>>11645664Wrong. Ill reiterate this very specifically so you can see the flaw in your reasoning.
A person can either be healthy, H, or infected, H'
A person can either test positive, +, or negative, +'
The test having a 99% accuracy means that, 99% of the time, it will return the correct result. Mathematically, this is:
Positive given the person is healthy, P(+|H) = .01
Negative given the person is healthy, P(+'|H) = .99
Positive given the person is infected, P(+|H') = .99
Negative given the person is infected, P(+'|H') = .01
You are told that you are tested positive. You want to find the probability that you are infected, GIVEN that you are positive, or P(H'|+)
P(H'|+) does NOT necesarilly equal P(+|H')
To find P(H'|+) you can use bayes formula (the derivation of which is left as an exercise to the reader)
P(H'|+) = [P(+|H') * P(H')] / [P(+)]
With P(+) being the probability that a person is infected, or [P(+|H') * P(H')] + [P(+|H) * P(H)]
[P(+|H') * P(H')] can also be rewritten as P(+ ^ H'), or the probability that a person is positive AND infected.
In English, this all means the following:
"The probability that a person is infected, given that they have tested positive, is equal to the proportion of the population that has tested positive and is infected, over the proportion of the population that has tested positive"
So then the answer is C. 99%
