Probability
Total Probability and Bayes Theorem
Grade 12
Question:
<p>HIV prevalence: 0.1%. Test: 90% sensitivity (true positive rate), 99% specificity. P(HIV positive | test positive) = ? <em>[JEE Advanced 2018]</em></p>
<p>\(\dfrac{90}{1089}\)</p>
<p>\(\dfrac{1}{11}\)</p>
<p>\(\dfrac{90}{1188}\)</p>
<p>\(\dfrac{10}{101}\)</p>
Step-by-Step Solution
Key Concept: Apply Bayes' theorem. P(+test) = P(+|HIV)P(HIV) + P(+|no HIV)P(no HIV) = 0.9 \times 0.001 + 0.01 \times 0.999.
<p>\(P(H)=0.001\), \(P(H^c)=0.999\).</p><p>\(P(+|H)=0.90\), \(P(+|H^c)=0.01\).</p><p>\(P(+)=0.90\times0.001+0.01\times0.999=0.0009+0.00999=0.01089\).</p><p>\(P(H|+)=\dfrac{0.0009}{0.01089}=\dfrac{0.0009}{0.01089}=\dfrac{90}{1089}=\dfrac{10}{121}\approx0.0826\).</p>
Correct Answer: A