Bayes' theorem

Higher Level

A thousand people. A disease, and a test that is right 99% of the time both ways. Every dot that tested positive is coloured. Slide the prevalence and watch which colour there is more of.

1 in 1000

1true positives

10false positives

9.0%P(disease | positive)

The test never changed. Only how many people actually have the disease.

Why a good test gives a bad answer

There are far more healthy people to get wrong. At 1 in 1000, one person in a thousand has it and the test finds them. The other 999 are healthy, and 1% of 999 is about 10 false alarms. So about 10 of the 11 positives are wrong, and a positive result means roughly a 9% chance, not 99%.

P(D)= 0.001 the base rate, or prevalence P(+ | D)= 0.99 sensitivity P(+ | D′)= 0.01 1 − specificity P(+)= 0.001(0.99) + 0.999(0.01) = 0.01098 P(D | +)= 0.001 × 0.990.01098 = 0.0902

The theorem

P(A | B) = P(A)P(B | A)P(B), where the bottom is found by total probability: P(B) = ∑ P(A_i)P(B | A_i) over the whole partition. At Higher Level you need it for up to three events.

With three causes A₁, A₂, A₃ that between them cover every possibility, the bottom is just every route to B added up:

given P(A₁)=0.5, P(A₂)=0.3, P(A₃)=0.2 and P(B|A₁)=0.02, P(B|A₂)=0.05, P(B|A₃)=0.10 P(B)= 0.5(0.02) + 0.3(0.05) + 0.2(0.10) = 0.045 P(A₃ | B)= 0.2 × 0.100.045 = 0.444

The posteriors must sum to 1. Here 0.222 + 0.333 + 0.444 = 1. It is the fastest check there is, and it catches a wrong denominator immediately.

Your turn

1. With the numbers above, find P(A₁ | B) to 3 significant figures.

2. Keeping the same 99% test, at what prevalence does a positive result become a 50/50 chance? Give your answer as a decimal.

3. A test is 99% accurate and someone tests positive for a disease affecting 1 in 1000. The best description of their situation is:

They almost certainly have it, because the test is 99% accurate. It is about 9% likely, and a second independent test would be worth doing. The test is useless and tells you nothing. It is exactly 50/50, since they either have it or they do not.

Where the marks go

Writing the denominator as a full total probability, every branch of it. Dividing by P(B | A) instead of P(B) is the error that loses the question.

Defining your events in words before any numbers. “D is the event that the person has the disease” costs one line and prevents the whole thing inverting.

Interpreting the answer in context. A question that sets up a medical test is asking you to say what a positive result actually means to the patient.

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