A disease has prevalence 1%. A test with sensitivity 99% and specificity 95%. If a person tests positive, what is P(D|Pos) approximately?

• A. Approximately 99%
• B. Approximately 1%
• C. Approximately 16.7%
• D. Approximately 84.3%

Answer: (C)

The key idea is that the chance a positive test truly means you have the disease depends on both the test’s accuracy and how common the disease is in the population. This is a Bayes’ theorem type calculation: P(D|Pos) = [P(Pos|D) × P(D)] / [P(Pos|D) × P(D) + P(Pos|not D) × P(not D)].

Plugging in the numbers: P(D) = 0.01, so P(not D) = 0.99. P(Pos|D) = 0.99 (sensitivity). P(Pos|not D) = 1 − specificity = 0.05 (false positive rate).

Numerator: 0.99 × 0.01 = 0.0099.

Denominator: 0.0099 + (0.05 × 0.99) = 0.0099 + 0.0495 = 0.0594.

P(D|Pos) = 0.0099 / 0.0594 ≈ 0.1667, i.e., about 16.7%.

So, even with a highly sensitive test and decent specificity, a positive result in a population with only 1% prevalence means roughly a 17% chance the person truly has the disease. The rest of the positives are false positives, about 83% in this scenario. The other options would misinterpret the balance between true positives and false positives or ignore the base rate.