Question

A certain genetic condition affects 8% of the population in a city of 10,000. Suppose there is a test for the condition that has an error rate of 1% (i.e., 1% false negatives and 1% false positives).

Consider the values that would complete the table below.

Has condition Does not have condition totals

Test positive

Test negative

Totals

What is the probability (as a percentage) that a person has the condition if he or she tests positive? (Round your answer to one decimal place.)

Answer 1

Let
`A` be the event that a randomly selected person has the condition, and
`B` the event that the test returns a positive result. We are given that

`P(A)=0.08`

`P(B\mid A^C)=0.01` (false positive; that is, the event that the test incorrectly returns a positive result)

`P(B^C\mid A)=0.01` (false negative; incorrectly returns a negative result)

We want to find
`P(A\mid B)`. By definition of conditional probability,

`P(A\mid B)=(P(A\cap B))/(P(B))`

By the same definition,

`P(A\cap B)=P(B\mid A)P(A)`

and by the law of total probability,

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

So we have (and this is known as Bayes' rule)

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

`\implies P(A\mid B)=(0.99*0.08)/(0.99*0.08+0.01*0.92)\approx0.9`