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Probability in Virus Testing

Date: 06/04/99 at 17:52:08
From: Bernie Warner
Subject: Stats (Bayes Theorem)

In a country, 1 person in 10,000 (.01%) has the EB virus. A test that 
identifies the disease gives a positive indication 99% of the time 
when an individual has the disease. However, 2% of the time it gives 
a positive indication when a person doesn't have the disease.

a) If a randomly selected individual is tested and the test turns out 
positive, what is the chance that this individual has the virus?

b) Construct a 2 by 2 table.

I am having trouble constructing the table. I am also not sure what I 
am supposed to do with the numbers to arrive at the answer for (a).

Date: 06/04/99 at 18:38:06
From: Doctor Anthony
Subject: Re: Stats (Bayes Theorem)

    Initial Prob.          Initial Prob. 
    Has the virus          Does not have virus
    Prob = 0.0001          Prob = 0.9999
    0.00001 x 0.99         0.9999 x 0.02         Test is positive
      = 0.0000099            = 0.019998

    0.00001 x 0.01         0.9999 x 0.98         Test is negative 
      = 0.0000001            = 0.979902

Since we are told that the test is positive, the sample space is 
confined to the top row.

   Prob has virus = --------------------  = 0.0004948
                    0.0000099 + 0.019998

With such a low probability of a person actually having the disease 
given a positive result, the test itself is practically worthless.

- Doctor Anthony, The Math Forum   
Associated Topics:
High School Probability

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