Question

A certain virus infects one in every 150 people. A test used to detect the virus in a person is positive 90​% 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

a) P(A|B)=0.108

b)
`P(A^c|B^c)`=0.999

Explanation:

Given the events:

A: the person is​ infected

B: the person tests​ positive

`A^c`: the person is​ not infected

`B^c`: the person tests​ negative

a) If we check the attached picture, we can see that:

`P(A)=(1)/(150)`

`P(A^c)=(149)/(150)`

P(B|A)=
`(90)/(100)`

`P(B|A^c)=(5)/(100)`

Bayes' Theorem:

P(B)= P(B∩A) + P(B∩
`A^c`)

P(B)= P(B|A)×P(A) + P(B|
`A^c`)×P(
`A^c`)

P(B)=
`(90)/(100) (1)/(150) +(5)/(100) (149)/(150)=(167)/(3000)`

We have to find the probability that the person is infected given that a person tests positive.

P(A|B)=
`(P(B|A)P(A))/(P(B))`=
`((90)/(100).(1)/(150)  )/((167)/(3000) )=(18)/(167)=0.108`

b) We have to find the probability that the person is not infected given that a person tests negative:

`P(A^c|B^c)=(P(B^c|A^c)P(A^c))/(P(B^c))=(P(B^c|A^c)P(A^c))/(1-P(B))`

=
`((95)/(100) . (149)/(150) )/(1-(167)/(3000))=(2831)/(2833)=0.999`