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# HW2S - S Therefore E i ∩ E j = for i ≠ j and S E n i i...

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ISE 220 PROBABILITY CONCEPTS IN ENGINEERING Solutions to Homework Problem Set # 2 Instructor: Dr. Ashok Patel Fall 2010 Problem 1 (a) Prove that A-B and A B form a partition of the set A. Solution: Note that ( A-B ) ( A B ) = ( A B c ) ( A B ) = A B c B = A φ = . Further A = A S = A ( B B c ) = ( A B ) ( A B c ) by the distributive law = ( A B ) ( A-B ) Thus ( A-B ) ( A B ) = and ( A-B ) ( A B ) = A . Hence A-B and A B form a partition of the set A. (b) Suppose that E 1 , E 2 , …, E n form a partition of sample space S . Let B S . Define A 1 , A 2 , … , A n such that A i = E i B . Then prove that A 1 , A 2 , … , A n form a partition of the set B . Solution: Note that E 1 , E 2 , …, E n form a partition of sample space

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Unformatted text preview: S. Therefore E i ∩ E j = for i ≠ j and S E n i i = = U 1 . We need to show that ( E i ∩ B ) ∩ ( E j ∩ B ) = for i ≠ j and U n i i B E 1 ) ( = ∩ = B . Note that ( E i ∩ B ) ∩ ( E j ∩ B ) = E i ∩ E j ∩ B = ∩ B since E i ∩ E j = for i ≠ j = for i ≠ j and U n i i B E 1 ) ( = ∩ = B E n i i ∩ = U 1 by the distributive law = S B ∩ since S E n i i = = U 1 = B . Problem 2 Problem 3 Problem 4 Important: Let ¬ S = S c , ¬ G = G c , ¬ F = F c , ¬ ( S ∪ G ∪ F ) = ( S ∪ G ∪ F ) c ....
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HW2S - S Therefore E i ∩ E j = for i ≠ j and S E n i i...

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