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%
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.
The probability that a person does not have the virus given that they have tested negative is approximately 99.9%.
To find the probability that a person does not have the virus given that they have tested negative, we need to use Bayes' theorem. Bayes' theorem is a formula that allows us to update the probability of an event based on new information.
Let's denote the event "the person does not have the virus" as A', and the event "the person tests negative" as B'. Using Bayes' theorem, we can calculate the probability of A' given B' as follows:
P(A' | B') = (P(B' | A') * P(A')) / P(B')
Now, let's calculate the individual probabilities required for this equation.
P(B' | A') represents the probability of testing negative given that a person does not have the virus. As given in the problem, this probability is 0.9.
P(A') represents the probability of a person not having the virus, which can be calculated as 1 - P(A). Since the virus infects one in every 500 people, P(A) is 1/500, so P(A') = 1 - 1/500 = 499/500.
P(B') represents the probability of testing negative. This can be calculated by summing the probabilities of testing negative in both infected and non-infected individuals. As given in the problem, this probability is 0.9 * P(A) + 0.1 * P(A'). Substituting the values, we get P(B') = 0.9 * 1/500 + 0.1 * 499/500 = 449/500.
Now, plugging these values into the Bayes' theorem equation, we get:
P(A' | B') = (0.9 * (499/500)) / (449/500) = 0.9 * (499/449) = 899/449 ≈ 1.999
Therefore, the probability that a person does not have the virus given that they have tested negative is approximately 1.999 or about 99.9%.