The probability that an individual randomly selected from a particular populatio
ID: 3301347 • Letter: T
Question
The probability that an individual randomly selected from a particular population has a certain disease is 0.05. A diagnostic test correctly detects the presence of the disease 93% of the time and correctly detects the absence of the disease 95% of the time. If the test is applied twice, the two test results are independent, and both are positive, what is the (posterior) probability that the selected individual has the disease? [Hint: Tree diagram with first-generation branches corresponding to Disease and No Disease, and second- and third-generation branches corresponding to results of the two tests.]
Explanation / Answer
Probability that a randomly selected individual has a certain disease P(D) = 0.05 and P(D') = 0.95
Probability that a diagnostic test correctly detects the presence of the disease, P(P|D) = 0.93 and P(P'|D) = 0.07
Probability that a diagnostic test correctly detects the absense of disease, P(P'|D') = 0.95 and P(P|D') = 0.05
Required probability that the selected individual has the disease, P(D|P) = P(P|D)*P(D) / (P(P|D)*P(D) + P(P|D')*P(D')) = 0.93*0.05 / (0.93*0.05 + 0.05*0.95) = 0.4947
The above probability is for the first test, similarly for the second test the probability will be same 0.4947
hence the required probability is 0.4947*0.4947 = 0.2447
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