A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 5%

virus-infects

#1

A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 5% of the time if 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) Find the probability that a person has the virus given that they have tested positive, i.e. find P(A|B). Round your answer to the nearest tenth of a percent and do not include a percent sign.
P(A|B)= %

b) Find the probability that a person does not have the virus given that they test negative, i.e. find P(A’|B’). Round your answer to the nearest tenth of a percent and do not include a percent sign.
P(A’|B’) = %

Answer:

We need to create a tree. The first branch will be person has virus or person doesn’t have the virus.

P(has virus) = 1/300
P(doesn’t have virus) = 299/300

Then from each option there we branch out again. The two options being a positive or negative test.

has virus
P(positive test) = 90/100
P(negative test) = 10/100

doesn’t have virus
P(positive test) = 5/100
P(negative test) = 95/100

A) P(has virus | positive test) = P(positive and has virus) / P(positive test)
P(positive and has virus) = 90/100 * 1/300 = 90/30000
P(positive test) = 1/300 * 90/100 + 5/100 * 299/300
= 90/30000 + 1495/30000
= 1585/30000
P(positive and has virus) / P(positive test) = (90/30000) / (1495/30000)
= 0.06

B) P( doesn’t have virus | negative test) = P(negative and doesn’t have virus) / P(negative test)

P(negative and doesn’t have virus) = 95/100 * 299/300 = 28405/30000
P(negative test) = 95/100 * 299/300 + 10/100 * 1/300
= 28405/30000 + 10/30000
= 28415 / 30000
P(negative and doesn’t have virus) / P(negative test) = (28405/30000) / (28415 / 30000)
= 0.9996 approximately 1.0