A certain virus infects one in every 500 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."

Find the probability that a person does not have the virus given that they have tested negative.

What is the probability that a person does not have the virus given that they have tested negative." Of the 180 people who are infected, .1(180)= 18 will text negative. Of the 99800 people who are not infected, .9(99800)= 89820 will test negative. Of the 89820+ 18= 89838 people who tested negative 89820 do not have the virus so the probability a person who tested negative does not have the virus is 89820/89832= 0.999 or about 99.9%

1 answer

The idea is sound, but You have the very first step incorrect. In probability problem, it is best to write out the assumptions, so that you AND others who read your text will be able to understand.

You have assumed 10000 people in the population, which is ok, but 1/500 of 10000 people is 200, not 180.
After that, your procedure and reasoning are both correct.
Also, when dealing with probability, it is not sufficient to give the answer 0.999 approximately because it is too vague.
It is preferable to give in fractions, which is not hard with your approach. If given in decimal, I would give at least two or three non-nine digits if the answer starts with a string of nine.