BAYES' FORMULA A test for a certain disease gives a positive result 95% of the time...

BAYES' FORMULA

A test for a certain disease gives a positive result 95% of the time if the person actually carries the disease. However, the test also gives a positive result 3% of the time when the individual is not carrying the disease. It is known that 10% of the population carries the disease. If a person tests positive, what is the probability that he or she has the disease?

Expert Solution

P(test positive) = 0.95 * 0.10 + 0.03 * (1 - 0.1) = 0.122

P(has the disease | test positive) = (P(test positive | has the disease) * P(has the disease))/P(test positive) = (0.95 * 0.1)/0.122 = 0.7787

milcah answered 2 years ago

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