ECE314hw2 - Tong Huang ECE 314 spring 2007 Homework 2 1....

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Unformatted text preview: Tong Huang ECE 314 spring 2007 Homework 2 1. Given P ( A ) = 0.3; Π ( Β 29 = 0.6; Π ( Χ 29 = 0.2; Π ( Α Χ 29 = 0; P ( B | C ) = 0.5; Π ( Α Β Χ 29 = 0.8. a) Because P ( A Χ 29 = 0 , which means A and C are independent events, so the intersection of A, B and C must be 0. b) P ( B | C ) = Π ( ΒΧ 29 / Π ( Χ 29 = Π ( ΒΧ 29 / 0.2 = 0.5 , therefore P ( B Χ 29 = 0.1 . P ( B Χ 29 = Π ( Β 29 + Π ( Χ 29 - Π ( Β Χ 29 = 0.6 + 0.2 - 0.1 = 0.7 c) From part b, we know that P ( B Χ 29 = 0.1 , we also know P ( B ) = 0.6 . The part of B that’s not intersecting with C is P ( B )- Π ( Β Χ 29 = 0.5 . P ( C c ) = 1 - 0.2 = 0.8. P ( B | C c ) = Π ( ΒΧ χ 29 / Π ( Χ χ 29 = 0.5 / 0.8 = 0.625 d) P ( A Β Χ 29 = Π ( Α 29 + Π ( Β 29 + Π ( Χ 29 - Π ( Α Β 29 - Π ( Α Χ 29 - Π ( Β Χ 29 - Π ( Α Β Χ 29 . From the previous part, we know that P ( A Χ 29 = 0, Π ( Β Χ 29 = 0.1, Π ( Α Β Χ 29 = 0. Therefore we know P ( A Β 29 = 0.2. P ( A | B ) = Π ( ΑΒ 29 / Π ( Β 29 = 0.2 / 0.6 = 1 / 3....
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This note was uploaded on 04/18/2008 for the course ECE 314 taught by Professor Idk during the Spring '07 term at UMass (Amherst).

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ECE314hw2 - Tong Huang ECE 314 spring 2007 Homework 2 1....

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