Question

9. On the basis of a physical examination, a physician assesses the probabilities that a patient has no tumor, a benign tumor, or a malignant tumor as 0.70, 0.20 and 0.10, respectively. A thermographic test is subsequently given to the patient. This test gives a negative result with probability 0.90 if there is no tumor, with probability 0.80 if there is a benign tumor, and with probability 0.20 if there is a malignant tumor. (a) Suppose the test gives a positive for this patient. Find the probability that this patient has i) no tumor ii) a benign tumor iii) a malignant tumor (b) Is the event “test is positive” independent from the event "patient has no tumor"? Justify.

Answer 1

Solution: Let us take A, B and c denotes respectively the events that a patient has no tumor, a begin tumos and a malignant tumor Let us N denotes the event that the test gives a negative result for a patient. Given Data is P(A) = 0.70 P(B) = 0.20 P() - OJO P(NA) : 0.90 P (NIB) 0.80 P(NIC) = 0.20 a, Suppose the test gives a positive for this patient, find the probability that this patient has de no tumor: P(A) = PICAO) U (AONE)] foi PCA) = P(AUB)= P(A)*P(R)-piano P(ANN) + P (ANC) - P(ANN). (ANC)] - P(ANN) + PLANNC) {:: PICANN)" (Anne)] for By using Dighibukie law} →PlanNONE) P(ANO), ø is the impossible Event. P(0) 20

P(A) - PAON) + P(Anne) P(A) = P(NA) P(A) + PLANNC) → P(Anwe) =P(A) - PNJA)P(A) (00) P(Anne) = 0.76 -0.90X0.70 (08) P(ANC) -0.07 -0.63 0.07 GPAONC) - 0,07 In the same way => PCB) - P(Box) +P (Bone) P(8) = P(NIB) P(B) + P (BONO)=P(Bowe) - PCB)–PWB)P (8) P(Bowe) : 0.20-0.80 X0.20 tett e P (BONC) = 0.20 0.16 A PlBowe) = 0.04) → PC) = P(CON) + P(enne) P(C) : P(N/C) P(() + Plenne) PC CONC) - PCC) - PCN/c)PCC) P (CANC) : 0.10 -0.20%0.10 P (CONC) : 0.10 0.02 SP(CoNe) : 0.08

Now, P(N) > PC (NOA)U(NOB) (NOC] P(N) : 0.81 P(NC) = 1-P(N) 21-0.8 = 0.19. I Required probability "no tumor" »P(AN). P(Anne) P(N) - 0.07 0.07 0.19 0.3684 are 1 Required probability in a begin tumas" P(B/Ne) P(Bone) P(Ne) boy 0.19 - 0.2105

" Required probability "a malignant tumor" => p (c/ves P(CONC) P(Ne) 0.08 0.4211 0.19 6. Note that P(Anne) = 0.07 P(NC) : 0.19 P(A) - 0.70 hen L = P(ANN) P(A) XPONE) P(NO = 0.70 x 0.19 e 0-133 . PLANNe) P (A) XP (NC) So, A and we are not independent Events.

We know that a events E, and Ez are indepent if [PCE,06) PCE) 4 PCE)) the event & test is positive" is not independent from the event in patient has no tumor"