In a certain country, 2% of the population have a certain disease. A diagnostic
ID: 2977656 • Letter: I
Question
In a certain country, 2% of the population have a certain disease. A diagnostic blood test has been developed for the disease. If a person has the disease, the result is positive 99% of the time while if he doesn't has the disease, the result is positive 1% of the time. If a person gets a positive blood test, what is the probability that he has the disease? The same test is used in another country, and it is found that 20% of a sample population tests positive. If we assume this sample is representative, what fraction of the population has the disease?Explanation / Answer
P(positive result) = P(positive result and not infected) + P(positive result and infected) = P(positive result|not infected)*P(not infected) + P(positive result| infected)*P(infected) = = .01*(1-.02)+.99*.02 = .01*.98+.99*.02 = .0296 Then, P(positive|positive result) = P(positive and positive result)/P(positive) = P(positive result| infected)*P(infected)/P(positive) = .99*.02/.0296 = .0198/.0296 = 99/148 is approximately .6689 In country 2, 20% of a population tests positive. Let x be the fraction that are actually positive. Then, from above, we know that P(positive test) = P(positive result|not infected)*P(not infected) + P(positive result| infected)*P(infected) = .01*(1-x)+.99*x = .2 .01 - .01x + .99x = .2 Thus, .98x = .19 x = 19/98 is approximately 0.1938 or 19.38%