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Suppose that for the general population, 1 in 5000 people carries the human immunodeficiency virus (HIV). A test for the presence of HIV yields either a positive (+) or negative (-) response. Suppose the test gives the correct answer 99% of the time. What is P[-|H], the conditional probability that a person tests negative given that the person does have the HIV virus? What is P[H|+], the conditional probability that a randomly chosen person has the HIV virus given that the person tests positive?

Respuesta :

Answer:

The answer is "0.019".

Explanation:

For HIV positive: [tex]P[HIV{+}] =\frac{1}{5000}[/tex]

For HIV negative: [tex]P[HIV{-}] =\frac{4999}{5000}[/tex]

calculating the proability for test gives right reslut: [tex]P[TV] =\frac{99}{100}[/tex]

calculating the proability for test gives wrong reslut: [tex]P[TX] =\frac{1}{100}[/tex]

For HIV negative: [tex]P[HIV{-}] = p[TX]= \frac{1}{100}[/tex]

calculating proability to have HIV:

[tex]P[HIV{+}] = \frac{P[HIV{-} \times HIV{-}]}{P{+}}[/tex]

               [tex]= \frac{\frac{1}{5000} \times \frac{99}{100}}{\frac{1}{5000} \times \frac{99}{100}+\frac{4999}{5000} \times\frac{1}{100}}\\\\= \frac{\frac{1}{5000} \times 0.99}{\frac{1}{5000} \times 0.99+\frac{4999}{5000} \times0.01}\\\\= \frac{\frac{0.99}{5000} }{\frac{0.99}{5000} +\frac{49.99}{5000}}\\\\= \frac{0.000198}{0.000198 +0.009998}\\\\= \frac{0.000198}{0.010196}\\\\=0.019[/tex]

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