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.

<h3>What is Conditional Probability?</h3>

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

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ 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:

$P(A) = 0.99(0.1) + 0.02(0.9) = 0.117$

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

$P(A \cap B) = 0.99(0.1)$

Hence, the conditional probability is:

$P(B|A) = \frac{0.99(0.1)}{0.117} = 0.8462$

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 brainly.com/question/14398287

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2 years ago