Answer:
about 29%
Step-by-step explanation:
You want the probability that a person who tests positive for a disease actually has it, if the incidence rate is 0.9%, the false negative rate is 8%, and the false positive rate is 2%.
Conditional probability
The probability of interest is ...
P(has|tests+) = P(has & tests+)/P(tests +)
Of the 0.9% who have the disease, 8% will test negative, and the other 92% will test positive:
P(has & tests+) = 0.9% × 92% = 0.00828
Of the 99.1% who do not have the disease, 2% will test positive:
P(has not & tests+) = 99.1% × 2% = 0.01982
Positive
Then the total fraction who will test positive is ...
P(tests+) = P(has & tests+) +P(has not & tests+)
P(tests+) = 0.00828 +0.01982 = 0.02810
Now we can fill in the values in the formula above:
P(has | tests+) = 0.00828/0.02810 ≈ 0.2947
P(has | tests+) ≈ 29%
