Question 1061916
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.

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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).