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Unformatted text preview: Chapter 7 P1 Let X (= H,T ) be the result of flipping coin, and Y (= 1 , 2 ,... 6) the value that appears on the die. The winnings g ( x,y ) is 2 y if X = H and 1 2 y if X = T . By Proposition 2.1 the expected winnings are E [ g ( X,Y )] = ∑ y ∑ x g ( x,y ) p ( x,y ) = 6 ∑ y =1 (2 y · 1 12 + 1 2 y · 1 12 ) = 5 2 · 1 12 6 ∑ y =1 y = 35 8 P4 E [ XY ] = 1 R y R xy 1 y dx dy = 1 R y 2 2 dy = 1 6 . E [ X ] = 1 R y R x 1 y dx dy = ... = 1 4 . E [ Y ] = 1 R y R y 1 y dx dy = ... = 1 2 . Note that E [ XY ] 6 = E [ X ] · E [ Y ]. P6 Let X i be the result of the i th roll, and S = 10 ∑ i =1 X i the sum. Then E [ X i ] = 6 ∑ n =1 n · 1 6 = 7 2 and E [ S ] = 10 ∑ i =1 E [ X i ] = 10 · 7 2 = 35. P7 Enumerate the objects 1 through 10, and let A n = 1 if object # n is chosen by A 0 otherwise ¬ A n = 1 if object # n is not chosen by A 0 otherwise B n = 1 if object # n is chosen by B 0 otherwise ¬ B n = 1 if object # n is not chosen by B 0 otherwise (a) Let X be the number of objects chosen by both A and B. Then X = 10 ∑ n =1 A n · B n and E [ X ] = E [ 10 ∑ n =1 A n · B n ] = 10 ∑ n =1 E [ A n · B n ] =(by independence of A n from B n ) = 10 ∑ n =1 E [ A n ] · E [ B n ] = 10 · 3 10 · 3 10 = 9 10 (b) Let Y be the number of of objects not chosen by either A or B. Then Y = ∞ ∑ n =1 ¬ A n · ¬ B n and E [ Y ] = 10 · 7 10 · 7 10 = 4 . 9....
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This note was uploaded on 10/27/2010 for the course MATH 351 taught by Professor Moumen,f during the Spring '08 term at George Mason.
 Spring '08
 Moumen,F

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