IV A diagnostic test is used by a clinic to determine whether or not a person ha

Question

IV A diagnostic test is used by a clinic to determine whether or not a person has a certain disease. If the test is positive (T) the person is assumed to have the disease, while if the test is negative (Tc) the person is assumed not to have the disease. Let E1 be the event that a person has the disease. Suppose P(E1) = 0.2, P(T|E1) = 0.95, and P(Tc|E2)= 0.90 where E2 is “not E1”. (1) Find the probability that the test is positive. (6 points)

(2) A mistaken diagnosis occurs when either (i) the test is negative and the person has the disease or (ii) the test is positive and the person does not have the disease. What is the probability that the clinic will make a mistaken diagnosis? (6 points)

(3) Find the probabilities P(E1|T), P(E1|Tc), P(E2|T), and P(E2|Tc). (8 points)

Explanation / Answer

(a)

By the complement rule we have

P(E2) = 1 - P(E1) = 0.8

P(T|E2) = 1- P(Tc|E2) = 0.10

By the law of total probability the probability that the test is positive is

P(T) = P(T|E1)P(E1) + P(T|E2)P(E2) = 0.95*0.2 + 0.10 * 0.8 = 0.27

(2)

P(Tc and E1) = P(Tc|E1)P(E1) = 0.90 * 0.20 = 0.18

P(T and E2) = P(T|E2)P(E2) = 0.10 * 0.80 = 0.08

So the probability that the clinic will make a mistaken diagnosis is

P(Tc and E1)+P(T and E2)=0.26

(3)

P(T) = P(T|E1)P(E1) + P(T|E2)P(E2) = 0.95*0.2 + 0.10 * 0.8 = 0.27

P(E1|T) = [P(T|E1)P(E1)] / P(T) = [0.95*0.2] / 0.27 = 0.7037

By the complement rule we have

P(Tc) = 1 - P(T)=1-0.27 = 0.73

P(E1|Tc) = [P(Tc|E1)P(E1)] / P(Tc) = [0.05*0.2] / 0.73 = 0.0137

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By the complement rule

P(E2|T)=1-P(E1|T) =1- 0.7037 = 0.2963

P(E2|Tc)=1-P(E1|Tc) =1- 0.0137 = 0.9863

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