Question

A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 85% 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 has the virus" and B be the event "the person tests positive".

1) 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 hundredth of a percent).

2 A person does not have the virus given that they test negative, i.e. find P(A'|B'). Round your answer to the nearest hundredth of a percent.

Hint: please answer this Questions with clear handwriting because I am not fluent im English .. I will be very thankfull for this

Answer 1

A: Event of the person has the virus

A' : Event of the person does not have the virus

A certain virus infects one in every 400 people.

P(A) = 1/400 = 0.0025

P(A') =1 -P(A) =1-0.0025 = 0.9975

B : Event of the person test positive

A test used to detect the virus in a person is positive 85% of the time if the person has the virus

i.e

Probability that person test positive given that the person has virus = P(B|A) = 85/100 =0.85

P(B'|A) = 1 - P(B|A) = 0.15

5% of the time if the person does not have the virus ;

Probability that person test positive given that the person does not have virus = P(B|A') = 5/100 =0.05

P(B'|A') = 1-P(B|A') = 1-0.05 =0.95

1) Find the probability that A person has the virus given that they have tested positive, i.e. find P(A|B).

The probability that A person has the virus given that they have tested positive =  P(A|B)

By bayes theroem

P(A)P(B|A) = 0.0025 x 0.85 = 0.002125

P(A')P(B|A') = 0.9975 x 0.05 = 0.049875

P(A)P(B|A) + P(A')P(B|A') = 0.002125+0.049875=0.052

The probability that A person has the virus given that they have tested positive = 4.0865% 4.09%

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2 . A person does not have the virus given that they test negative, i.e. find P(A'|B').

A person does not have the virus given that they test negative : P(A'|B')

By bayes theroem

P(A')P(B'|A') = 0.9975 x 0.95 = 0.947625

P(A)P(B'|A) = 0.0025 x 0.15 = 0.000375

P(A')P(B'|A') + P(A)P(B'|A) = 0.947625+0.000375=0.948

The probability that A person has the virus given that they have tested positive = 99.9604% 99.96%