This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ....
View Full
Document
 '10
 FROHMADER

Click to edit the document details