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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 2 - Problem 26e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 2 - Problem 26e

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# A certain system can experience three different types of ISBN: 9780321629111 32

## Solution for problem 26E Chapter 2

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 26E

A certain system can experience three different types of defects. Let Ai(i= 1,2,3) denote the event that the system has a defect of type ?i?. Suppose that P(A1)= .12 P(A2) = .07 P(A3) =.05 P(A1 ?A2)= .13 P(A1 ?A3)= .14 P(A2 ?A3) = .10 P(A1 ?A2 ?A3) = .01

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Problem 26E Answer: Step1: We have A certain system can experience three different types of defects. Let A(i= 1,2,3) denote i the event that the system has a defect of type i . Suppose that P(A 1= 0.12 P(A 2= 0.07 P(A 3= 0.05 P(A 1 ) =2.13 P(A A ) = 0.14 1 3 P(A 2 ) =3.10 P(A 1 A2 = 0.3 Our goal is to find, a.What is the probability that they system does not have a type 1 defect b.What is the probability that the system has both type 1 and type 2 defects c.What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect d.What is the probability that the system has at most two of these defects Step2: a). The probability that they system does not have a type 1 defect is given by P(system does not have a type 1 defect) = 1 - P(A ) 1 = 1 - 0.12 = 0.88 Therefore, P(system does not have a type 1 defect) = 0.88. b). The probability that the system has both type 1 and type 2 defects is given by P(A 1A ) =2(A )+ P(A1) – P(A2A ) 1 2 = 0.12 + 0.07 - 0.13 = 0.06 Therefore, P(the system has both type 1 and type 2 defects) = 0.06. Step3: c). The probability that the system has both type 1 and type 2 defects but not a type 3 defect is given by P(A 1 A 2= P(A3A ) - P(A1A A2) 1 2 3 = 0.06 - 0.01 = 0.05 d). The probability that the system has at most two of these defects is given by P(at most 2 errors) = 1 - P(all three errors) = 1 - P(A A A ) 1 2 3 = 1 - 0.01 = 0.99

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