 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-221/3/b/iv/solution

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

Yes. Conditioning on

D

blocks

C \leftarrow D \to A

, while

C \to M \leftarrow A

is blocked at the collider

M

. Therefore

C ⊥ A ∣ D

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