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| 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.
 Found 2 solutions by  Shin123, math_tutor2020:
 Answer by Shin123(626)
      (Show Source): 
You can put this solution on YOUR website! 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). 
 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.
 
 P(A) is the probability that a person has the disease, which is given in the problem as 0.8%=0.008.
 
 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.
 
 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%.
Answer by math_tutor2020(3817)
      (Show Source): 
You can put this solution on YOUR website! Terms to know:
 Positive = the test claims the person has the disease (the test's claim may be true or false)Negative = the test claims the person does NOT have the disease (the test's claim may be true or false)False negative = when the test says "negative", but the person actually has the diseaseFalse positive = when the test says "positive", but the person does NOT actually have the disease
 
 Here's a chart to help remember
 
 
|  | Tests Positive | Tests Negative |  | Has Disease | Correct Outcome | False Negative |  | Does Not Have Disease | False Positive | Correct Outcome |  
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 Consider a population of 100,000 people.
 
 "A certain disease has an incidence rate of 0.8%" will mean that:
 0.8% of 100,000 = (0.8/100)*100000 = 800 people have the disease.
 The remaining 100000 - 800 = 99,200 people do not have the disease.
 
 We are told that "the false negative rate is 6%".
 Of the 800 people who have the disease, 6% of them will get a false negative.
 The test mistakenly says to these unfortunate people "no you don't have the disease" when it should say "yes you do have the disease".
 6% of 800 = 0.06*800 = 48 people will get a false negative when it should say "positive".
 The other 800-48 = 752 people with the disease get the proper positive test result.
 
 We are also told that "the false positive rate is 2%"
 It means 2% of the 99,200 people who do not have the disease, will get back erroneous results of "positive" when instead it should say "negative".
 2% of 99200 = 0.02*99200 = 1984 people will get false positives and 99200 - 1984 = 97216 people will get correct negative test results.
 
 Here's a chart summarizing the values.
 
 
|  | Tests Positive | Tests Negative | Total |  | Has Disease | 752 | 48 | 800 |  | Does Not Have Disease | 1984 | 97,216 | 99,200 |  | Total | 2736 | 97,264 | 100,000 |  
 Based on the chart, there are 2736 people who tested positive.
 Of this subgroup, 752 have the disease.
 752/2736 = 0.27485380117 is the approximate probability of someone actually having the disease if they test positive.
 
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