A certain system can experience three different types of defects. Let A_i (i = 1

Question

A certain system can experience three different types of defects. Let A_i (i = 1, 2, 3) denote the event that the system has a defect of type i. Suppose that P(A_1) = 0.33 P(A_2) = 0.39, P(A_3) = 0.47, P(A_1 union A_2) = 0.64 P(A_1 union A_3) = 0.67, P(A_2 union A_3) = 0.72, P(A_1 Intersection A_2 Intersection A_3) = 0.04 (a) Find the probability that the system has exactly 2 of the 3 types of defects. (b) Find the probability that the system has a type 1 defect given that it does not have a type 2 or type 3 defect. 0.23 0.3372

Explanation / Answer

P(A1 A2) = P(A1) + P(A2) - P(A1 U A2) = 0.33 + 0.39 - 0.64 = 0.08

P(A1 A3) = P(A1) + P(A3) - P(A1 U A3) = 0.33 + 0.47 - 0.67 = 0.13

P(A2 A3) = P(A2) + P(A3) - P(A2 U A3) = 0.39 + 0.47 - 0.72 = 0.14

a) P(only A1 and A2) = P(A1 A2) - P(A1 A2 A3) = 0.08 - 0.04 = 0.04

P(only A1 and A3) = P(A1 A3) - P(A1 A2 A3) = 0.13 - 0.04 = 0.09

P(only A2 and A3) = P(A2 A3) - P(A1 A2 A3) = 0.14 - 0.04 = 0.10

P(exactly 2 defects) = P(only A1 and A2) + P(only A1 and A3) + P(only A2 and A3)

                                = 0.04 + 0.09 + 0.10

                                = 0.23

b) P(A1 A2' A3') = P(A1) - [P(only A1 and A2) + P(only A1 and A3) + P(A1 A2 A3) ]

                                 = 0.33 - [0.04 + 0.09 + 0.04]

                                 = 0.16

P(A2' A3') = 1 - P(A2 U A3) = 1 - 0.72 = 0.28

P(A1 | A2' A3') = P(A1 A2' A3') / P(A2' A3')

                           = 0.16/0.28

                           = 0.5714 (ans)

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