GingerSnaps
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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.)

Respuesta :

LammettHash

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

[tex]P(A)=0.08[/tex]

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

[tex]P(B^C\mid A)=0.01[/tex] (false negative; incorrectly returns a negative result)

We want to find [tex]P(A\mid B)[/tex]. By definition of conditional probability,

[tex]P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}[/tex]

By the same definition,

[tex]P(A\cap B)=P(B\mid A)P(A)[/tex]

and by the law of total probability,

[tex]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)[/tex]

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

[tex]P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}[/tex]

[tex]\implies P(A\mid B)=\dfrac{0.99\times0.08}{0.99\times0.08+0.01\times0.92}\approx0.9[/tex]

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