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Probability
In a certain population, a rare disease affects 1% of individuals. There is a diagnostic test for this disease, which correctly identifies the disease in 95% of cases when a person has it and incorrectly identifies the disease in 2% of cases when a person does not have it. If a person tests positive for the disease, what is the probability that they actually have the disease?
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1 Reply
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To find the probability that a person actually has the disease given that they tested positive, we can use Bayes’ theorem. Let’s define the following probabilities:
P(D) = Probability of having the disease = 0.01 (1% of individuals)
P(Pos|D) = Probability of testing positive given that a person has the disease = 0.95 (95% accuracy)
P(Pos|¬D) = Probability of testing positive given that a person does not have the disease = 0.02 (2% false positive rate)
We want to calculate P(D|Pos), which is the probability of having the disease given that the person tested positive. According to Bayes’ theorem:
P(D|Pos) = (P(Pos|D) * P(D)) / P(Pos)
To calculate P(Pos), the probability of testing positive, we need to consider both scenarios: the person having the disease and testing positive, and the person not having the disease but testing positive.
P(Pos) = P(Pos|D) * P(D) + P(Pos|¬D) * P(¬D)
P(¬D) = Probability of not having the disease = 1 – P(D) = 0.99 (99% of individuals)
Now we can substitute these values into the equation:
P(D|Pos) = (P(Pos|D) * P(D)) / (P(Pos|D) * P(D) + P(Pos|¬D) * P(¬D))
P(D|Pos) = (0.95 * 0.01) / (0.95 * 0.01 + 0.02 * 0.99)
Simplifying the equation:
P(D|Pos) = 0.0095 / (0.0095 + 0.0198)
P(D|Pos) ≈ 0.324
Therefore, the probability that a person actually has the disease given that they tested positive is approximately 0.324 or 32.4%.
