mdm12Section6_1_OddSolutionsFinal

mdm12Section6_1_OddSolutionsFinal - Q1 Q7-9 Q3-5 Q11...

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Q1 Q3-5 Q7-9 Q11 CHAPTER 6 Section 6.1, pp. 312–313 Solutions to Odd Number Problems 1. a) () 1 2 m Px n = = b) 1 4 m n = = c) (at least one head)= 1 (three tails) 1 =1 8 7 8 PP = d) There are two composite numbers, 4 and 6, among the faces of a die. 2 (composite number) 6 1 3 P = = e) There are three ways to roll a 4, and four ways to roll a 9 with two dice. 7 (not a perfect square) = 1 36 29 36 P = f) There are 12 face cards in a deck of 52. 12 (face card)= 52 3 13 P = Top
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3. A A Sum 2 3 4 Prod 2 3 4 B 2 4 5 6 B 2 4 6 8 3 5 6 7 3 6 9 12 4 6 7 8 4 8 12 16 a) i) (2,3), (3, 2), (2, 4), (4, 2), (3, 3), (4, 4), (3, 4), (4, 3) ii) no outcomes iii) (2, 2) b) Player B has the advantage because the probability of the product being greater than the sum of the numbers drawn is 8 9 . 5. a) b) 6 (2 ) 16 3 8 Px == = c) 1 (4 ) 16 d) Assumptions: each team is equally likely to win; no tie is allowed. Top
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7. a) n (sum=5) = 4, n (sum=7) = 6, n (sum=9) = 4 464 (Player A wins) 36 7 18 P ++ = = 7 (Player B wins) =1 18 11 18 P = b) n (sum = doubles) = 6 466 (Player A wins) = 36 4 9 P = 4 (Player B wins) = 1 9 5 9 P = c) Answers may vary. Player A wins if sum is 5, 7, 11, or doubles.
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This note was uploaded on 01/23/2012 for the course MATHS MDM-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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mdm12Section6_1_OddSolutionsFinal - Q1 Q7-9 Q3-5 Q11...

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