markosheehan Posted May 15, 2016 Share 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). Link to comment Share on other sites More sharing options...
andrewcellini Posted May 15, 2016 Share 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 Link to comment Share on other sites More sharing options...
andrewcellini Posted May 18, 2016 Share Posted May 18, 2016 Did you end up figuring it out? Link to comment Share on other sites More sharing options...
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