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The probability that an individual randomly selected from a particular population has a certain disease is...

The probability that an individual randomly selected from a particular population has a certain disease is 0.06. A diagnostic test correctly detects the presence of the disease 94% of the time and correctly detects the absence of the disease 97% of the time. If the test is applied twice, the two test results are independent, and both are positive, what is the (posterior) probability that the selected individual has the disease? [Hint: Tree diagram with first-generation branches corresponding to Disease and No Disease, and second- and third-generation branches corresponding to results of the two tests.]

Math   Statistics-And-Probability   
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Answer #1

A probability tree for the given condition is given below. Bayes' Theorem can be used to calculate the given conditional probability

P(individual has the disease | both results are positive) = P(individual has the disease and both results re positive) / P(both results are positive)

= 0.06x0.94x0.94/(0.06x0.94x0.94 + 0.94x0.03x0.03)

= 0.9843

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