 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2/8/solution

For independent fair dice, the probability of avoiding one on each die is

5/6

. Therefore

P (neither is one) = (\frac{5}{6})^{2} = \frac{25}{36} .

P (at least one is one) = 1 - \frac{25}{36} = \frac{11}{36} .

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