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You should play around with some numbers to see what these four probabilities actually mean. We could have P1 = 100% by simply saying,"YES, you have the disease." to everybody. In this case P2 = 0 and the test provides no information about your probability of having the disease . At the other extreme, theoretically it might be possible to devise a test that is always accurate, so that P1 = P2 = 100% That means if you test positive, then you must have the disease (and P3 = 100%). In the same way, in this case p4 = 100% as well. Of course, things in our imperfect world are seldom certain. So how about when P1 = P2 = 99%. This means that 99% of those with the disease are told that they have the disease, while 1% of those without the disease are told they have the disease. Should 50% of the people being tested have the disease, then this translates to the probability of being tested positive is (.99)(.5) + (.01)(.5) = .5; the probability of both having the disease and being tested positive is (.99)(.5) = .495. Thus, the probability of having the disease among those who have tested positive would be (.495)/(.5) = .99. Thus there would be a P3 = 99% chance that someone who tested positive would have the disease with this scenario. How about a disease that is more rare, say 0.1% of those who are being tested have the disease (and, therefore 99.9% do not have it). Then the probability of being tested positive is .99(.001) + (.01)(.999) = .01098; the probability of both having the disease and being tested positive is (.99)(.001)=.00099. Thus, the probability of having the disease among all those who have tested positive would be (.00099)/(.01098) = .0901639, so that P3 is only about 1%. The short answer to your question is that the accuracy of the test depends on both the probability of a false positive and the probability of a false negative, while the chance that somebody tests positive actually has the disease depends also on the distribution of the disease among those who are being tested. Chris  


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