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ID: 3353604 • Letter: U
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ur best submission for each question part is used for your score. +0.5/1 points | Previous Answors Devorestat 2.E.048. My Notes Ask Your Teacher A certaln system can experience three different types of defects. Let A, (i1,2,3) denote the event that the system has a defect of type i. Suppose that the following probablitles are true. P(41) 0.10 P(Az) = 0.07 A(A3) = 0.06 P(A2 UA.) = 0.11 r(Al n A, n A.) = 0.01 (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.) 0.6000 (b) 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.) .1 (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.) (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? (Round your answer to four decimal places.) Need Help? Read It ILTakto a Tutol Submit Answer Save Progress |Practice Another Version View Previous Question Question 3 of 16 View Next Question Home My Assignments Extension Request 5:05 PM Type here to search 1/27/20181Explanation / Answer
as we know that P(AnB)=P(A)+P(B)-P(AUB)
therefore from above:
a)
P(A2|A1) =P(A1nA2)/A1 =0.06/0.1 =0.60
b)P(A1nA2nA3|A1) =0.01/0.1 =0.1
c)P(system has at least type of defect) =
P(exactly one type of defect) =
hence probability =0.04/0.130=.3077
d) Probability =(P(A1nA2)-P(A1nA2nA3))/P(A1nA2) =(0.06-0.01)/0.06 =0.8333
P(A1)= 0.1 P(A2)= 0.07 P(A3)= 0.06 P(A1nA2)= 0.06 P(A2nA3)= 0.02 P(A1nA3)= 0.03 P(A1UA2)= 0.11 P(A2UA3)= 0.11 P(A1UA3)= 0.13 P(A1nA2nA3) = 0.01Related Questions
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