Question 1178623: A certain disease has an incidence rate of 0.2%. If the false negative rate is 8% and the false positive rate is 2%, compute the probability that a person who tests positive actually has the disease.
Give your answer accurate to at least 3 decimal places
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem using Bayes' Theorem:
1. Define the Events:
D: The person has the disease.
: The person tests positive.
2. Given Probabilities:
P(D) = 0.002 (incidence rate)
P(D') = 1 - 0.002 = 0.998 (probability of not having the disease)
P(-|D) = 0.08 (false negative rate, probability of testing negative given the disease)
P(+|D) = 1 - 0.08 = 0.92 (true positive rate, probability of testing positive given the disease)
P(+|D') = 0.02 (false positive rate, probability of testing positive given no disease)
3. Apply Bayes' Theorem:
We want to find P(D|+), the probability of having the disease given a positive test.
Bayes' Theorem: P(D|+) = [P(+|D) * P(D)] / P(+)
We also need to find P(+), the probability of testing positive:
P(+) = P(+|D) * P(D) + P(+|D') * P(D')
4. Calculate:
P(+|D) * P(D) = 0.92 * 0.002 = 0.00184
P(+|D') * P(D') = 0.02 * 0.998 = 0.01996
P(+) = 0.00184 + 0.01996 = 0.0218
P(D|+) = 0.00184 / 0.0218 ≈ 0.08440367
5. Round to Three Decimal Places:
P(D|+) ≈ 0.084
Therefore, the probability that a person who tests positive actually has the disease is approximately 0.084.
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