Reconsider the system defect situation described in Exercise 26 (Section 2.2). a. Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? b. Given that the system has a type 1 defect, what is the probability that it has all three types of defects? c. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? d. Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect? Reference exercise- 26 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

Problem 48E Answer: Step1: We have 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(A )= 0.12 1 P(A 2) = 0.07 P(A 3) =0.05 P(A A )= 0.13 1 2 P(A1A )30.14 P(A A ) = 0.10 2 3 P(A 1A 2 3) = 0.01 Our goal is to find, a. Given that the system has a type 1 defect, what is the probability that it has a type 2 defect b. Given that the system has a type 1 defect, what is the probability that it has all three types of defects c. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect d. Given that the system has both of the first two types of defects, what is the probability that it does not have the third type of defect Step2: a). Given that the system has a type 1 defect,Then the probability that it has a type 2 defect is given by P(A | A ) = P(A A )/P(A ) 2 1 0.06 1 2 1 = 0.12 = 0.5 Where, P(A1 A2) = P(A1)+ P(A2) – P(A1 U A2) = 0.12 + 0.07 – 0.13 = 0.06 b). P(A1 A2 A3) P(A1 A2 A3|A1) = P(A1) = 1 12 =0.084 Step3: c). Given that the system has at least one type of defect,Then the probability that it has exactly one type of defect is given by {P(A n1t others)P(A not2others)P(A not o3hers)} = P(A 1A 2A )3 P((A A A )) + P((A A A )) + P((A A A )) 1 1 2 3 1 2 3 1 2 2 = P(A 1A 2 ) 3 {P(A 1P(A 1A )P2A A 1} + 3P(A ) P(2 A )P1A 2 )} + {1( A 3 P(A A 3P(A 1 )} 2 1 3 = P(A 1+P(A )+2(A )P3A A )1P(A2A )P2A 3 )+P(A1 A3 A ) 1 2 3 = {0.12(0.12+0.070.13)(0.12+0.050.14)}+{0.07 (0.12+0.070.13)(0.12+0.050.14)} + {0.05 (0.12+0.070.14)(0.12+0.050.14 0.12+0.07+0.05(0.06)(0.02)(0.03) + 0.01 0.04 + 0 + 0.01 = 0.14 0.05 = 0.14 = 0.3571 d). Given that the system has both of the first two types of defects,Then the probability that it does not have the third type of defect is given by 1 P(A 3 A1 A2) P(A |A A 1 = 2 3 P(A1 A2) P(A1A2)P(A 1A 2A 3 = P(A1 A2) = 0.06 0.01 0.06 = 0.05 0.06 = 0.84