 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/8/a/i/solution

Put

q = 1 - p

G_{X} (s) = E [s^{X}] = p k = 0 \sum \infty (q s)^{k} = \frac{p}{1 - q s}, ∣ s ∣ < \frac{1}{q} .

This is the zero-based geometric distribution, counting failures before the first success.

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