Screening Test: One Positive, Two Positives, and the Base Rate
A clinic uses a screening test for a condition that affects 6% of the population being screened. The test is positive in 92% of people who have the condition and negative in 85% of people who do not. A patient with no symptoms tests positive and asks whether the result means there is a 92% chance of having the condition. The clinic plans to repeat the test once, and for planning purposes assumes the two test results are conditionally independent given the person's true status. The patient wants to know the probability of having the condition after one positive result and after two positive results. The clinic also wants to know how many positive results would be needed before the posterior probability first exceeds 90%, but the staff is unsure whether that calculation is appropriate without additional information. All probabilities should be reported to three decimal places, and the answer must distinguish P(positive | condition) from P(condition | positive).
1. Compute the probability that the patient has the condition after one positive test result. Explain why it is not 0.92.
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Let D be the event that the patient has the condition, and let + be a positive test. The given information is P(D) = 0.06, P(+ | D) = 0.92, and P(− | not D) = 0.85, so P(+ | not D) = 0.15.
By the law of total probability: P(+) = P(+ | D)P(D) + P(+ | not D)P(not D) = 0.92(0.06) + 0.15(0.94) = 0.0552 + 0.1410 = 0.1962.
Then P(D | +) = P(+ | D)P(D)/P(+) = 0.0552/0.1962 ≈ 0.281.
The value 0.92 is P(+ | D), the sensitivity of the test. The patient needs P(D | +), which also depends on the low base rate: most positive results come from the much larger group without the condition. Hence the posterior probability is only about 0.281, not 0.92.
2. Under the stated conditional-independence assumption, compute the probability of having the condition after two positive results.
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If the two results are conditionally independent given D, then P(++ | D) = 0.92^2 = 0.8464 and P(++ | not D) = 0.15^2 = 0.0225.
The numerator is P(++ | D)P(D) = 0.8464(0.06) = 0.050784. The denominator is 0.050784 + 0.0225(0.94) = 0.050784 + 0.021150 = 0.071934.
Thus P(D | ++) = 0.050784/0.071934 ≈ 0.706. Two positives raise the probability substantially, but it is still about 0.706, not near certainty.
3. How many positive results would be needed before the posterior probability first exceeds 0.900? Discuss the appropriateness of the calculation.
Read the worked answer
Using the likelihood ratio for one positive result, LR = P(+ | D)/P(+ | not D) = 0.92/0.15 ≈ 6.1333. The prior odds are P(D)/P(not D) = 0.06/0.94 ≈ 0.06383. After n positive results, the posterior odds are about 0.06383(6.1333)^n.
We need posterior probability > 0.900, so posterior odds > 0.9/0.1 = 9. Thus (6.1333)^n > 9/0.06383 ≈ 141.0. Solving gives n > ln(141.0)/ln(6.1333) ≈ 4.95/1.81 ≈ 2.73, so the first integer is n = 3.
Check: after 3 positives, P(D | +++) = 0.06(0.92^3) / [0.06(0.92^3) + 0.94(0.15^3)] = 0.046721/0.049894 ≈ 0.936. After 2 positives the value was about 0.706, so 3 is indeed the first count above 0.900.
The calculation is appropriate only under the stated model: the same 6% prevalence applies to this patient, and the repeated results are conditionally independent given disease status. If the patient has symptoms, the prior is higher; if the repeat test is not truly independent, the posterior after three positives may be overstated.