 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/10/b/solution

The complement of an event here is that all three dice avoid six, with probability

(5/6)^{3}

. Therefore

P (at least one six) = 1 - (5/6)^{3} = \frac{91}{216} .

 New to topics? Read the docs here!