 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability/probability/7/b/ii/solution

For both boxes to be empty, each ball must avoid two specified boxes. The individual avoidance probability is

(n - 2) / n

, and the choices of different balls are independent. Therefore

P (M_{1} = 0, M_{2} = 0) = (1 - \frac{2}{n})^{n} .

This also covers

n = 2

, when the answer is zero.

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