 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iii/4/a/i/solution

Put

q = 1 - p

. Since

X

P (X > 0) = 1 = q^{0}

. Taking differences of the given tail probabilities therefore yields, also for

k = 1

P (X = k) = P (X > k - 1) - P (X > k) = p q^{k - 1}, k \geq 1.

{1, 2, \dots}

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