A rapid field immunoassay screens groundwater samples for total coliform. Historically, 6% of samples drawn from the well field are contaminated. The assay returns a positive result for 95% of contaminated samples and for 4% of uncontaminated samples. The probability that a sample returning a positive result is actually contaminated is most nearly:
- (A)0.057
- (B)0.095
- (C)0.603
- (D)0.950
Show worked solution
Answer: (C)
About 60 percent of positive results are true detections, because at 6 percent prevalence the uncontaminated 94 percent of samples yields nearly as many false positives as there are true ones.
Positive results arise from two disjoint populations, so the conditioning denominator is the total positive rate, not the false-positive rate alone.
Bayes' theorem reweights the contaminated branch against that total to reverse the direction of conditioning.
Prevalence dominates the result: the clean fraction contributes 0.0376 of false-positive mass against only 0.0570 of true-positive mass, so nearly two of every five positives are spurious.
FE Reference Handbook — Engineering Probability and Statistics: Bayes' Theorem
Why the other choices appear
- (A)This is the joint probability P(C and +) = (0.95)(0.06) = 0.0570, reported without dividing by the total positive rate.
- (B)This is the marginal probability of a positive result, P(+) = 0.0946, which is only the denominator of the Bayes expression.
- (D)This is the sensitivity P(+|C) = 0.95, obtained by reversing the direction of conditioning.