3.8% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely accurate. 93.9% of those who have the disease test positive. However 4.1% of those who do not have the disease also test positive (false positives). A person is randomly selected and tested for the disease. What is the probability that the person has the disease given that the test result is positive?

Respuesta :

The conditional probability that the person has the disease given that the test result is positive is of 0.4750 = 47.50%.

What is Conditional Probability?

Conditional probability is the probability of one event happening, considering a previous event. The formula is:

[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, the events are:

  • Event A: Positive test.
  • Event B: Has the disease.

The percentages associated with a positive test is:

  • 93.9% of 3.8%(has the disease).
  • 4.1% of 100 - 3.8 = 96.2%(does not have the disease).

Hence:

[tex]P(A) = 0.939(0.038) + 0.041(0.962) = 0.075124[/tex]

The probability of both a positive test and having the disease is given by:

[tex]P(A \cap B) = 0.939(0.038) = 0.035682[/tex]

Hence the conditional probability is given by:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.035682}{0.075124} = 0.4750[/tex]

More can be learned about conditional probability at https://brainly.com/question/14398287

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