1. Bayes’ Theorem Bayes’ Theorem allows us to calculate the conditional probability one way (e.g., P(B|A) when we know the conditional probability the other way (P(A|B).

2. Bayes’ Theorem We use the following logic: P(A ∩ B) = P(A) * P(B|A) P(A ∩ B) = P(B) * P(A|B) Thus, P(A) * P(B|A) = P(B) * P(A|B) And P(B|A) = P(B) * P(A|B) P(A)

3. Bayes’ Theorem All that remains is to figure out P(A) for the denominator. In the simplest case, there are two ways A could happen: A and B, or A and B’. P(A) = P(A ∩ B) + P(A ∩ B’) By the formula for conditional probability: P(A) = P(B)*P(A|B) + P(B’)*P(A|B’)

4. Bayes’ theorem P(B|A) = P(B) * P(A|B) P(A) P(A) = P(B)*P(A|B) + P(B)*P(A|B’) This leads us to Bayes’ Theorem: P(B|A) =P(B) * P(A|B) P(B)*P(A|B) + P(B)*P(A|B’)

5. Bayes’ Theorem – Example It is known that at any one time, one person in 200 has a given disease. A new test for this disease is developed and a study of this test’s reliability is performed, using samples of people known to have or known not to have the disease. Of 1000 people known to have the disease, 990 test positive. Of 5000 people known not to have the disease, 250 test positive. What is the probability that a randomly-selected person from the general population has the disease, given that they test positive?

6. Bayes’ Theorem – Example What we know from the question: IllNot Ill Test +990 (99%) 250 (5%) Test – 10 (1%)4750 (95%) 10005000

7. Bayes’ Theorem – Example What is the probability that a randomly- selected person from the general population has the disease, given that they test positive? P(Ill) = 1/200 =.005P(Not Ill) = 1 –.005 =.995 P(T+ |Ill) =.99P(T+|Not Ill) =.05 P(Ill | T+) = P(Ill)* P(T+|Ill) P(Ill)*P(T+|Ill) + P(Not Ill)* P(T+|Not Ill)

8. Bayes’ Theorem – Example P(Ill|T+) = (.005)(.99) (.005)(.099) + (.995)(.05) =.00495 (.00495) + (.04975) =.0905

9. Bayes’ Theorem – Example We can generalize this approach to cases where B consists of 3 or more mutually exclusive, exhaustive events: P(Bi|A) = P(Bi) P(A|Bi) P(B1)P(A|B1)+P(B2)P(A|B2)+ …+ P(Bk)P(A|Bk)