Question 1097027: A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 85% 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 by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! Assume a population of 300000. 1 in 300 has the virus, or 1000 people. Out of those 1000, 85% or 850 will test positive, and 150 will test negative. Out of the 299,000 who don't have the virus, 269,100 will test negative, while 29,900 will test positive. So:
a)Given that someone has tested positive, they have a 850/850+29,900, or 0.02764227642276 probability of actually having the virus
b)Given the obverse of a, the person who tests negative has a 269,100/269,100+150, or 0.9994428969 probability of not having the disease
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