# hw1sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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ORIE 3500/5500 Fall Term 2008 Assignment 1-Solution 1. (a) P ( B ) = 1 - P ( B c ) = 1 - 0 . 35 = 0 . 65. Since P ( A B ) = P ( A ) + P ( B ) - P ( A B ), we have P ( A B ) = 0 . 55 + 0 . 65 - 0 . 75 = 0 . 45 . P ( A c B ) = P ( B ) - P ( A B ) = 0 . 65 - 0 . 45 = 0 . 2 . P ( A B c ) = P ( A ) - P ( A B ) = 0 . 55 - 0 . 45 = 0 . 1 . (b) Note that x ( A B ) ( A B ) c x ( A B ) and x ( A B ) c x ( A B ) and x 6∈ ( A B ) either ( x A and x 6∈ B ) or ( x 6∈ A and x B ) x ( A B c ) or x ( A c B ) x ( A c B ) ( A B c ) . Conversely if x ( A c B ) ( A B c ), then either x ( A c B ) or x ( A B c ). In the ﬁrst case x 6∈ A and x B . So x ( A B ) and x 6∈ ( A B ). In the second case x A and x 6∈ B . So then also x ( A B ) and x 6∈ ( A B ). Hence in both the cases x ( A B ) ( A B ) c . This shows the desired result.

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## This note was uploaded on 12/04/2008 for the course STSCI 6000 taught by Professor Turnbull during the Fall '08 term at Cornell.

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hw1sol - ORIE 3500/5500 Fall Term 2008 Assignment...

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