Question 1061916:  Hi Tutors,
 
I have another math problem that I have solved but is asking whether it's correct. I would appreciate it! 
 
 
A certain 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(A|B). Round your answer to the nearest tenth of a percent and do not include a percent sign.
 
P(A|B)= 0.0045
 
 
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') = 0.896 
 
 Answer by math_helper(2461)      (Show Source): 
You can  put this solution on YOUR website! I tried the problem and my answers differ from yours.  You can compare to mine to see where you (or I) went wrong.  I'm a little rusty on Bayes' Theorem problems. 
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Bayes' Theorem: 
 P( A|B ) = P(B|A) * P(A)/P(B)  
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P(A) = Pr{person is infected} 
P(B) = Pr{person tests positive} 
P(B|A) = 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(B|A) = 0.90 
 
P(A|B) = 0.90 * 0.005 / 0.104 = 0.0433  
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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 false-positive 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     ( = 1-P(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 
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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).
 
 
 
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