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

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.

ii

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.

The negative-test probability is

P (-) = 1 - 0.05094 = 0.94906

. Applying Bayes' theorem again,

P (G ∣ -) = \frac{P ( - ∣ G ) P ( G )}{P ( - )} = \frac{0.01 ( 0.001 )}{0.94906} = 1.05 \times 1 0^{- 5} (3 s.f.) .

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