Question

A test for a genetic disorder can detect the disorder with 96% accuracy. However, the test will incorrectly report positive results for 3% of those without the disorder. If 14% of the population has the disorder, find the probability that a person testing positive actually has the genetic disorder. (Round your answer to two decimal places.)

Answer 1

Let A be the event of a positive test result and B the event that a given person has the disorder. We're told

`P(A \mid B) = 0.96`

`P(A \mid B^c)=0.03`

`P(B)=0.14`

and we want to find
`P(B\mid A)`. This follows directly from Bayes' theorem:

`P(B\mid A)=(P(A\cap B))/(P(A))=(P(A\cap B))/(P(A\cap B)+P(A\cap B^c))`

`P(B\mid A)=(P(A\mid B)P(B))/(P(A\mid B)P(B)+P(A\mid B^c)P(B^c))`

`P(B\mid A)=(0.96\cdot0.14)/(0.96\cdot0.14+0.03\cdot0.86)\approx\boxed{0.84}`