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218 Revising judgments in the light of new information
Events
0.7
Component isOK
Component is defective
New information
Prior probabilities Conditional probabilities
0.8
Component passes test
Component failstest
0.2
0.3 0.1
Component passes test
Component fails test
0.9
Joint probabilities
0.7 × 0.2 = 0.14
0.3 × 0.9 = 0.27
p(component fails test) = 0.41
Figure 8.2 – Applying Bayes’ theorem to the components problem
that a component will fail the test given that it is ‘OK’ is 0.2 (i.e.
p(fails test|‘OK’) = 0.2). Given that our component did fail the test, we
are not interested in the branches labeled ‘component passes test’ and in
future diagrams we will exclude irrelevant branches.
We now calculate the probability of a component failing the test. First
we determine the probability that a component will be ‘OK’ and will
fail the test. This is, of course, a joint probability and can be found by
applying the multiplication rule:
p(OK and fails test) = 0.7 × 0.2 = 0.14
Next we determine the probability that a component will be defective
and will fail the test:
p(defective and fails test) = 0.3 × 0.9 = 0.27
Now a component can fail the test either if it is ‘OK’ or if it is defective,
so we add the two joint probabilities to obtain:
p(fails test) = 0.14 + 0.27 = 0.41
The posterior probability is then found by dividing the appropriate
joint probability by this figure. Since we want to determine the posterior