1.8% of a population is infected with a certain disease.There is a test for the disease; however,the test is not completely accurate.94.2% of those who have the disease will test positive.However 4.8% of those who do not have the disease will also test positive (false positives) .What is the probability that a person who tests positive actually has the disease?

A) 0.360
B) 0.265
C) 0.942
D) 0.018
E) 0.735

Question

Matthew McKendrick · Accepted Answer

Final Answer : B, Explanation :  We can use Bayes' theorem to solve this problem.Let A be the event that a person has the disease, and B be the event that the person tests positive.We want to find P(A|B), the probability that a person actually has the disease given that they test positive.From Bayes' theorem, we have:P(A|B) = P(B|A)P(A) / [ P(B|A)P(A) + P(B|not A)P(not A) ]P(B|A) = 0.942 (given)P(A) = 0.018 (given that 1.8% of the population has the disease)P(B|not A) = 0.048 (given)P(not A) = 0.982 (complement of A)Substituting these values, we get:P(A|B) = 0.942 x 0.018 / [ 0.942 x 0.018 + 0.048 x 0.982 ]P(A|B) = 0.265Therefore, the probability that a person who tests positive actually has the disease is 0.265 or 26.5%. Option B is correct.

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