Question

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Q2. Consider a disease Z and a test T for the disease. Assume that 1 in 1000 persons has the disease Assume that, for a person having the disease there is a 95% probability that the test is positive; however, for a person not having the disease, there is a 1% probability that the test is positive. Assume the test result is pos itive for a new patient. What is the probability that the patient has the disease? Hint: Use the Bayes rule.

Answer 1

Answer As giveni in 1000 persory has the disease in question Let Z denttes the event thot person has thu diseare O.001 Then Z P(Z): 1000 Prolability e (hobability hoviny the disease P(Z 1-P(2) not 0.00T PIZ)0.999 Let T dinotes the event that thu test &a person comes out to b pesi tive uven in 9uetion potobioity ferson having tert is poitivt. So, we wl ue the conupt o cenditienal probabtlity 951 that the the diseae thun here auady knowiny given that tut pos tire phobabiliy Occurrtd uhch is the 0s We know et e thu evertt hos aving th we need to pussen the ousear and no ind the o wt are 95 So, it w be riten like this P(T/Z) 95/. 0.95 as 1/. given get the f(T/Z) whi ch is we Can atablity which i the whun we know that porson is not having the cisease then hew mich art the thances of teat Pontit teing PIT/Z) 0 0

robabisity that the patrent nas Noto, we neeodl to find the he disease given that thu test comes aut to be poxhve P(Z/T) ? Ne, using Rayes rule ut can uTit P(z/T) belou PIT/2) PIZ) P(Z/T) r(T/2) P(2)+ PT/2) z) previou Calculationo and Mfertncs Now just Rut the valus rom the f(2/1)= D.95 X0.00 1 O195X0.001+ 0.01X0.999 Or00095 0-D00 95 PIZIT) D.O109니 0O0095 t0-0b999 0.00095 PCZ/T) 0.0868 0.01094 PIZIT) 3980.0 &68.1 fhobability reguired thi is the