{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math414 hw2

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

This preview shows pages 1–2. Sign up to view the full content.

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 ( 5 3 ) = ( 5 2 ) = 5! · 4 2!3!

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online