markosheehan Posted May 15, 2016 Posted May 15, 2016 hi i cant work this out Its is estimated that 0.3% of a large population have a particular disease. a test developed to detected the disease gives a false positive in 4% of tests and a false negative in 1% of tests. a person is tested positive for the disease. what is the probability that the person actually has the disease? i was told my teacher that the answer is (probability of having the disease)/(probability the test is positive)i can work out the probability of having the disease but i cant work out the probability of the test being positive the probability of having the disease is .003. what is the probability of the test being positive i cant work this out. orginally i added .04 (probability the test is positive) and .99(probability test is positive, i got this from taking 1-.01) but that leaves you with a probability of the test being positive of 1.03 and i dont think this is possible. my teacher told me you get the answer by (probability of having the disease)/(probability the test is positive).
andrewcellini Posted May 15, 2016 Posted May 15, 2016 (edited) Did you guys learn bayes theorem? edit: though you may not need it after all. your teacher is talking about a conditional probability where the | or / in your notation refers to "given __" where after the | are some conditions, P(having disease| positive test), you're given P(not having disease| positive) = .04, this is not the probability of having a positive test. Edited May 16, 2016 by andrewcellini
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