A single fair die is rolled twice. a) Determine the number of points in the sample space. b) Construct a tree diagram and determine the sample space. c) Determine the probability that a double (a 1,1 or 2,2 , etc.) is rolled. d) Determine the probability that a sum of 12 is rolled. e) Determine the probability that a sum of 8 is rolled. f) Are you as likely to roll a sum of 12 as you are of rolling a sum of 8 ? Use the fundamental counting principle to determine the answer to part a). Assume that each event is equally likely to occur.

When you roll a single fair die, there are 6 possible outcomes for each face of the die—1 through 6. Since the die is rolled twice, by the fundamental counting principle, you multiply the number of outcomes for the first roll by the number of outcomes for the second roll. So, \(6 \times 6\) results in 36 points in the sample space. To visualize the sample space more clearly, a tree diagram can be used. Start with the first branch for the first roll with 6 options: 1, 2, 3, 4, 5, 6. Then, from each branch, create 6 subsequent branches for the second roll. The resulting pairs, like (1,1), (1,2), ..., (6,6), give us all 36 combinations which form our complete sample space. Next, let’s calculate probabilities. The probability of rolling a double (like (1,1), (2,2)...(6,6)) is found since there are 6 favorable outcomes and 36 total outcomes. Therefore, \(P(\text{double}) = \frac{6}{36} = \frac{1}{6}\). For a sum of 12, the only combination is (6,6). So there’s just 1 favorable outcome, leading to \(P(\text{sum of 12}) = \frac{1}{36}\). To find the probability of rolling a sum of 8, the combinations that produce this are (2,6), (3,5), (4,4), (5,3), and (6,2), resulting in 5 favorable outcomes. Thus, \(P(\text{sum of 8}) = \frac{5}{36}\). Finally, comparing the probabilities, it turns out that rolling a sum of 12 is significantly less likely than rolling a sum of 8, since \(\frac{1}{36} < \frac{5}{36}\).