The correct answer is: C. Pr(r = 8 | r/4 is an integer) = (5/9)
The probability of event A happening, given that event B has already happened, is called the conditional probability of A given B, and is denoted by $P(A|B)$.
In this case, event A is “the sum of the numbers turned up is 8” and event B is “the sum of the numbers turned up is a multiple of 4”.
There are 36 possible outcomes when two fair dice are rolled. The number of outcomes where the sum is 8 is 5: (2, 6), (3, 5), (4, 4), (5, 3), and (6, 2).
The number of outcomes where the sum is a multiple of 4 is 12: (2, 2), (2, 6), (3, 3), (3, 6), (4, 4), (4, 6), (5, 5), (5, 6), (6, 6), (1, 5), (5, 1), and (6, 4).
Therefore, the probability of the sum of the numbers turned up being 8, given that the sum is a multiple of 4, is $\frac{5}{12}$.
The other options are incorrect because:
- Option A is the probability of the sum of the numbers turned up being greater than 6, which is $\frac{1}{6}$.
- Option B is the probability of the sum of the numbers turned up being a multiple of 3, which is $\frac{5}{6}$.
- Option D is the probability of the sum of the numbers turned up being 6, given that the sum is a multiple of 5, which is $\frac{1}{18}$.