Conditional Probability: Diagram-Based Questions (5)

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Conditional Probability: diagram-based questions for JEE+CBSE

doubts.ai · Accepted Answer

Solution — Step by Step Let D = "has disease", T = "tests positive". P(D) = 0.01, P(D^c) = 0.99. P(T D) = 0.95 (true positive). P(T \mid D^c) = 0.05 (false positive). We want P(D T). P(T) = P(T D)P(D) + P(T \mid D^c)P(D^c) = 0.95 0.01 + 0.05 	imes 0.99 = 0.0095 + 0.0495 = 0.059. P(D T) = \dfrac{0.0095}{0.059} 0.161. Final answer: P(D T) \approx \mathbf{16.1\%}. Why This Works The result is counter-intuitive — only 16% chance of actually having the disease despite a "95% accurate" test! This is the classic base-rate fallacy . When the disease is rare, false positives outnumber true positives even with a high test accuracy. Imagine a population of 10{,}000. About 100 have the disease (95 test positive, 5 miss). 9{,}900 are healthy (495 false positive, 9{,}405 correctly negative). Of the 590 positive tests total, only 95 are real — 16\%. Alternative Method Build a 2 2 tree diagram or contingency table with 10{,}000 people: Test + Test − Total --- --- --- --- Disease 95 5 100 Healthy 495 9405 9900 Total 590 9410 10000 P(D T) = 95/590 \approx 0.161. Identical answer, no formulas. Common Mistake Students often confuse P(D T) with P(T \mid D). They are different quantities! "Test accuracy" is P(T \mid D) — given disease, probability of positive test. "Predictive value of a positive test" is P(D \mid T) — given positive test, probability of disease. These differ wildly when the disease is rare. Always write down which conditional you're computing.

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