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

Question

A certain system can experience three different types of defects. Let A, (i = 1, 2, 3) denote the event that the system has a defect of type Suppose that the following probabilities are true. P(A_1) = 0.10 P(A_2) = 0.07 P(A_3) = 0.06 P[A_1 union A_2) = 0.11 P(A_1 union A_1) = 0.13 p[A_2 union A_3) = 0.11 p(a, Intersection A_2 Intersection A_3) = 0.01 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.) Given that the system has a type 1 defect, v/hat is the probability that it has all three types of defects? (Round your answer to four decimal places.) 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.) 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? (Round your answer to four decimal places.)

Explanation / Answer

here P(A1nA2)=P(A1)+P(A2)-P(AUA2)=0.1+0.07-0.11=0.06

P(A2nA3)=0.07+0.06-0.11=0.02

P(A1nA3)=0.1+0.06-0.13=0.03

hence a) P(A2|A1)=P(A1nA2)/P(A1) =0.06/0.1=0.6

b) P(A1nA2nA3|A1) =P(A1nA2nA3)/P(A1) =0.01/0.1=0.1

c)probabilty of at least one defect=P(A1UA2A3)=P(A1)+P(A2)+P(A3)-P(A1nA2)-P(A2nA3)-P(A1nA3)+P(A1nA2nA3)

=0.1+0.07+0.06-0.06-0.02-0.03+0.01=0.13

probability of exactly one defect =P(A1UA2UA3)-(P(A1nA2)+P(A2nA3)+P(A1nA3)-2P(A1nA2nA3))

=0.13-(0.06+0.02+0.03-2*0.01)=0.04

hence probabilty of exactly one defect given at least one defect =0.04/0.13=0.3077

d)probabilty of first two types of defect =P(A1nA2nA3)+P(A1nA2nA3c) =0.06

hence P(A1nA2nA3c) =0.06-0.01=0.05

hence probabilty that it does not have third type of defect given has first two type of defects =0.05/0.06=0.8333

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