Probability
Bayes' Theorem
Grade 12
Question:
<p>A laboratory blood test is 99% effective in detecting a certain disease, when it is in fact present. However, the test also yields a false positive result for 0.5% of the healthy person tested (that is, if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?</p>
<p>\(\dfrac{22}{133}\)</p>
<p>\(\dfrac{22}{123}\)</p>
<p>\(\dfrac{11}{133}\)</p>
<p>\(\dfrac{1}{133}\)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem to find P(Disease|Positive) by accounting for both the test's sensitivity and the base rate of the disease in the population. The key is recognizing that a positive test can come from two sources: truly diseased people (99% detected) or healthy people (0.5% false positive).
<p><strong>Step 1: Define events and probabilities</strong></p><p>Let D = person has disease, T⁺ = positive test result</p><p>P(D) = 0.001 (0.1% have disease)</p><p>P(D') = 0.999 (99.9% healthy)</p><p>P(T⁺|D) = 0.99 (sensitivity: 99% detection when diseased)</p><p>P(T⁺|D') = 0.005 (false positive: 0.5% when healthy)</p><p><strong>Step 2: Find total probability of positive test</strong></p><p>P(T⁺) = P(T⁺|D)·P(D) + P(T⁺|D')·P(D')</p><p>P(T⁺) = (0.99)(0.001) + (0.005)(0.999)</p><p>P(T⁺) = 0.00099 + 0.004995 = 0.005985</p><p><strong>Step 3: Apply Bayes' theorem</strong></p><p>P(D|T⁺) = [P(T⁺|D)·P(D)] / P(T⁺)</p><p>P(D|T⁺) = (0.99 × 0.001) / 0.005985</p><p>P(D|T⁺) = 0.00099 / 0.005985</p><p>P(D|T⁺) = 0.1653 ≈ <strong>16.53%</strong> or <strong>99/597</strong></p><p>∴ Answer: A</p>
Correct Answer: A