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

$P(A)=\dfrac1{200}=0.005\implies P(A^C)=0.995$

$P(B\mid A)=0.7$

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

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

$P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$

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

$P(A\mid B)=\dfrac{P(B\mid A)P(A)}{P(B\mid A)P(A)+P(B\mid A^C)P(A^C)}\approx0.0657$

(the first equality is Bayes' theorem)

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

$P(A^C\mid B^C)=\dfrac{P(A^C\cap B^C)}{P(B^C)}=\dfrac{P(B^C\mid A^C)P(A^C)}{1-P(B)}\approx0.9984$

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