Lecture 5: Bayes' rule

My favorite example: suppose the probability of being ill (I) with some deadly but rare disease is 10^-4. There is a test for this disease which has no false negatives: if you have the disease, you will test positive (P(+|I) = 1). However, there are occasionally false positives; 1 person in 1000 who doesn't have the disease (is healthy, H) will test positive anyway (P(+|not I) = 10^-3). We want to know the probability that someone who has a positive test is actually ill.

Let's replace B in Bayes' rule with ``is ill'' (I) and A with ``tests positive'' (+). Then

P(I|+) = P(+|I) P(I) P(+) .

(2)

We know P(+|I) (=1) and P(I) ( = 10^-4), but we have to figure out P(+), the overall probability of testing positive. This is

P(+) = (p(ill È+)) + (p(not ill È+)),

(3)

according to the rule that if A and B are mutually exclusive (you can't be both ill and not ill) P(A ÈB) = P(A)+P(B). We can then say

P(+) = P(I) P(+|I) + (1-P(I)) P(+|not I)

(4)

by the rule that P(A ÈB) = P(A) P(B|A). Putting it all together,

P(I|+) = P(+|I) P(I) P(I) P(+|I) + (1-P(I))P(+|not I) = 1 ×10-4 1 ×10-4 + (1-10-4) ×10-3 » 10-4 10-3 = 1 10 .

Even though false positives are rare, the chance of being ill if you test positive is still only 10%!