Homework 11 Solutions

Homework 11 Solutions - a Øh4Ô Ö iÒg6ÓÑ eÛ Ó Ök Ó...

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Unformatted text preview: a Øh4Ô Ö iÒg6ÓÑ eÛ Ó Ök# Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 5.1 Problem 5: The probability that each die gives an even number is 1 / 2. The probability that they both give an even number is (1 / 2) × (1 / 2) = 1 / 4. The probability that the both give an odd number is also 1 / 4. Thus the probability that they sum to an even number is 1 / 4 + 1 / 4 = 1 / 2. § 5.1 Problem 13: The number of hands is ( 52 5 ) . The number of hands without an ace is ( 48 5 ) . Thus the number of hands with at least one ace is ( 52 5 ) − ( 48 5 ) . The probability of getting such a hand is ( 52 5 ) − ( 48 5 ) ( 52 5 ) = 18 472 54 145 ≈ . 34 . § 5.1 Problem 24 ( a ): 1 ( 30 6 ) = 1 593 775 ≈ 1 . 68 · 10- 6 . ( b ): 1 ( 36 6 ) = 1 1 947 792 ≈ 5 . 13 · 10- 7 . ( c ): 1 ( 42 6 ) = 1 5 245 786 ≈ 1 . 91 · 10- 7 . ( d ): 1 ( 48 6 ) = 1 12 271 512 ≈ 8 . 15 · 10- 8 . § 5.2 Problem 5: There are 6 ways to roll a 7: 1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2 and 6 + 1. The probability for each of these is (1 / 7) 2 , (1 / 7) 2 , (1 / 7) 2 , (2 / 7) 2 , (1 / 7) 2 and (1...
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This note was uploaded on 07/12/2008 for the course MAT 243 taught by Professor Callahan during the Spring '06 term at ASU.

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Homework 11 Solutions - a Øh4Ô Ö iÒg6ÓÑ eÛ Ó Ök Ó...

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