Question

Answer parts c,d, and e

2. Suppose two doctors, A and B, test all patients coming into a clinic for syphilis. Let events A = Doctor A makes a positive diagnosis and B+ = Doctor B makes a positive diagnosis. Suppose doctor A diagnoses 10% of all patients as positive, doctor B diagnoses 17% of all patients as positive, and both doctors diagnose 8% of all patients as positive. a. Are the events A*, B* independent of each other? Show your work. b. Suppose a patient is referred for further lab tests if either doctor A or B makes a positive diagnosis What is the probability that a patient will be referred for further lab tests? Hint: You are trying to find where "at least one doctor will get a positive diagnosis" = P(AUB). Round your final answer to two decimal places. c. Find the conditional probability that Doctor B makes a positive diagnosis of syphilis given that Doctor A makes a positive diagnosis. In other words, what is the probability of P(B+14+)? Round your final answer to one decimal places. d. Find the conditional probability that Doctor B makes a positive diagnosis of syphilis given that Doctor A makes a negative diagnosis. In other words, what is the probability of P(B+|A")? Round your final answer to one decimal places. e. What is the relative risk? Interpret your results in 1 - 2 sentences. Round your final answer to one decimal place.

Answer 1

the solutions are described below

Solution P 6+) - -ooogo P(B+) = he has a 0.17 P (Ato B+) = 08 © PCB+/A+)- P(BYPAS. 0.8 P (A+) ) P(pt/A-) - P(BOA) P CA-) PCBtna-) - PBM) - PlBtn AT) = 0.17 – 0.08 = 0.09 P(A+) - 1-PCA+) = 1-0:10 = 0.90 p(/A-) = 0;&= 0.1 Ponit the A+ = doctor A maken diagonina Bt = doctor B make pritine diagonatin. the me Docter AB.1 0.10 | Alo1 Doctor B 0.17 0.83 0.27 1.13 A+B CAD - 0.71 RR = A/AB C/C+ - o.fl times 0.1/0. 2 7 0.90/3.73 the diagonist in then doctore B Doctor A gines 0.77/0.27 - RR BAB D/CAD = 31.31 0.83/1.73 the diagonina thon doctor A. Doctor B ginest 31 timen