Math 1431 – Spring 2003 – Test #3 – Practice
Question 1d Answer

The sample space is:

1 + 1 = 21 + 2 = 31 + 3 = 41 + 4 = 51 + 5 = 61 + 6 = 7
2 + 1 = 32 + 2 = 42 + 3 = 52 + 4 = 62 + 5 = 72 + 6 = 8
3 + 1 = 43 + 2 = 53 + 3 = 63 + 4 = 73 + 5 = 83 + 6 = 9
4 + 1 = 54 + 2 = 64 + 3 = 74 + 4 = 84 + 5 = 94 + 6 = 10
5 + 1 = 65 + 2 = 75 + 3 = 85 + 4 = 95 + 5 = 105 + 6 = 11
6 + 1 = 76 + 2 = 86 + 3 = 96 + 4 = 106 + 5 = 116 + 6 = 12

This is a conditional probability. The cases where the first roll is a 6 are highlighted below:

1 + 1 = 21 + 2 = 31 + 3 = 41 + 4 = 51 + 5 = 61 + 6 = 7
2 + 1 = 32 + 2 = 42 + 3 = 52 + 4 = 62 + 5 = 72 + 6 = 8
3 + 1 = 43 + 2 = 53 + 3 = 63 + 4 = 73 + 5 = 83 + 6 = 9
4 + 1 = 54 + 2 = 64 + 3 = 74 + 4 = 84 + 5 = 94 + 6 = 10
5 + 1 = 65 + 2 = 75 + 3 = 85 + 4 = 95 + 5 = 105 + 6 = 11
6 + 1 = 76 + 2 = 86 + 3 = 96 + 4 = 106 + 5 = 116 + 6 = 12

Given these 6 cases, there are 4 in which the sum is greater than 8. So, P(sum ≥ 8 | first die is 6) = 4/6 = 2/3.