(1) The Food and Drug Administration (FDA) states that the Wellcome Elisa test for HIV (human immunodeficiency virus, which causes AIDS) is positive 99.3 percent of the time among those persons who carry the HIV virus (called the sensitivity of the test). Furthermore, FDA has concluded that the test is negative 99.99 percent of the time on those persons who do not have the HIV virus (called the specificity of the test). State these results as conditional probabilities. Let T+ stand for the event that the test is positive and let T, stand for the event that the test is negative. Similarly, Let HIV+ stand for the event that the person carries the HIV virus and HIV- stand for the event that the person does not carry the HIV virus. With this notation, the following answers are proposed. (a) P(T + |HIV+) = 0.993, (b) P(T + |HIV–) = 0.993, (c) P(T – |HIV+) = 0.993, (d) P(T + |HIV+) = 0.993, (e) P(T + |HIV+) = 0.993, The correct answer is P(T – |HIV–) = 0.9999. P(T – |HIV–) = 0.9999. P(T – |HIV–) = 0.9999. P(T + |HIV-) = 0.9999. P(T – |HIV+) = 0.9999. (a) (b) (c) (d) (e) N/A
(1) The Food and Drug Administration (FDA) states that the Wellcome Elisa test for HIV (human immunodeficiency virus, which causes AIDS) is positive 99.3 percent of the time among those persons who carry the HIV virus (called the sensitivity of the test). Furthermore, FDA has concluded that the test is negative 99.99 percent of the time on those persons who do not have the HIV virus (called the specificity of the test). State these results as conditional probabilities. Let T+ stand for the event that the test is positive and let T, stand for the event that the test is negative. Similarly, Let HIV+ stand for the event that the person carries the HIV virus and HIV- stand for the event that the person does not carry the HIV virus. With this notation, the following answers are proposed. (a) P(T + |HIV+) = 0.993, (b) P(T + |HIV–) = 0.993, (c) P(T – |HIV+) = 0.993, (d) P(T + |HIV+) = 0.993, (e) P(T + |HIV+) = 0.993, The correct answer is P(T – |HIV–) = 0.9999. P(T – |HIV–) = 0.9999. P(T – |HIV–) = 0.9999. P(T + |HIV-) = 0.9999. P(T – |HIV+) = 0.9999. (a) (b) (c) (d) (e) N/A
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 11ECP: A manufacturer has determined that a machine averages one faulty unit for every 500 it produces....
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