Definition D1
For P(B) > 0, define P(A∣B) = P(A ∩ B) / P(B).
Proof.
(1)This definition and its warrant were established in Proof 164. Here it is taken as given.convention
(2)Write D for the event of having the disease, and + for the event of a positive test.convention
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Lemma L1
If P(X) > 0, then P(Y ∩ X) = P(Y∣X) · P(X)
Proof.
(1)P(Y∣X) = P(Y∩X) / P(X)assumption, D1
(2)Multiply both sides by P(X). P(Y∣X)·P(X) = (P(Y∩X)/P(X))·P(X)(1), E
(3)P(X) ≠ 0, so the right side is P(Y∩X).(2), A5
(4)P(Y ∩ X) = P(Y∣X) · P(X)(3), E
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Lemma L2
If P(D) > 0 and P(Dᶜ) > 0, then P(+) = P(+∣D)P(D) + P(+∣Dᶜ)P(Dᶜ)
Proof.
(1)+ = (+∩D) ∪ (+∩Dᶜ), and the two events are disjoint.A2
(2)P(+) = P(+∩D) + P(+∩Dᶜ)(1), A1
(3)P(+∩D) = P(+∣D)P(D) and P(+∩Dᶜ) = P(+∣Dᶜ)P(Dᶜ)assumption, L1
(4)P(+) = P(+∣D)P(D) + P(+∣Dᶜ)P(Dᶜ)(2), (3), E
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Theorem T1
If P(D) > 0, P(Dᶜ) > 0, P(+) > 0, then P(D∣+) = P(+∣D)P(D) / [ P(+∣D)P(D) + P(+∣Dᶜ)P(Dᶜ) ]
Proof.
(1)P(D∣+) = P(D∩+) / P(+)assumption, D1
(2)P(D∩+) = P(+∩D)A3
(3)P(+∩D) = P(+∣D) · P(D)assumption, L1
(4)P(D∣+) = P(+∣D)P(D) / P(+)(1), (2), (3), E
(5)Expand the denominator by total probability. P(D∣+) = P(+∣D)P(D) / [ P(+∣D)P(D) + P(+∣Dᶜ)P(Dᶜ) ](4), L2, E
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Corollary C
P(D∣+) = 11/122, which is about 0.09.
Proof.
(1)P(D) = 1/1000 > 0 and P(Dᶜ) = 1 − 1/1000 = 999/1000 > 0A6, A4, N
(2)Numerator: P(+∣D)P(D) = (99/100)(1/1000) = 99/100000A7, A6, N
(3)Second term of the denominator: P(+∣Dᶜ)P(Dᶜ) = (1/100)(999/1000) = 999/100000A8, (1), N
(4)P(+) = 99/100000 + 999/100000 = 1098/100000 > 0(2), (3), L2, N
(5)P(D∣+) = (99/100000) / (1098/100000) = 99/1098T1, (2), (4), N
(6)99 = 9 × 11 and 1098 = 9 × 122, so 99/1098 = 11/122(5), N
(7)11/122 = 0.0901…, so P(D∣+) is about 0.09. Of a hundred positive results, about nine have the disease.(6), N
(8)Both numerator and denominator carry P(D) as a factor. The performance figures P(+∣D), P(+∣Dᶜ) alone do not determine this value.T1, (5)
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