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In a simulation of a virus transmission model in Sydney, the probability that a person who is infected will transmit the virus to a close contact is 0.7. The probability that a person who is not infected will be falsely identified as infected in a test is 0.03.
In a population of 2000 people, 40 are initially infected. A random person from this population is selected and tested positive.
What is the exact probability that this person is actually infected? Express your answer as a simplified fraction.
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Worked solution
- Given: 40 infected people out of 2000, so the prior probability of being infected is 40/2000 = 0.02.
- The probability of a true positive (infected and tested positive) is 0.7, so P(positive | infected) = 0.7.
- The probability of a false positive (not infected but tested positive) is 0.03, so P(positive | not infected) = 0.03.
- Calculate the number of infected people who test positive: 40 × 0.7 = 28.
- Calculate the number of uninfected people who test positive: (2000 - 40) × 0.03 = 1960 × 0.03 = 58.8.
- Total people who test positive = true positives + false positives = 28 + 58.8 = 86.8.
- The probability that a person who tested positive is actually infected is P(infected | positive) = true positives / total positives = 28 / 86.8.
- Simplify 28 / 86.8 by multiplying numerator and denominator by 10 to eliminate decimal: 280 / 868.
- Find the greatest common divisor (GCD) of 280 and 868. Factor 280 = 2^3 × 5 × 7; 868 = 2^2 × 7 × 31. GCD = 2^2 × 7 = 28.
- Divide numerator and denominator by 28: 280 ÷ 28 = 10, 868 ÷ 28 = 31. So 280/868 = 10/31.
- The simplified fraction is 10/31. This is the exact probability that a person who tested positive is actually infected.
- Check: 10/31 ≈ 0.3226; 28 / 86.8 ≈ 0.3226 — matches. Final answer is 10/31.
Reference answer: 10/31
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