(a)
Given type 1 defect, probability that it has a type 2 defect = P(A2 | A1)
=
(b)
Given type 1 defect, probability that it has all type 2 defect =
(c)
Probability that the system has at least one type of defect =
= 0.1 + 0.08 + 0.06 - 0.06 - 0.03 - 0.02 + 0.01
= 0.14
By Venn diagram, Probability that it has exactly one type of defect =
= 0.14 - 0.06 - 0.03 - 0.02 + 2 * 0.01 = 0.05
Given that the system has at least one type of defects, probability that it has exactly one type of defect
= P(exactly one type of defect | at least one type of defects)
= P(exactly one type of defect and at least one type of defects) / P(at least one type of defects)
= 0.05 / 0.14 = 0.3571
(d)
Given that the system has both of the first two types of defects, probability that it does not have third type of defect
=
A certain system can experience three different types of defects. Let A, i1,2,3) denote the event...
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