 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/12f/ii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Let

I_{i}

indicate that

v_{i}

is isolated. All its

n - 1

incident edges must be absent, so

E I_{i} = P (I_{i} = 1) = (1 - p)^{n - 1} .

Since

N = \sum_{i = 1}^{n} I_{i}

E (N) = n (1 - p)^{n - 1} .

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