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solution pset 1

solution pset 1 - 1Solution A1 When A1 A2 A3 are disjoint...

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1 Solution: . (2) Solution: (a) m a x (() ,() ) ( ) PA PB PA B ≤∪ ; and any probability should be a number between 0 and 1. 0.7 ( ) 1 ∪≤ ( ) m i n ) ∩≤ ; and the Bonferroni’s inequality gives the lower boundary of () . ()1 ( ) + −≤ . So 0.1 ( ) 0.4 (b)The Boole’s inequality gives the upper boundary of . ( ) PA B PA PB + , so 0.6 ( ) 0.8 A1 A2 A3 A1 A2 A3 When A1, A2, A3 are disjoint from each other, obviously, (1 2 3 ) ) (2 ) (3 ) PA A A PA PA PA ∪∪ = + + When A1, A2 and A3 are not disjoint from each other, 2 3 ) ) ) ) A < + +
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Similar to part (a), but here the boundary given by Bonferroni’s inequality is trivial ( 0.2 ( ) PA B −< ) 0( ) 0 . 2 ∩≤ (c)The Boole’s inequality is () ()1 ( ) PA PB +− .Only when ()1 0 > , the Boole’s inequality provides the non-trivial information. Otherwise, the lower boundary of got from Boole’s inequality is just a non-positive number which is already known from the definition of probability. (d)The Bonferroni’s inequality is ( ) PA B PA PB ∪≤ + . Only when () 1 +< , the Bonferroni’s inequality provides the non-trivial information.
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solution pset 1 - 1Solution A1 When A1 A2 A3 are disjoint...

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