math414 hw2 - SOLUTIONS FOR HWK # 2 1.3-7 The sample space...

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SOLUTIONS FOR HWK # 2 1.3-7 The sample space consists of all possible ordered sample of size 4 from 10 digits. N = 10 4 = 10 , 000. (a) Each permutation of the different digits 6, 7, 8, 9 is favorable. Therefore N ( A ) = P (4) = 4! = 24; P ( A ) = 24 10 , 000 = 0 . 0024. (b) We consider distinguishable permutations of three kinds of digits; here n 6 = 1 ,n 7 = 1 ,n 8 = 2 ,n = 4. Therefore N ( A ) = 4! 1!1!2! = 24 2 = 12; P ( A ) = 12 10 , 000 . (c) n 7 = 2 ,n 8 = 2 ,n = 4; N ( A ) = 4! 2!2! = 24 4 = 6; P ( A ) = 6 10 , 000 . (d) n 7 = 1 ,n 8 = 3; N ( A ) = 4! 1!3! = 24 6 = 4; P ( A ) = 4 10 , 000 . 1.3-11(c) The series stops at the sixth game in the following two (mutually exclusive) cases: C 1 = ” AL wins 4 games in the first 5 games and wins the 6th game”; C 2 = ” NL wins 4 games in the first 5 games and wins the 6th games. Let us mark a game by ”AL” (resp. ”NL”) if it won by AL (resp. by NL). In case C 1 the 6th game is chosen for ”AL” and from the previous 5 games 3 ones have to be chosen for the ”AL” and 2 games for ”NL”. So in this case the number of choices is
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This note was uploaded on 10/13/2010 for the course MATH 414 at Pennsylvania State University, University Park.

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math414 hw2 - SOLUTIONS FOR HWK # 2 1.3-7 The sample space...

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