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Unformatted text preview: Will Landau September 12, 2011 Problem Set 3 Solutions Exercise 3.1 (Devore 2.29) . a. Domain names with just 2 letters: this is like sampling from a bag of 26 unique blocks with replacement. # names = (# letters) (# letters) = 26 26 = 676 . Domain names of length 2 with digits and letters: this time we have 36 things to choose from (26 letters, 10 digits. # names = 36 36 = 1296 b. This is exactly like part (a) except that now, domain names are 3 characters long instead of 2. Without digits: # names = 26 3 = 17,576 . With digits: # names = 36 3 = 46,656 . c. 26 4 = 456,976 ;36 4 4 = 1,679,616 d. probability that a 4character name is already taken = 1 probability that a name is available = 1 # available names # allowable names = 1 97786 36 4 = .942 Exercise 3.2 (Devore 2.32) . a. 5 for receiver 4 for disc player 3 for speakers 4 for turntable = 240 b. 1 1 3 4 = 12 c. 4 3 3 3 = 108 d. # ways with at least on Sony component = (total # ways)  (# with no Sony component) = 240 4 3 3 3= 132 e. P(at least one Sony component) = # ways to select at least one Sony component total # possible selections = 132 240 = .55 P(exactly one sony component) = P(the only Sony component is the receiver) + P(the only Sony component is the compact disc player) + P(the only Sony component is the turntable) = 1 3 3 3 240 + 1 4 3 3 240 + 4 3 3 1 240 = .413 1 Exercise 3.3 (Devore 2.34) . a. ( 25 5 ) = 25! 20!5! = 53 , 130 b. # ways = (# ways to select 2 with electrical defects) (# ways to select 3 without electrical defects) = ( 6 2 ) ( 19 3 ) = 6! 4!2! 19! 16!3! = 14535 c. P(at least 4) = P(exactly 4) + P(exactly 5) = ( 19 4 ) ( 6 1 ) ( 25 5 ) + ( 19 5 )( 6 ) ( 25 5 ) = 0.6565 Exercise 3.4 (Devore 2.38) . a. P(select 2 75W bulbs) = (# ways to select 2 75W bulbs)(# ways to select one non75W bulb) total # ways to select 3 bulbs = ( 6 2 )( 9 1 ) ( 15 3 ) = .2967 b. P(same rating) = P(select all 40W) + P(select all 60W) + P(select all 75W) = ( 4 3 ) ( 15 3 ) + ( 5 3 ) ( 15 3 ) + ( 6 3 ) ( 15 3 ) = .0747 c....
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