A certain system can experience three different types of defects. Let A1 (i = 1,

Question

A certain system can experience three different types of defects. Let A1 (i = 1,2,3) denote the event that the system has a defect of type i. Suppose that the following probabilities are true. P(A1) = 0.10 P(A2) = 0.08 P(A3) = 0.05 P(A1 A2) = 0.12 P(A1 A3) = 0.12 P(A2 A3) = 0.11 P(A1 A2 A3) = 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, what 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

a) P(A2/A1)=P(A1intersection A2)/P(A1)=0.12/0.1=0.0120

b)P(all three defects/A1)=P(A1intersection A2 intersection A3 intersection A1)/P(A1)=0.01/0.1=0.0010

c)P(system has atleast 1 type of defect)=

 P(A1 union A2  union A3)=P(A1)+P(A2)+P(A3)-P(A1 INTERSCTION A2)-P(A1 INTERSCTION  A3)-P(A2 INTERSCTION A3)-2(A1 INTERSCTION A2 INTERSCTION A3)=0.01+0.08+0.05-0.12-0.12-0.11-2x(0.01)=0

P((system has atleast 1 type of defect)/A1)=0

d)P(system has 3rd defect)=0.05

p(A1 intersection A2)=0.12

P(A1/(A1intersection A2))=P(A1intersection A2 intersection A3 intersection )/P(A1)=0.01/0.12=0.0833

P(not A1 defect/(A1intersection A2))=1-P(A1/(A1intersection A2))=1-0.0833=0.9166

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