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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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 Fall '05
 Ross
 Trigraph, Natural number, sample space S., Dr. Ashok Patel, Ei ∩Ej ∩B

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