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Example · Example 7

Q.In a certain population, 0.5%0.5\% of people have a particular disease. A diagnostic test correctly identifies the disease in 99%99\% of people who actually have it, but also incorrectly shows a positive result in 2%2\% of healthy people. A person selected at random from the population tests positive. Find the probability that this person actually has the disease.

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By Bayes' theorem, P(D∣Pos)=P(D)P(Pos∣D)P(D)P(Pos∣D)+P(D′)P(Pos∣D′)P(D\mid\text{Pos})=\dfrac{P(D)P(\text{Pos}\mid D)}{P(D)P(\text{Pos}\mid D)+P(D')P(\text{Pos}\mid D')}. The numerator is 0.005×0.99=0.004950.005\times0.99=0.00495. The denominator adds the false-positive contribution: 0.995×0.02=0.01990.995\times0.02=0.0199, giving 0.00495+0.0199=0.024850.00495+0.0199=0.02485. So $P(D\mid\text{Pos})=\ …

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