Suppose that we wish to determine whether a rare, but very costly, flaw is prese

Question

Suppose that we wish to determine whether a rare, but very costly, flaw is present (event F). Let us assume that P(F) = 0.01; thus, P(F) = 0.99. A fairly simple procedure is proposed to test for this flaw. However, the test is preliminary, as the probabilities of reaching the wrong conclusion are large. About 5 percent of the time, the test indicates a flaw (TF) when no flaw is present; and about 3 percent of the time, it indicates the absence of a flaw when a flaw is present. That is, P(TF | F) = 0.05, but P(TF | F) = 0.03. Show that the probability that the test indicates a flaw is P(TF) = 0.0592. Show that the posterior probability that there is no flaw, given that the test has indicated a flaw, is P(F|TF) = 0.8361.

Explanation / Answer

Solution : 2.2-15) Given -

P(F) = 0.01

P(F') = 0.99

P(TF|F') = 0.05

P(TF'|F) = 0.03

Proof -

a) P(TF|F') = P(TF)/P(F')

0.05 = P(TF)/ 0.99

P(TF) = 0.05 x 0.99 = 0.0495

Hence the value is near to the flaw of test we can say that P(TF) value is true.

b) P(F'|TF) = P(F') / P(TF)

= 0.99 / 0.0495 = 20 > 0.8361

Hence we can say that there is no flaw.

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