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Sagot :
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
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