 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-4/9h/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

π

π_{j} = \sum_{i \in S} π_{i} p_{ij}

for every

j

, equivalently

π P = π

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