document.write( "Question 1207328: A certain disease has an incidence rate of 0.8%. If the false negative rate is 6% and the false positive rate is 2%, compute the probability that a person who tests positive actually has the disease. \n" ); document.write( "
Algebra.Com's Answer #845139 by Shin123(626)\"\" \"About 
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Let A be the event that a person has the disease. Let B be the event that a person tests positive for the disease. Then, we are trying to calculate P(A|B). By Baye's Theorem, this equals P(B|A)*P(A)/P(B).

\n" ); document.write( "P(B|A) is the probability that a person tests positive for the disease given they have it. Note that P(~B|A), the probability that person tests negative for the disease given that they have it is 6%, the false negative rate. Therefore, P(B|A)=1-P(~B|A)=1-0.06=0.94.

\n" ); document.write( "P(A) is the probability that a person has the disease, which is given in the problem as 0.8%=0.008.

\n" ); document.write( "P(B) is the probability that a person tests positive for the disease. There is a 0.008 chance that a person actually has the disease. In that case, there is a 0.94 chance that they also test positive (1-false negative rate). There is a 0.992 chance that a person doesn't doesn't actually have the disease. In that case, there is a 0.02 chance the person tests positive (false positive rate). In total, since the events are mutually exclusive, it is 0.008*0.94+0.992*0.02=0.02736.

\n" ); document.write( "Putting all of this together, we have P(A|B)=P(B|A)*P(A)/P(B)=0.94*0.008/0.02736, which is approximately 27.4854%.
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