 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/8/a/i/solution

p

N

counts failures before the first success. The event

N = n

requires

n

failures followed by one success, giving

(1 - p)^{n} p

. This is the zero-based geometric distribution, rather than the count of trials including the success.

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