The probability that an individual randomly selected from a particular populatio
ID: 3069076 • Letter: T
Question
The probability that an individual randomly selected from a particular population has a certain disease is 0.04. 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 pos tenor probability that the selected individual has the disease? Hint: Tree diagram with first-dene ration branches comes ondina to sease and No Disease, and second and th generation branches corresponding to results of the two tests.Explanation / Answer
p(both time positive)=P(have disease and both time posiitve)+P(not have disease and both time posiitve)
=0.04*0.93*0.93+(1-0.04)*(1-0.95)*(1-0.95)=0.036996
therefore P(have disease|both test posiitve)
=P(have disease and both time posiitve)/P(both test positive)
=0.04*0.93*0.93/0.036996=0.935128
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