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A certain system can experience three different types of defects. Let A, ( 1,2,3) denote the event that the system has a defect of type .Suppose that the following probabilities are true PA,) 0.11 PA2)-0.08 PA) -0.06 P(A, UA2) -0.13 PIA, UA) -0.1 (a) Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? (Round your answer to four decimal places.) (b) Given that the system has a type 1 defect, what is the probablity that it has all three types of defects? (Round your answer to four decimal places.) (c) Given that the system has at least one type of defect, what is the probability that it has exactly one type of defect? (Round your answer to four decimal places.) first two types of defects, what is the probability that it does not have the third type of defect? (Round your answer to four (d) Given that the system has both of the decimal places.) Need Help?
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Answer #1

small P(A_1 cap A_2) =P(A_1)+P(A_2)-P(A_1cup A_2) = 0.11+0.08-0.13 =0.06

small P(A_1 cap A_3) =P(A_1)+P(A_3)-P(A_1cup A_3) = 0.11+0.06-0.14 =0.03

PlA2N Аз) = P(Ap) + P(As)-PLA2 U Aa) 0.084 0.06-0.12 = 0.02

a) Probability that the system has a type 2 defect given that it has a type 1 defect:

P ( P(A21A1) = P(A1) P

= 0.1-= 0.5455 0.11 .45

b) Probability that the system has all three type of defect given that it has a type 1 defect:

P(A1

0.01 0.11 0.0909

c) small P(A_1 only)= P(A_1)-P(A_1cap A_2)-P(A_1cap A_3)+P(A_1cap A_2cap A_3)

0.11-0.06-0.03 + 0.01 0.03

small P(A_2 only)= P(A_2)-P(A_1cap A_2)-P(A_2cap A_3)+P(A_1cap A_2cap A_3)

0.08-0.06-0.02 + 0.01 0.01

small P(A_3 only)= P(A_3)-P(A_1cap A_3)-P(A_2cap A_3)+P(A_1cap A_2cap A_3)

0.06-0.03-0.02 + 0.01 0.02

small P(A_1cup A_2 cup A_3 )=P(A_1)+P(A_2)+P(A_3)-P(A_1cap A_2 )-P(A_1cap A_3)-P(A_2cap A_3 )+P(A_1cap A_2 cap A_3)

0.11 + 0.08 0.06-0.06-0.03-0.02 + 0.01 0.15

Probability that the system has exactly one defect given that it has atleast one type of defect:

small P(Exactly one|Atleast one)=rac{P(A_1 only )+P(A_2 only )+P(A_3 only )}{P(A_1cup A_2cup A_3)}

0.03+ 0.010.02 0.15 0.4000

d) Probability that the system does not have type 3 defect given that it has both 1 and 2 defect:

small P(A'_3|A_1cap A_2) = rac{P(A_1cap A_2cap A'_3)}{P(A_1cap A_2)}

PLA 1 กั A2)-PlA1 กั An A3) PlA1 กั A2) 0.06-0.01 0.06 0.8333

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