Bayes' Theorem: Conditional Probabilities
Source:
http://faculty.vassar.edu/lowry/bayes.html
| P(A) = .005 | the probability that the disease will be present in any particular person
| P(~A) = 1.005 = .995
| the probability that the disease will not be present in any particular person
| P(B|A) = .99
| the probability that the test will yield a positive result [B] if the disease is present [A]
| P(~B|A) = 1.99 = .01
| the probability that the test will yield a negative result [~B] if the disease is present [A]
| P(B|~A) = .05
| the probability that the test will yield a positive result [B] if the disease is not present [~A]
| P(~B|~A) = 1.05 = .95
| the probability that the test will yield a negative result [~B] if the disease is not present [~A]
| |
| P(B) = [P(B|A) x P(A)] + [P(B|~A) x P(~A)]
= [.99 x .005]+[.05 x .995] = .0547 | the probability of a positive test result [B], irrespective of whether the disease is present [A] or not present [~A]
| P(~B) = [P(~B|A) x P(A)] + [P(~B|~A) x P(~A)]
| = [.01 x .005]+[.95 x .995] = .9453 the probability of a negative test result [~B], irrespective of whether the disease is present [A] or not present [~A]
| |
which in turn allows for the calculation of the four remaining conditional probabilities
| P(A|B) = [P(B|A) x P(A)] / P(B)
= [.99 x .005] / .0547 = .0905 | the probability that the disease is present [A] if the test result is positive [B] (i.e., the probability that a positive test result will be a true positive)
| P(~A|B) = [P(B|~A) x P(~A)] / P(B)
| = [.05 x .995] / .0547 = .9095 the probability that the disease is not present [~A] if the test result is positive [B] (i.e., the probability that a positive test result will be a false positive)
| P(~A|~B) = [P(~B|~A) x P(~A)] / P(~B)
| = [.95 x .995] / .9453 = .99995 the probability that the disease is absent [~A] if the test result is negative [~B] (i.e., the probability that a negative test result will be a true negative)
| P(A|~B) = [P(~B|A) x P(A)] / P(~B)
| = [.01 x .005] / .9453 = .00005 the probability that the disease is present [A] if the test result is negative [~B] (i.e., the probability that a negative test result will be a false negative)
| |
Statistics - Conditional Probability
Kaye, D.H. (1999). Burdens of persuasion: what Baysian decision rules do and do not do. International Journal of Evidence and Proof. 3(1):1-28.
Redmayne, M. (1998). Bayesianism and Apriorism. ICE Journal.