Bayes' Theorem
posted by MILLY
I tried the problem and my answers are still coming up wrong?. You can compare to mine to see where you (or I) went wrong. I'm a little rusty on
Virus infects one in every 200 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. (This 10% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive".
a) Find the probability that a person has the virus given that they have tested positive, i.e. find P(AB). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(AB)= .0045
b) Find the probability that a person does not have the virus given that they test negative, i.e. find P(A'B'). Round your answer to the nearest tenth of a percent and do not include a percent sign. P(A'B')
Bayes' Theorem problems.

Bayes' Theorem:
P( AB ) = P(BA) * P(A)/P(B)
—
P(A) = Pr{person is infected}
P(B) = Pr{person tests positive}
P(BA) = Pr{person tests positive given that they are infected}
P(A) = 1/200 = 0.005
P(B) = (1/200)*0.90 + (199/200)*0.10 = 0.104
P(BA) = 0.90
P(AB) = 0.90 * 0.005 / 0.104 = 0.0433
—
Comment: This answer is somewhat surprising. A 4.33% chance that the person has the virus given that they've tested positive. That seems way too low, but that's probably because the 10% false positive rate is pretty high. I think real world tests strive for a much lower falsepositive rate.
——
A' = event 'person is not infected'
B' = event 'person tests negative'
P(B'A') = probability of person testing negative given that they are not infected
P(A') = 199/200 = 0.995
P(B') = 0.896 ( = 1P(B))
P(B'A') = (199/200)(0.896) + (1/200)*0.104) = 0.89204
P(A'B') = 0.89204*0.995/0.896 = 0.991
—
Comment: This answer makes sense. If a person tests negative, it is highly unlikely that they have the virus. Still, I suspect that 9/1000 is still too many false negatives for a real world scenario. I guess the importance of minimizing false negatives depends on how dangerous the virus is (its one thing not to detect if someone has a cold virus, its a different story to not detect they have Ebola, for example).

MathMate
Agree with the first part, except that with probability calculations, I prefer to work with fractions. If a decimal answer is required, it will be rounded at the end.
First Part:
P(A)=1/200
P(+A)=9/10
P(A')=1/10
P(+)=P(+A)+P(+A')
=(1/200)(9/10)+(199/200)(1/10)
=13/125 ~0.104
P(A+)=P(+A)*P(A)/P(+)
=(9/10)(1/200)/(13/125)
=9/208 ~0.043269
as you had it.
Agree completely with your comments. The important factors are
1. high false positive rates, compared to
2. low incidence.
Second Part:
P(A')=(9/10) [given]
P(A')=1P(A)=199/200
P()=P(A)+P(A')
=(1/200)(19/10)+(199/200)(9/10)
=112/125 ~ 0.896
P(A')=P(A')P(A')/P()
=(9/10)(199/200)/(112/125)
=1791/1792 ~0.999442
Agree also with your comments.
False negative here is actually 1/1792~0.00055, which is reasonable. 
Joe
Here is my solution:
Given probabilities: P(A) = 1/200 = 0.005 P(B) = 199/200 = 0.995 P(BA) = 0.9 P(B~A) = 0.1 Infer: P(~BA) = 0.1 P(~B~A) = 0.9 Then:
a.) Find P(AB) P(AB) = P(A)P(BA) = 0.005×0.9 = 0.0045 rounds to 0.0
b.) Find P(~A~B) P(~A~B) = P(~A)P(~B~A) = 0.995×0.9 = 0.896 rounds to 0.9 Note: the symbol ~A means A' in your notation. QED 
Joe
Please ignore my entry: there is an error in the above calc. that I offered.
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