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Unformatted text preview: EEE 350: Homework 2 Solutions Problem Solutions : Yates and Goodman, 1.6.3 1.7.4 1.8.5 1.9.4 1.10.2 1.11.4 and 1.11.6 Problem 1.6.3 Solution (a) Since A and B are disjoint, P [ A ∩ B ] = 0. Since P [ A ∩ B ] = 0, P [ A ∪ B ] = P [ A ] + P [ B ] P [ A ∩ B ] = 3 / 8 . (1) A Venn diagram should convince you that A ⊂ B c so that A ∩ B c = A . This implies P [ A ∩ B c ] = P [ A ] = 1 / 4 . (2) It also follows that P [ A ∪ B c ] = P [ B c ] = 1 1 / 8 = 7 / 8. (b) Events A and B are dependent since P [ AB ] 6 = P [ A ] P [ B ]. (c) Since C and D are independent, P [ C ∩ D ] = P [ C ] P [ D ] = 15 / 64 . (3) The next few items are a little trickier. From Venn diagrams, we see P [ C ∩ D c ] = P [ C ] P [ C ∩ D ] = 5 / 8 15 / 64 = 25 / 64 . (4) It follows that P [ C ∪ D c ] = P [ C ] + P [ D c ] P [ C ∩ D c ] (5) = 5 / 8 + (1 3 / 8) 25 / 64 = 55 / 64 . (6) Using DeMorgan’s law, we have P [ C c ∩ D c ] = P [( C ∪ D ) c ] = 1 P [ C ∪ D ] = 15 / 64 . (7) (d) Since P [ C c D c ] = P [ C c ] P [ D c ], C c and D c are independent. Problem 1.7.4 Solution The tree for this experiment is A 1 / 2 X X X X X X B 1 / 2 H 1 / 4 T 3 / 4 H 3 / 4 X X X X X X T 1 / 4 • AH 1 / 8 • AT 3 / 8 • BH 3 / 8 • BT 1 / 8 The probability that you guess correctly is P [ C ] = P [ AT ] + P [ BH ] = 3 / 8 + 3 / 8 = 3 / 4 . (1) 1 Problem 1.8.5 Solution When the DH can be chosen among all the players, including the pitchers, there are two cases: • The DH is a field player. In this case, the number of possible lineups, N F , is given in Problem 1.8.4. In this case, the designated hitter must be chosen from the 15 field players. We repeat the solution of Problem 1.8.4 here: We can break down the experiment of choosing a starting lineup into a sequence of subexperiments: 1. Choose 1 of the 10 pitchers. There are N 1 = ( 10 1 ) = 10 ways to do this.= 10 ways to do this....
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This note was uploaded on 04/03/2008 for the course EEE 350 taught by Professor Duman during the Fall '08 term at ASU.
 Fall '08
 Duman

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