Question

A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. Let A be the event "the person is infected" and B be the event "the person tests positive."

(a) Find the probability that a person has the virus given that they have tested positive.



(b) Find the probability that a person does not have the virus given that they have tested negative.

Answer 1

Solution:

Given:

A certain virus infects one in every 400 people.

A test used to detect the virus in a person is positive 90% of the time if the person has the virus and

10% of the time if the person does not have the virus.

Let A = "the person is infected" and B = "the person tests positive."

Thus we have:

P(A) = 1/400=0.0025

then P("the person is NOT infected") = P(A^c) = 1 - P(A) = 1 - 0.0025 = 0.9975

P(B|A) = 90% = 0.90

P(B^c|A) = 1- P(B|A) = 1 - 0.90 = 0.10

P(B|A^c) = 10% = 0.10

and

P(B^c|A^c) =1-P(B|A^c) = 1 - 0.10 = 0.90

Part a) Find  the probability that a person has the virus given that they have tested positive.

P( A | B) =.............?

Using Bayes rule:

Part b) Find the probability that a person does not have the virus given that they have tested negative

P( A^c | B^c) =.............?

Using Bayes rule: