Test for TuberculosisDate: 07/28/97 at 14:21:33 From: Rachel Strickland Subject: Probability A test for tuberculosis was given to 1000 subjects, 8 percent of whom were known to have tuberculosis. For the subjects who had tuberculosis, the test indicated tuberculosis in 90 percent of the subjects,was inconclusive for 7 percent and negative for 3 percent. For the subjects who did not have tuberculosis, the test indicated tuberculosis in 5 percent, was inconclusive for 10 percent, and was negative for the remaining 85 percent. What is the probability of a randomly selected person having tuberculosis given that the test indicates tuberculosis? Of not having tuberculosis given that the test was inconclusive? Date: 07/28/97 at 17:11:57 From: Doctor Anthony Subject: Re: Probability This is a Bayesian probability problem, in that we are given that the test result is known. i.e. positive test result in first instance. For this situation, we draw up a table as shown: Have T.B.(8%) Do not have T.B.(92%) ---------------------------------------------- .08 x .90 .92 x .05 Test is positive ---------------------------------------------- .08 x .07 .92 x .10 Test inconclusive ---------------------------------------------- .08 x .03 .92 x .85 Test negative ---------------------------------------------- Now we can read off probabilities knowing that the sample space is restricted to just one line of the table. (1) Probability that person has T.B. given that test is positive .08 x .90 .072 36 = --------------------- = ------------- = ---- = 0.6102 .08 x .90 + .92 x .05 .072 + .046 59 (2) Probability of having T.B. given that test is inconclusive .08 x .07 .0056 7 = -------------------- = -------------- = ----- = 0.057377 .08 x .07 + .92 x .10 .0056 + .092 122 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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