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Bayes' Rule. Suppose B1, . . . , Bn are disjoint events each of which with positive probability and whose union is the entire sample space. Then for any event A and any j {1, . . . , n} we have
 

Summary: Bayes' Rule. Suppose B1, . . . , Bn are disjoint events each of which with positive probability and whose
union is the entire sample space. Then for any event A and any j {1, . . . , n} we have
P(Bi|A) =
P(A|Bj)P(Bj)
n
i=1 P(A|Bi)P(Bi)
.
Proof. We have already shown that the denominator of the right hand side equals P(A). Then numerator
equals P(A Bi).
So you don't get prize for proving Bayes' Rule. It's interesting because of the way it can be applied.
Example 3d, page 72. A laboratory blood test is 95 percent effective in detecting a certain disease when
it is, in fact, present. However, the test also yields a "false positive" result for 1 percent of the healthy
persons tested. If .5 percent of the population actually has the disease, what is the probability that a person
has the disease given that the test is positive?
Solution. Let B1 be the event that a person has the disease, let B2 be the event that a person does not
have the disease and let A be the event that the test result is positive. Then
P(B1) = .005, P(B2) = 1 - P(B1) = .995,
P(A|B1) = .95, P(A|B2) = .01
so
P(B1|A) =

  

Source: Allard, William K. - Department of Mathematics, Duke University

 

Collections: Mathematics