Bayes' Theorem: Conditional Probabilities

Bayes' theorem describes the relationships that exist within an array of simple and conditional probabilities. For example: Suppose there is a certain disease randomly found in one-half of one percent (.005) of the general population. A certain clinical blood test is 99 percent (.99) effective in detecting the presence of this disease; that is, it will yield an accurate positive result in 99 percent of the cases where the disease is actually present. But it also yields false-positive results in 5 percent (.05) of the cases where the disease is not present. The following table shows (in red) the probabilities that are stipulated in the example and (in blue) the probabilities that can be inferred from the stipulated information:
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]

Given this information, Bayes' theorem allows for the derivation of the two simple probabilities
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