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<math>P (+ \cap HIV) = P (+ | HIV) P (HIV)\!</math> and <math>P(+ \cap no HIV) = P (+ | no HIV) P (not HIV)\!</math>
 
<math>P (+ \cap HIV) = P (+ | HIV) P (HIV)\!</math> and <math>P(+ \cap no HIV) = P (+ | no HIV) P (not HIV)\!</math>
 +
 +
 +
Thus, we can conclude that:
 +
 +
 +
<math>P(+) = P (+ | HIV) P (HIV) + P (+ | no HIV) P (not HIV)\!</math>
 +
 +
 +
This can easily be solved with the given information.
 +
 +
 +
<math>P (+ | HIV) P (HIV) = 0.9 (0.005)\!</math> and <math>P (+ | no HIV) P (not HIV) = 0.1 (0.995)\!</math>

Revision as of 07:12, 5 October 2008

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)\! $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva