# hw1soln - ORIE 3500/5500, Fall 10 HW 1 Solutions Assignment...

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Unformatted text preview: ORIE 3500/5500, Fall 10 HW 1 Solutions Assignment 1 Solutions Problem 1 (a) A = { TTH ; THT ; HTT } B = { TTH ; THT ; HTT ; TTT } C = { HHH ; HTH ; HHT ; HTT } D = { THH ; TTH ; THT ; TTT } (b) A c = { HHH ; THH ; HTH ; HHT ; TTT } A ( C D ) = A = A = { TTH,THT,HTT } A D c = { HTT } Problem 2 We have P ( A B ) = P ( A ) + P ( B )- P ( A B ) = 1 . 3- P ( A B ). Since P ( A B ) 1, we have 0 . 3 P ( A B ). Moreover, ( A B ) A , so that P ( A B ) P ( A ) = 0 . 6 . Thus we conclude that 0 . 3 P ( A B ) . 6 . Problem 3 Since each of the 4 houses can call any of the 4 plumbers, there are 4 4 possible outcomes. We assume that each house picks a plumber randomly and indepen- dently from the other houses, so that each outcome is equally likely. There are ( 4 1 ) ways of picking a single plumber out of 4, and for any fixed plumber, the number of outcomes where exactly that plumber (and none of the others) is called to all 4 houses is n 1 = 1 ....
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## hw1soln - ORIE 3500/5500, Fall 10 HW 1 Solutions Assignment...

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