hw02soln

# hw02soln - ECE 310 Spring 2005 HW 2 solutions Problem E2.1...

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ECE 310, Spring 2005, HW 2 solutions Problem E2.1 a) Using sampling with replacement formula (page 59): μ 26 , 2 = 26 2 = 676. b) Let A be the set of all Scrabble words of length 2 and Ω be all possible pairs of letters. Then, k A k = 96 and k Ω k = 26 2 . = P ( A ) = 96 26 2 = . 142 c) Let C be the set of all Scrabble words of length 3 that begin with c and Ω as before. Then, k C k = 37 and k Ω k = 26 2 . = P ( C ) = 37 26 2 0 . 055 d) Let A be the set of all words of length 4, and B be the set of all possible words of length 4 that contain exactly three vowels. Then: k A B k = 62, k B k = 4 × 21 × 5 3 = 10500, where 4 - positions of a consonant, 21 - possible consonants, 5 3 - possible triples of vowels. = P ( A | B ) = 62 10500 5 . 905 × 10 - 3 Problem E2.3 Let A be the event that M produces a byte, then: P ( A ) = 2 8 3 8 = 0 . 039 Problem E2.7 For the given length n, there are 4 n different messages, so we want: 4 n > 1000 = n > 4 . 982 , thus the minimum number required is 5. Problem E2.12 Ω is the set of all possible 64-bit sequences and A is the set of all possible 64-bit sequences that start with 00 and have exactly 4 zeros. Then, k A k = ( 62 2 ) and k Ω k = 2 64 . = P ( A ) = ( 62 2 ) 2 64 1

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Problem E2.15 d) Let D be the event of at least one bit has a value 1, then D c is the event of receiving  . Therefore: P ( D ) = 1 - P ( D c ) = 1 - 1 2 8 e) Let E be the event that b 1 + b 2 = 1, then k E k = 2 × 2 6 = 2 7 , where 2 - possible outcomes for b 1 b 2 , either  or  ; 2 6 - possible outcomes for the remaining 6 bits.
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• Spring '05
• HAAS
• Trigraph, The Gates, Possible world

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