Revision as of 07:19, 5 October 2008 by Vhsieh (Talk)

The Problem:

There is a new test that will test for the HIV virus, but we are not sure of whether this test's results are usually correct or not. We are given the following information:


$ P (+ | HIV) = 0.9, P (- | HIV) = 0.1\! $ $ P (+ | no HIV) = 0.1, P (- | no HIV) = 0.9\! $


We are also given that only $ 0.5%\! $ of the population has the HIV virus. The rest do not. Are the results usually correct, and what can you tell from the results?


The Solution:

First, we determine the probability that a random person within the population tests positive. Remember, this person will be selected at random, so we have no clue whether or not he/she actually has the virus or not.


$ P(+) = P (+ \cap HIV) + P(+ \cap no HIV)\! $


From previous notes in this class, we know that the following is true:


$ P (+ \cap HIV) = P (+ | HIV) P (HIV)\! $ and $ P(+ \cap no HIV) = P (+ | no HIV) P (not HIV)\! $


Thus, we can conclude that:


$ P(+) = P (+ | HIV) P (HIV) + P (+ | no HIV) P (not HIV)\! $


This can easily be solved with the given information.


$ P (+ | HIV) P (HIV) = 0.9 (0.005)\! $ and $ P (+ | no HIV) P (not HIV) = 0.1 (0.995)\! $


We can add both of those products together to obtain the probability that our random person tests positive, which turns out to be a $ 0.104\! $ probability. From this, we can already tell that our test is not that great because $ 0.5%\! $ of the people have HIV, but a much larger number of people will test positive.


But just because the test results are not always good, doesn't mean we can't tell anything from the results. Let us calculate the probability that, given a selected individual has the HIV virus, they will test positive. Similarly, we will find the probability that, given the person does not have the virus, that they will test negative.

<math<P (HIV | +) = (P (HIV \cap +)) / (P (+)) = (P (+ | HIV) P (HIV)) / (P (+)) = (0.9 (0.005)) / 0.104 = 0.0005\!</math>

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood