Question

RBV testing Suppose that 1% of all people are infected with the rare banana virus (RBV). There is a test to detect the RBV: if you do have the RBV, then the test will correctly detect this 99% of the time; if you do not have RBV, then the test will correctly indicate this 97% of the time. We assume that if the RBV test is given repeatedly to the same person, then the test results are independent of cach other (i.e. if the person has RBV, then each test has an independent 99% chance to confirm this). True RBV status Infected Not infected 99% 1% 3% 97% Positive Negative 1. Suppose that conditional probability that they are actually infected? randomly chosen person tests positive for RBV. Given this test result, what is the a 2. We ask the same question as above, but now conditioning that the person has tested positive twice in a row Result

Answer 1

9. ut D: Pesson has the disease E The testTesult is positve @Giuen P(D)= 1 'A0.01, PCD): 0.99 P(E)0.99 P (E/D) 0.01 According ts Baye's theoseu P(E) PCD PlE PE) PC D)+ P(6) PCD) P (E)-0.03 PCE/B) 0.97 0.990.0 0.99)0.01+(0.03) (o.99) ut P(EE): Probability that the test mesult is positive tuice P (E PCD) P(EE).PCD)+ P(E) PCD) Q.990. (0.99) 0.01+ 0.03) (o.39) O-25 PE) O.01 p.9167