8. A certain system can experience three different types of defects. Let A -1,2,

Question

8. 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 i, Suppose that the following probabilities are true. PA1) 0.10, PA:)-0.08 PAs)-0.05 P (Ai U A:)-0.12 P (A1 UAs)-0.12 Given that the system has a type 1 defect, what is the probability that it has a type 2 defect? a. b. Given that the system has a type 1 defeet, what is the probability that it has all three types of defect? c. Given that the system has at least one type of defect, what is the probability that it has exactly one type of defeet?

Explanation / Answer

P(A1intersectionA2)=P(A1)+P(A2)-P(A1UA2)= 0.10+0.08-0.12= 0.06

P(A1intersectionA3)=P(A1)+P(A3)-P(A1UA3)= 0.10+0.05-0.12= 0.03

P(A2intersectionA3)=P(A2)+P(A3)-P(A2UA3)= 0.08+0.05-0.11= 0.02

a) P(it has type 2 defect)=P(A1intersectionA2)/P(A1)= 0.06/0.10=0.6

b) P(all three type of defect)= P(A1intersectionA2intersectionA3)/P(A1)= 0.01/0.10=0.1

C) P(exactly one type of defect)=P(A1UA2)+P(A3)-P(A1intersectionA2)-2P(A1intersectionA3)-2P(A2intersectionA3)+3P(A1intersectionA2intersectionA3)=

=0.12+0.05-0.06-2(0.03)-2(0.02)+3(0.01)

=0.04

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