Question

A screening test for a rare from of TB has a 7% false positive rate (i.e. indicates the presence of the disease in people who do not have it). The test has an 18% false negative rate (i.e. indicates the absence of the disease in people who do have it). Suppose that 3% of the population have the disease. (i) Partition the sample space into those who have the disease, B1, and those who don’t have the disease, B2. Find P(B1) and P(B2). (ii) Let A be the event that some tests positive for the disease. Find P(A | B1), P(A | B2) and P(A) (using the Total Probability Theorem). (iii) What is the probability that someone who tests positive has the disease?

Answer 1

sat % of fahre positive state = 0.07 on 7% taf fabre negative scate = 18% ox One It is given that 3% of the population have disease Let B1 srepresents the partition of sample space containing those who have the disease B2 sepmesents the partition of sample space contain those who haven't the disease Now, P(81) = 0.03 P132) - 1-0.63 20.97 (16) Let A be the disease event that someone tests positive for the disease. PIAI8.) = 1-P ( false negadive late) PLAIBI) = 1.0.18 PIA|B1)= 0.82 m , PIA18.) = Pl pulse +ve sicude) | PlAlB2) = 0.07 mg PLA) =P(AB)x PCB.) P[A162) P182) pla)=0-03 86-82)+(091X0-00 PIA) = 6.0925

(iii) Pl Someone who fest's the has the disease) P(BIA) = P(ALB) X PIB) in Para PIA) P1B11A) = 0.82X0.03 0.0925 | PL BIIA) = 0.2659 my