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Exercise: Total Probability and Bayes... · Q24

Q.A disease affects 1%1\% of a population. A diagnostic test correctly identifies the disease 95%95\% of the time in people who have it, but incorrectly indicates the disease in 3%3\% of healthy people. A person selected at random tests positive. Find the probability that this person actually has the disease.

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Numerator: P(D)P(Pos∣D)=0.01×0.95=0.0095P(D)P(\text{Pos}\mid D)=0.01\times0.95=0.0095. Denominator: add the false-positive term, 0.99×0.03=0.02970.99\times0.03=0.0297, giving 0.0095+0.0297=0.03920.0095+0.0297=0.0392. So $P(D\mid\text{Pos})=\dfrac{0.0095}{0.0392}=\dfrac{95}{39 …

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