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:

$P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B)}{P(A\cap B)+P(A\cap B^c)}$

$P(B\mid A)=\dfrac{P(A\mid B)P(B)}{P(A\mid B)P(B)+P(A\mid B^c)P(B^c)}$

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