 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/18/a/ii/solution

Attending none requires a tail on every independent toss. Thus

P (N = 0) = (\frac{1}{4})^{24} = 2^{- 48} \approx 3.55271 \times 1 0^{- 15} .

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