document.write( "Question 1061916: Hi Tutors,\r
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document.write( "I have another math problem that I have solved but is asking whether it's correct. I would appreciate it! \r
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document.write( "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\". \r
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document.write( "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.\r
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document.write( "P(A|B)= 0.0045\r
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document.write( "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.\r
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document.write( "P(A'|B') = 0.896
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Algebra.Com's Answer #676691 by math_helper(2461)![]() ![]() 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. \n" ); document.write( "-- \n" ); document.write( "Bayes' Theorem: \n" ); document.write( " P( A|B ) = P(B|A) * P(A)/P(B) \n" ); document.write( "—\r \n" ); document.write( "\n" ); document.write( "P(A) = Pr{person is infected} \n" ); document.write( "P(B) = Pr{person tests positive} \n" ); document.write( "P(B|A) = Pr{person tests positive given that they are infected}\r \n" ); document.write( "\n" ); document.write( "P(A) = 1/200 = 0.005 \n" ); document.write( "P(B) = (1/200)*0.90 + (199/200)*0.10 = 0.104 \n" ); document.write( "P(B|A) = 0.90 \r \n" ); document.write( "\n" ); document.write( "P(A|B) = 0.90 * 0.005 / 0.104 = 0.0433 \n" ); document.write( "—--\r \n" ); document.write( "\n" ); document.write( "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.\r \n" ); document.write( "\n" ); document.write( "——\r \n" ); document.write( "\n" ); document.write( "A' = event 'person is not infected' \n" ); document.write( "B' = event 'person tests negative' \n" ); document.write( "P(B'|A') = probability of person testing negative given that they are not infected\r \n" ); document.write( "\n" ); document.write( "P(A') = 199/200 = 0.995 \n" ); document.write( "P(B') = 0.896 ( = 1-P(B)) \n" ); document.write( "P(B'|A') = (199/200)(0.896) + (1/200)*0.104) = 0.89204\r \n" ); document.write( "\n" ); document.write( "P(A'|B') = 0.89204*0.995/0.896 = 0.991 \n" ); document.write( "—\r \n" ); document.write( "\n" ); document.write( "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).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |