When a Positive Result Is Not the Whole Story
A screening test can be good at detecting a disease and still produce many positive results among people who do not have it. This can happen when the disease is rare: even a small false-positive rate applied to a large group without the disease may create more false positives than true positives.
In Reversing the Condition with Tree Diagrams, you learned to find a probability such as \(P(D\mid +)\) by dividing the disease-and-positive path by the probability of all positive results. Here, \(D\) means that a person has the disease, and \(+\) means that the test result is positive. Medical screening makes the importance of that reverse conditional probability especially clear.
The sensitivity of a test is \(P(+\mid D)\), the probability of a positive result among people who have the disease. The false-positive rate is \(P(+\mid D^c)\), the probability of a positive result among people who do not have the disease. The specificity is \(P(-\mid D^c)\), the probability of a negative result among people without the disease. Since positive and negative are complementary test outcomes for someone without the disease, the false-positive rate is one minus specificity.
A test with high sensitivity detects a large proportion of people who have the disease. That fact alone does not tell us what proportion of positive testers have the disease. To find that probability, the denominator must include all positive results: true positives and false positives.
Use Natural Frequencies to See the Counts
A useful way to reason about a screening scenario is to imagine a large group, such as 10,000 or 100,000 people. First count how many people are expected to have the disease and how many are not. Then apply the test’s conditional probabilities to each group. This is the frequency version of following both paths through a tree diagram.
Prevalence is the proportion of the population that has the disease. Use it to estimate the numbers with and without the disease.
Apply sensitivity to the group with disease and the false-positive rate to the group without disease.
Add true positives and false positives. Both kinds of result belong in the denominator.
Divide the number of true positives by all positive results, then interpret the result among positive testers.
In probability notation, if the population consists of people with or without the disease, the total probability of a positive result is the sum of the two positive paths. The reverse conditional probability is the disease-and-positive path divided by that total.
Here \(P(D)\) is the prevalence and \(P(D^c)=1-P(D)\). The two groups, disease and no disease, are mutually exclusive and exhaustive. Within each group, positive and negative are complementary test outcomes. As in the earlier tree-diagram tutorials, the paths leading to a positive result are combined to find \(P(+)\).
Worked Example: A Rare Disease Screening Test
Worked Example: A Rare Disease Screening Test
Consider an invented screening model in which 0.5% of people have a certain disease. The test gives a positive result for 98% of people who have the disease. Among people without the disease, 4% receive a positive result. Given a positive result, find the probability that a person has the disease.
State: Let \(D\) mean that a person has the disease and \(+\) mean that the screening result is positive. The question asks for \(P(D\mid +)\), the probability of disease among people who tested positive. The given 98% is \(P(+\mid D)\), not the requested probability.
Plan: Imagine 10,000 people following this model. The disease and no-disease groups have proportions \(0.005\) and \(1-0.005=0.995\), which add to 1. The positive and negative probabilities within each group are complementary: among people with disease, they are \(0.98\) and \(0.02\); among people without disease, they are \(0.04\) and \(0.96\). Count positives in both groups, then divide the true-positive count by all positive results. The total positive probability is greater than 0.
Do: Of 10,000 people, \(10{,}000(0.005)=50\) have the disease, and \(10{,}000(0.995)=9{,}950\) do not. The expected positive counts are:
There are \(49+398=447\) positive results in all. Of these, 49 are true positives. Therefore:
The same result follows from the path probabilities. The disease-and-positive path has probability \(0.005(0.98)=0.0049\). The no-disease-and-positive path has probability \(0.995(0.04)=0.0398\). Thus \(P(+)=0.0049+0.0398=0.0447\), and \(0.0049/0.0447\approx0.1096\).
Conclude: Under this invented model, given a positive screening result, the probability that a person has the disease is about \(0.1096\), or 10.96%. Although the test detects 98% of people with the disease, most positive results in this model are false positives: 398 false positives compared with 49 true positives.
Why Rarity Changes the Interpretation
The false-positive rate in the example is only 4%, but it applies to 9,950 people without the disease. The sensitivity applies to only 50 people with the disease. A modest rate applied to a very large group can produce a substantial count. This is why the test’s sensitivity alone cannot answer the question, “If my result is positive, how likely is it that I have the disease?”
The calculation also shows that a false positive does not necessarily mean the test is unusually poor. It describes a positive result for a person without the disease. With a rare disease, even a relatively small false-positive rate can make up a large share of all positive results. The relevant conditional probability depends on the population being screened as well as on the test.
Worked Example: The Same Test in a Different Population
Worked Example: The Same Test in a Different Population
Suppose the test from the previous example is used in a population where 10% of people have the disease, rather than 0.5%. Its sensitivity remains 98%, and its false-positive rate remains 4%. Find the probability of disease given a positive result.
State: With \(D\) meaning disease and \(+\) meaning a positive result, find \(P(D\mid +)\). The test properties are unchanged; only the prevalence is different.
Plan: Imagine 10,000 people. The disease and no-disease groups are 10% and 90% of the population, so they are exhaustive and mutually exclusive. Use the 98% sensitivity for the disease group and the 4% false-positive rate for the no-disease group. Add the two positive counts for the denominator.
Do: There are \(10{,}000(0.10)=1{,}000\) people with the disease and \(10{,}000(0.90)=9{,}000\) without it. The number of true positives is \(1{,}000(0.98)=980\), and the number of false positives is \(9{,}000(0.04)=360\). Thus, there are \(980+360=1{,}340\) positive results.
For a probability check, the two positive path probabilities are \(0.10(0.98)=0.098\) and \(0.90(0.04)=0.036\). Their sum is \(0.134\), and \(0.098/0.134\approx0.7313\), matching the frequency calculation.
Conclude: In this invented population, given a positive result, the probability of disease is about 73.13%. The test’s sensitivity and false-positive rate did not change, but disease was more common in the screened population, so a larger proportion of positive results came from people with the disease.
Worked Example: A Lower False-Positive Rate
Worked Example: A Lower False-Positive Rate
Return to a population where 0.5% of people have the disease. Consider a different invented test with the same 98% sensitivity but a 0.5% false-positive rate. In a group of 200,000 people, find the probability of disease given a positive result.
State: Let \(D\) mean disease and \(+\) mean a positive result. The requested probability is \(P(D\mid +)\).
Plan: The disease and no-disease groups are mutually exclusive and exhaustive, with proportions 0.005 and 0.995. The positive probability is 0.98 among people with disease and 0.005 among people without it. Apply those rates to the corresponding counts, then divide true positives by all positives.
Do: Of 200,000 people, \(200{,}000(0.005)=1{,}000\) have the disease and 199,000 do not. There are \(1{,}000(0.98)=980\) true positives and \(199{,}000(0.005)=995\) false positives. There are \(980+995=1{,}975\) positive results.
Using probabilities gives the same result: \(P(D\cap +)=0.005(0.98)=0.0049\), while \(P(D^c\cap +)=0.995(0.005)=0.004975\). Therefore, \(P(+)=0.0049+0.004975=0.009875\), and \(0.0049/0.009875\approx0.4962\).
Conclude: Under this model, given a positive result, the probability of disease is about 49.62%. The false-positive rate is much lower than in the first example, which reduces false positives and raises the proportion of positive results that are true positives. It does not make every positive result proof of disease.
Common Mistakes and AP Exam Tips
- Reversing the conditional probability. Sensitivity is \(P(+\mid D)\), the probability of a positive result among people with disease. The probability asked for after a positive screen is often \(P(D\mid +)\). These use different reference groups.
- Leaving false positives out of the denominator. For \(P(D\mid +)\), count every positive result. Divide true positives by true positives plus false positives, not by the total number of people with disease.
- Calling sensitivity the chance that a positive result is correct. Sensitivity describes test results among people who have the disease. The chance of disease among positive testers is a separate conditional probability.
- Ignoring prevalence. A test’s sensitivity and false-positive rate do not, by themselves, determine the probability of disease after a positive result. State which population is being screened or use the prevalence supplied in the problem.
- Giving a number without context. A complete interpretation identifies the condition group: “Among people who test positive, the probability of having the disease is about 10.96% under this model.” Avoid saying simply that “the test is 10.96% accurate.”
- Treating a screening result as a diagnosis. A probability calculation describes the stated model. It does not replace medical evaluation or establish an individual’s diagnosis.
For a full-credit solution, define the events, identify the requested conditional probability, show how the true-positive and false-positive groups contribute to the denominator, and interpret the result in context. Natural frequencies are often the clearest way to show why a rare condition can lead to many false positives even when a test detects most cases.
Check Your Understanding
For each question, identify the relevant groups and distinguish the test’s positive rate from disease probability among positive testers.
- A disease affects 2% of a population. A test has 95% sensitivity and a 3% false-positive rate. In a group of 10,000, how many true positives and false positives are expected? Find the probability of disease given a positive result.
- In your own words, distinguish sensitivity \(P(+\mid D)\) from \(P(D\mid +)\).
- A test has specificity 97%. What is its false-positive rate? Explain what that rate describes.
- Suppose a test’s sensitivity and false-positive rate stay the same, but prevalence increases. How would you expect the proportion of positive testers who have the disease to change? Explain using the group counts.
- Why does the number of false positives depend on how many people without the disease are tested, as well as on the false-positive rate?