Question

10. A particular disease has an incidence rate of 1 in 1000 adults. A test shows a positiveresult of 99% of the time when an individual has the disease and a positive result 4% of the time when an individual is healthy. For a randomly selected adult, what is the probability of a negative result?

Answer 1

Let D be the disease, D' be the healthy and + and - for the test results. So,

P(D) = 1/1000

= 0.001

P(D') = 1-P(D)

=0.999

P(+|D) = 0.99

P(+|D') = 0.04

As we know,

P(+) = P(+|D)P(D) + P(+|D')P(D')

= 0.99 * 0.001 + 0.04 * 0.999

= 0.04095

P(-) = 1-P(+)

= 1-0.04095

= 0.95905

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