   SEARCH HOME Math Central Quandaries & Queries  Question from baaba, a student: Assume that the test for some disease is 99% accurate. If somebody tests positive for that disease, is there a 99% chance that they have the disease? There is not enough information: There are four probabilities that are relevant to the discussion. They are pretty much unrelated:

Among all those people who are known to have the disease, the test has a probability P1 of correctly providing a positive result.

Among all those people who are known not to have the disease, the test has a probability P2 of correctly giving a negative result.

The probabilities P1 and P2 apply to completely different groups of people and are therefore completely unrelated -- the goal of developing accurate tests is to have both P1 and P2 close to 100%. If P1 is too low, then many people who have the disease will be given a false sense of security; if P2 is too low, then many people will be needlessly worried (or worse, given unneeded medical treatment). When speaking about the accuracy of a test you must separately give both P1 and P2. (Otherwise the word "accuracy" would have no meaning.)

Among all those people who test positive, there is a probability P3 of having the disease.

Among all those people who test negative, there is a probability P4 of not having the disease.

To determine P3 and P4 from a knowledge of P1 and P2, one generally must also know the proportion of those having the disease among everybody who is tested.

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     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.