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In a simulation of a medical test for a rare disease, 1 in 1000 people in a population have the disease. The test is 99% accurate, meaning that if a person has the disease, the test correctly identifies it 99% of the time, and if a person does not have the disease, the test incorrectly identifies it as positive 1% of the time.
A person is selected at random from the population and tests positive. What is the exact probability that this person actually has the disease?
Express your answer as a simplified fraction.
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Worked solution
- Assume a population of 1000000 people to make calculations easier and avoid fractions.
- 1 in 1000 people have the disease, so the number of people with the disease is 1000000 / 1000 = 1000.
- The test is 99% accurate for people with the disease, so 99% of 1000 people with the disease test positive: 0.99 * 1000 = 990.
- The number of people without the disease is 1000000 - 1000 = 999000.
- The test incorrectly identifies 1% of people without the disease as positive, so 1% of 999000 test positive falsely: 0.01 * 999000 = 9990.
- Total number of people who test positive is 990 (true positives) + 9990 (false positives) = 10980.
- The probability that a person who tests positive actually has the disease is the number of true positives divided by total positives: 990 / 10980.
- Simplify the fraction 990 / 10980 by dividing numerator and denominator by 30: 990 ÷ 30 = 33, 10980 ÷ 30 = 366.
- Simplify 33 / 366 by dividing both by 3: 33 ÷ 3 = 11, 366 ÷ 3 = 122.
- The simplified fraction is 11 / 122.
- Check: 11 / 122 ≈ 0.09016, and 990 / 10980 ≈ 0.09016, so the fraction is correct.
- Therefore, the exact probability that a person who tests positive actually has the disease is 11/122.
Reference answer: 11/122
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