Can someone please answer this question I already asked once and the other perso
ID: 3275783 • Letter: C
Question
Can someone please answer this question I already asked once and the other person couldnt give me the correct answer. I have no idea how to do this problem.
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.]
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Explanation / Answer
P(has disease | 2 positive tests) = P(has diseas and 2 positive test)/P(two positive test)
p(two positive tests) = P(has disease and 2 positive tests) + P(no disease and two positive tests)
= (0.05x0.93x0.93) + (0.95x0.05x0.05)
= 0.043245 + 0.002375
= 0.04562
P(has disease | 2 positive tests) = 0.043245/0.04562
= 0.9479