Suppose that a patient is being tested for a disease and it is known that 1%...
Suppose that a patient is being tested for a disease and it is known that 1% of population have the disease. Suppose also that the patient tests positive and that the test is 95% accurate. Let D be the event that the patient has the disease and T the event that the tests positive. Then we know P(T|D) = P(T’|D’) = 0.95. Using Baye’s theorem and the Law of Total Probability, determine the prbability that the patient actually has the disease, i.e. , P(D|T)
Homework Answers
here P(T) =P(D)*P(T|D)+P(D')*P(T|D') =0.01*0.95+(1-0.01)*(1-0.95)=0.059
hence P(D|T) =P(D)*P(T|D)/P(T) =0.01*0.95/0.059=0.161017
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