Respuesta :
Using conditional probability, it is found that the probability a person who tests positive actually has the disease is 0.087 = 8.7%.
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.
For this problem, the events are given as follows:
- Event A: Positive test.
- Event B: Person has the disease.
The percentages associated with a positive test is:
- 95% of 0.5%(person has the disease).
- 5% of 99.5%(person does not have the disease).
Hence:
P(A) = 0.95 x 0.005 + 0.05 x 0.995 = 0.0545.
The probability of both a positive test and having the disease is:
[tex]P(A \cap B) = 0.95 \times 0.005 = 0.00475[/tex]
Hence the conditional probability is:
P(B|A) = 0.00475/0.0545 = 0.087.
The probability a person who tests positive actually has the disease is 0.087 = 8.7%.
More can be learned about conditional probability at https://brainly.com/question/14398287
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