Question

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.

Answer 1

We're told that

`P(A)=\frac1{200}=0.005\implies P(A^C)=0.995`

`P(B\mid A)=0.7`

`P(B\mid A^C)=0.05`

a. 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 law of total probability,

`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)`

Then

`P(A\mid B)=(P(B\mid A)P(A))/(P(B\mid A)P(A)+P(B\mid A^C)P(A^C))\approx0.0657`

(the first equality is Bayes' theorem)

b. We want to find
`P(A^C\mid B^C)`.

`P(A^C\mid B^C)=(P(A^C\cap B^C))/(P(B^C))=(P(B^C\mid A^C)P(A^C))/(1-P(B))\approx0.9984`

since
`P(B^C\mid A^C)=1-P(B\mid A^C)`.