 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/i/solution

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

Let

G

denote carrying the gene and

+

a positive result. The law of total probability gives

P (+) = P (+ ∣ G) P (G) + P (+ ∣ G^{c}) P (G^{c}) = 0.99 (0.001) + 0.05 (0.999) = 0.05094 ≃ 0.0509.

The last value is to three significant figures. The much larger non-carrier population contributes most of the positive tests in this mathematical model.

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