A system may be in four possible states: excellent, good, fair, or poor. You per

Question

A system may be in four possible states: excellent, good, fair, or poor. You perform a diagnostic test on the system. The test results can be positive, negative or inconclusive. Here are the relevant probabilities: The probability that the test result is inconclusive is 0.2, regardless of the state. For example, if the actual state is fair, then there is a 70% chance of a negative test result and a 20% chance of an inconclusive test result. Before the diagnostic test is performed, the probability that the state is excellent is 35%, the probability that the state is good is 40%, and the probability that the state is fair is 15%. Answer the following three questions: A diagnostic test is performed and the test result is negative. Find the probability that the state is good or excellent. Suppose that the diagnostic test is performed again. The second test result is positive. Find the (revised) probability that the state is good or excellent. Repeat part (a) for the case when the test result is inconclusive. Prove that the probabilities of the state after the test are the same as the probabilities before the test. Use symbols rather than numbers as much as possible - this makes the proof easier to understand.

Explanation / Answer

Let E shows the event that state is excellent, G shows the event that state is good, F shows the event that state is fair and P shows the event that states is poor. So we have

P(E) = 0.35, P(G) = 0.40, P(F) =0.15, P(P)= 1 -0.35 -0.40 -0.15 = 0.10

Let A shows the event that test gives positive result , I shows the event that test gives inconclusive result and N shows the event that test gives negative result. Here we have

P(N|E) = 0.10, P(N|G) = 0.35, P(N|F) = 0.70, P(N|P) = 0.75

and

P(I|E) = 0.20, P(I|G) = 0.20, P(I|F) = 0.20, P(I|P) = 0.20

By the complement rule we have

P(A|E) =1- 0.10-0.20=0.30, P(A|G) =1- 0.35-0.20=0.45, P(A|F) =1- 0.70-0.20=0.10, P(A|P) =1- 0.75-0.20=0.05

By the law of total probability we have

P(N)=P(N|E)P(E) +P(N|G)P(G)+P(N|F)P(F)+P(N|P)P(P)= 0.10*0.35+ 0.35*0.40+ 0.70*0.15+ 0.75*0.10 = 0.355

(a)

By the Baye's theorem we have

P(G|N) = [P(N|G)P(G)] / P(N)= [0.35*0.40]/0.355 = 0.3944

P(E|N) = [P(N|E)P(E)] / P(N)= [0.35*0.10]/0.355 = 0.0986

So required probability is

P(G or E |N) = P(G|N)+P(E|N) = 0.3944 +0.0986 = 0.4930

(b)

By the law of total probability we have

P(A)=P(A|E)P(E) +P(A|G)P(G)+P(A|F)P(F)+P(A|P)P(P)= 0.30*0.35+ 0.45*0.40+ 0.10*0.15+ 0.05*0.10 = 0.305

By the Baye's theorem we have

P(G|A) = [P(A|G)P(G)] / P(A)= [0.45*0.40]/0.305 = 0.5902

P(E|A) = [P(A|E)P(E)] / P(A)= [0.35*0.30]/0.355 = 0.3443

So required probability is

P(G or E |A) = P(G|A)+P(E|A) = 0.5902+.3443 = 0.9345

(c)

By the complement rule we have

P(I) = 1 - P(A)-P(E) = 1-0.355-0.305 = 0.34

Now

P(G|I) = [P(I|G)P(G)] / P(I)= [0.20*0.40]/0.34 = 0.2353

P(E|I) = [P(I|E)P(E)] / P(I)= [0.35*0.20]/0.34 = 0.2059

So required probability is

P(G or E |I) = P(G|I)+P(E|I) = 0.2353 +0.2059 = 0.4412

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