The conditional probability that the person has the disease given that the test result is positive is of 0.4750 = 47.50%.
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:
In this problem, the events are:
The percentages associated with a positive test is:
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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