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Consider the probabilities of people taking pregnancy tests. Assume that the true probability of pregnancy for all people who take the test is 10%. If pregnancy tests give a positive result in 99% of the cases where the woman is pregnant and give a negative result in 98% of the cases in which she is not pregnant, what is the probability that a woman who gets a positive test result is truly pregnant

Respuesta :

Using conditional probability, it is found that there is a 0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

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 result.
  • Event B: Pregnant.

The probability of a positive test result is composed by:

  • 99% of 10%(truly pregnant).
  • 2% of 90%(not pregnant).

Hence:

[tex]P(A) = 0.99(0.1) + 0.02(0.9) = 0.117[/tex]

The probability of both a positive test result and pregnancy is:

[tex]P(A \cap B) = 0.99(0.1)[/tex]

Hence, the conditional probability is:

[tex]P(B|A) = \frac{0.99(0.1)}{0.117} = 0.8462[/tex]

0.8462 = 84.62% probability that a woman who gets a positive test result is truly pregnant.

You can learn more about conditional probability at https://brainly.com/question/14398287