A certain virus infects one in every 200 people. a test used to detect the virus in a person is positive 70% of the time when the person has the virus and 5% of the time when the person does not have the virus. (this 5% result is called a false positive.) let a be the event "the person is infected" and b be the event "the person tests positive." (a) using bayes' theorem, when a person tests positive, determine the probability that the person is infected. (b) using bayes' theorem, when a person tests negative, determine the probability that the person is not infected.

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LammettHash LammettHash · Answer 1 · 2019-05-11T18:57:10+02:00

We're told that

[tex]P(A)=\dfrac1{200}=0.005\implies P(A^C)=0.995[/tex]

[tex]P(B\mid A)=0.7[/tex]

[tex]P(B\mid A^C)=0.05[/tex]

a. 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 law of total probability,

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

Then

[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)}\approx0.0657[/tex]

(the first equality is Bayes' theorem)

b. We want to find [tex]P(A^C\mid B^C)[/tex].

[tex]P(A^C\mid B^C)=\dfrac{P(A^C\cap B^C)}{P(B^C)}=\dfrac{P(B^C\mid A^C)P(A^C)}{1-P(B)}\approx0.9984[/tex]

since [tex]P(B^C\mid A^C)=1-P(B\mid A^C)[/tex].