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**Unformatted text preview: **Twelfth Homework Set — Solutions Chapter 7 Problem 5 If ( X,Y ) is the location of the accident, then X and Y are uniform random variables on ( − 3 2 , 3 2 ). Let D = | X | + | Y | . Then E [ D ] = E [ | X | ] + E [ | Y | ] = 2 E [ | X | ] = 2 integraldisplay 3 2- 3 2 | x | 3 dx = 4 3 integraldisplay 3 2 xdx = 4 3 · 9 8 = 3 2 . Problem 6 Let X i be the outcome of the i-th roll of the die, for i = 1 ,..., 10, and note that E [ X i ] = 7 2 . Let X = X 1 + ··· + X 10 . Now E [ X ] = E [ X 1 ] + ··· + E [ X 10 ] = 10 E [ X 1 ] = 35. Problem 7 (a) Let X i be one if both A and B choose the i-th object, for i = 1 ,..., 10 . Then E [ X i ] = P { X i = 1 } = ( 3 10 ) 2 = 9 100 . Now, the expected number of objects chosen by both A and B is E [ X 1 ] + ··· + E [ X 10 ] = 10 E [ X 1 ] = 0 . 9. (b) Let Y i be one if neither A nor B choose the i-th object. Then E [ Y i ] = P { Y i = 1 } = ( 7 10 ) 2 = 49 100 , so that E [ Y 1 + ··· + Y 10 ] = 10 E [ Y 1 ] = 4 . 9 . (c) Let Z i be one if either A or B (but not both) chooses the i- th object. Then E [ Z i ] = P { Z i = 1 } = 2 3 10 7 10 = 21 50 . Now, E [ Z 1 + ··· + Z 10 ] = 10 E [ Z 1 ] = 21 5 = 4 . 2 . Problem 8 Following the hint, let X 1 be one if the i-th arrival sits at a previously unoccupied table. Then E [ X i ] = P { X i = 1 } = (1 − p ) i- 1 , so that E [ X 1 + ··· + X N ] = N summationdisplay i =1 (1 − p ) i- 1 = 1 − (1 − p ) N 1 − (1 − p ) = 1 − (1 − p ) N p . Problem 11 Let X i be one if the i-th outcome differs from the ( i − 1)-th outcome, for i = 2 ,...,n . We have E [ X i ] = P { X i = 1 } = 2 p (1 − p ), so that E [ X 2 + ··· + X n ] = 2( n − 1) p (1 − p ) . 1 Problem 18 Let X i be one if the i-th card is a match, for i = 1 ,..., 13, and let X = X 1 + ··· + X 52 . Then P { X i = 1 } = 1 13 , so that E [ X ] = 52 E [ X 1 ] = 52 13 = 4 . Problem 19 (a) If X is the number of insects caught before a type 1 catch, then ( X + 1) is geometric with parameter P 1 , so that E [ X ] = 1 P 1 − 1....

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