Question 1207328
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).<br>
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. <br>
P(A) is the probability that a person has the disease, which is given in the problem as 0.8%=0.008.<br>
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.<br>
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%.