 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-1/7/a/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

P (G ∣ +) = \frac{P ( + ∣ G ) P ( G )}{P ( + )} = \frac{0.00099}{0.05094} = 0.0194 (3 s.f.) .

This is about

1.94%

; it is a different conditioning direction from the test's

99%

positive probability among carriers.

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