ..... 2−1 ..... 3−1 ..... 4−1 ..... 5−1 ..... 6−1
1−2 ..... 2−2 ..... 3−2 ..... 4−2 ..... 5−2 ..... 6−2
1−3 ..... 2−3 ..... 3−3 ..... 4−3 ..... 5−3 ..... 6−3
1−4 ..... 2−4 ..... 3−4 ..... 4−4 ..... 5−4 ..... 6−4
1−5 ..... 2−5 ..... 3−5 ..... 4−5 ..... 5−5 ..... 6−5
1−6 ..... 2−6 ..... 3−6 ..... 4−6 ..... 5−6 ..... 6−6
Of these, 11 combinations have at least one 6
1−6, 2−6, 3−6, 4−6, 5−6,
6−1, 6−2, 6−3, 6−4, 6−5, 6−6
Of these 11, only 1 has sum = 12
P(sum = 11 | one of the numbers = 6) = 1/11
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NOTE:
If you had asked to find the probability that the sum was 12 given that the FIRST number was a 6, then answer would indeed be 1/6, since there are only 6 combinations where FIRST number = 6:
6−1, 6−2, 6−3, 6−4, 6−5, 6−6
and of these 6 combinations, only 1 has sum = 12
If you had asked to find the probability that the sum was 12 given that the SECOND number was a 6, then answer again would be 1/6, since there are only 6 combinations where SECOND number = 6:
1−6, 2−6, 3−6, 4−6, 5−6, 6−6
and of these 6 combinations, only 1 has sum = 12
So in both these cases, we are starting out with 6 combinations, and the probability that sum is 12 from these 6 combinations is 1/6
BUT, since you do not specify which number is 6, only that one of the numbers is a 6, we must look at all combinations where there is a 6. There are 11 such combinations (as I've shown above), and the probability that sum is 12 from these 11 combinations is 1/11
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