Solution3

# Solution3 - SOLUTIONS FOR HOMEWORK 3 15. (a) @ 4 1 1 A @ 13...

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Unformatted text preview: SOLUTIONS FOR HOMEWORK 3 15. (a) @ 4 1 1 A @ 13 5 1 A @ 52 5 1 A = 33 16660 ≈ . 0020. (b) 13 @ 4 2 1 A @ 12 3 1 A ... 4 · 4 · 4 @ 52 5 1 A = 352 833 ≈ . 4226. (c) @ 12 2 1 A @ 4 2 1 A @ 4 2 1 A · 44 @ 52 5 1 A = 396 8330 ≈ . 0475. (d) 13 · @ 4 3 1 A @ 12 2 1 A · 4 · 4 @ 52 5 1 A = 176 8330 ≈ . 0211. (e) @ 13 1 1 A @ 4 4 1 A @ 48 1 1 A @ 52 5 1 A = 1 4165 ≈ . 00024. 25. Let E n denote the event that a 5 occurs on the n th roll and no 5 or t occurs on the first n- 1 rolls. Then { E n } are mutually disjoint, and S ∞ n =1 E n is the event that 5 occurs before 7. So the desired probability is P ( ∞ [ n =1 E n ) = ∞ X n =1 P ( E n ) . We need to calculate P ( E n ). Note that E 1 = { (1 , 4) , (2 , 3) , 3 , 2) , (4 , 1) } . So P ( E 1 ) = 4 36 = 1 9 . Note also that E 2 = { ( x 1 , x 2 , x 3 , x 4 ) | x 3 + x 4 = 5 , x 1 + x 2 6 = 5 , x 1 + x 2 6 = 7 } . So E 2 has (36- 10) × 4 = 26 × 4 outcomes, and thus P ( E 2 ) = 26 × 4 (36) 2 = 1 9 26 36 ....
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## This homework help was uploaded on 04/13/2008 for the course M 362k taught by Professor Berg during the Spring '08 term at University of Texas.

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Solution3 - SOLUTIONS FOR HOMEWORK 3 15. (a) @ 4 1 1 A @ 13...

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