Question

A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 10% of the time if the person does not have the virus. (This 10% 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 1

a) We can use Bayes' theorem to find the probability that a person has the virus given that they have tested positive:

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

where P(B|A) is the probability of testing positive given that the person has the virus, P(A) is the prior probability of a person having the virus (1/300), and P(B) is the total probability of testing positive.

P(B|A) = 0.8 (given in the problem)

P(A) = 1/300

P(B) = P(B|A) * P(A) + P(B|A') * P(A')

= 0.8 * 1/300 + 0.1 * 299/300 (using the law of total probability)

≈ 0.1033

Therefore,

P(A|B) = 0.8 * 1/300 / 0.1033

≈ 0.0245

So the probability that a person has the virus given that they have tested positive is about 2.5%.

b) Similarly, we can find the probability that a person does not have the virus given that they test negative:

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

where P(B'|A') is the probability of testing negative given that the person does not have the virus, P(A') is the prior probability of a person not having the virus (299/300), and P(B') is the total probability of testing negative.

P(B'|A') = 0.9 (given in the problem)

P(A') = 299/300

P(B') = P(B'|A') * P(A') + P(B'|A) * P(A)

= 0.9 * 299/300 + 0.2 * 1/300 (using the law of total probability)

≈ 0.9977

Therefore,

P(A'|B') = 0.9 * 299/300 / 0.9977

≈ 0.9978

So the probability that a person does not have the virus given that they test negative is about 99.8%.