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In a particular country, 3% of the population is HIV positive. A test for HIV correctly
diagnoses 95% of those who have the condition, but incorrectly diagnoses 4% of those
who are not HIV positive as being HIV positive. A patient obtains a positive test result,
apparently indicating the presence of HIV. Determine the probability that the patient
really is HIV positive.

Respuesta :

Using conditional probability, it is found that there is a 0.4235 = 42.35% probability that the patient  really is HIV positive.

Conditional Probability

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

  • P(B|A) is the probability of event B happening, given that A happened.
  • [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
  • P(A) is the probability of A happening

In this problem:

  • Event A: positive test.
  • Event B: HIV positive.

The percentages involving a positive test are:

  • 95% of 3%(positive)
  • 4% of 100 - 3 = 97%(not positive).

Hence:

[tex]P(A) = 0.95(0.03) + 0.04(0.97) = 0.0673[/tex]

The probability of both having a positive test and being HIV positive is:

[tex]P(A \cap B) = 0.95(0.03)[/tex]

Then, the conditional probability is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.95(0.03)}{0.0673} = 0.4235[/tex]

0.4235 = 42.35% probability that the patient  really is HIV positive.

A similar problem is given at https://brainly.com/question/14398287

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