Question

# A certain virus infects one in every 400 people. A test used to detect the virus...

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.

We are given here: P( A ) = 1/400 = 0.0025

P( B | A ) = 0.9 , therefore P( not B | A) = 0.1
P( B | not A ) = 0.1, therefore P( not B | not A) = 0.9

Using law of total probability, we get:

P(B) = P(B | A)P(A) + P(B | not A) P( not A) = 0.9*0.0025 + 0.1*(1 - 0.0025) = 0.1020

P( not B ) = 0.1*0.0025 + 0.9*(1 - 0.0025) = 0.898

a) The probability that a person has the virus given that they have tested positive is computed using bayes theorem as:

P(A | B) = P(B | A)P(A) / P(B)

P(A | B) = 0.9*0.0025 / 0.102

P(A | B) = 0.0221

Therefore 0.0221 is the required probability here.

b) The probability that a person does not have the virus given that they have tested negative is computed here as:

P( not A | not B) = P( not B | not A)P(not A) / P( not B) = 0.9*(1 - 0.0025) / 0.898 = 0.9997

Therefore 0.9997 is the required probability here.