How can the faces of two cubes be numbered so that when they are rolled, the resulting sum is a number 1 to 12 inclusive and each sum has the same probability? Choose the correct answer below. A. Number them so that one cube has the numbers \( 0,0,0,2,2,2 \), and the other cube has the numbers \( 1,3,5 \), \( 6,8,10 \). B. Number them so that one cube has the numbers \( 0,0,0,3,3,3 \), and the other cube has the numbers \( 1,2,3 \), \( 7,8,9 \). C. Number them so that one cube has the numbers \( 0,1,2,3,4,5 \) and the other cube has the numbers \( 1,2,3 \), \( 4,5,6 \). D. Number them so that each cupe has the numbers \( 1,2,3,4,5,6 \).

We want to assign numbers to the faces of two cubes such that, when rolled together, there are 36 equally-likely outcomes and the resulting sums cover exactly the numbers 1 through 12 with each sum having 3 outcomes (i.e. probability 3⁄36 = 1⁄12). Let’s examine option B: • Cube 1: {0, 0, 0, 3, 3, 3} • Cube 2: {1, 2, 3, 7, 8, 9} When these cubes are rolled: 1. If Cube 1 shows 0, then the sum is determined solely by Cube 2. The possible sums are:   0 + 1 = 1   0 + 2 = 2   0 + 3 = 3   0 + 7 = 7   0 + 8 = 8   0 + 9 = 9 Each of these sums comes from 3 outcomes (since Cube 1 has three faces showing 0). 2. If Cube 1 shows 3, then the sums are:   3 + 1 = 4   3 + 2 = 5   3 + 3 = 6   3 + 7 = 10   3 + 8 = 11   3 + 9 = 12 Again, each of these sums again appears with 3 outcomes (since Cube 1 has three faces showing 3). Thus, every sum from 1 through 12 is obtained exactly 3 times, giving an equal probability of 1⁄12 for each sum. None of the other options yield exactly 12 uniformly-likely sums from 1 to 12. For example, option D (standard dice) produces sums from 2 to 12 with varying probabilities, and option C doesn’t cover the sum 12. Therefore, the correct answer is B.