Problem
A certain virus infects one in every 200 people. A test used to detect the virus in a person is positive $80 \%$ of the time if the person has the virus and $8 \%$ of the time if the person does not have the virus. (This $8 \%$ 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 $P(A \mid B)$. Round your answer to a percent rounded to the nearest fourth decimal place and do not include a percent sign. \[ P(A \mid B)= \] b) Find the probability that a person does not have the virus given that they test negative, i.e. find $P\left(A^{\prime} \mid B^{\prime}\right)$. Round your answer as a percent rounded to the nearest fourth decimal place and do not include a percent sign. \[ P(\bar{A} \mid \bar{B})= \]
Solution
Step 1 ：Define the probabilities: \(P(A) = 0.005\), \(P(B|A) = 0.80\), and \(P(B|\neg A) = 0.08\).
Step 2 ：Calculate the probability of not having the virus, \(P(\neg A) = 1 - P(A) = 0.995\).
Step 3 ：Calculate the probability of testing positive, \(P(B) = P(B|A) \cdot P(A) + P(B|\neg A) \cdot P(\neg A) = 0.0836\).
Step 4 ：Calculate the probability that a person has the virus given that they have tested positive using Bayes' theorem: \(P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\).
Step 5 ：Round the result to the nearest fourth decimal place: \(P(A|B) = 0.0478\).
Step 6 ：Final Answer: The probability that a person has the virus given that they have tested positive is \(\boxed{0.0478}\) or \(\boxed{4.78\%}\).
