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Unformatted text preview: Eleventh Homework Set — Solutions Chapter 6 Problem 11 Let A be the number of people buying an ordinary set, B the number of people buying a plasma set, and C the number of people who are just browsing. Then P { A = 2 ,B = 1 ,C = 2 } = 5! 2!1!2! . 45 2 · . 15 · . 4 2 = . 1458. Problem 13 Let X be uniform on ( 15 , 15), and let Y be uniform on ( 30 , 30). Nobody waits longer than five minutes if  Y X  < 5. P { Y X  < 5 } = P { 5 < Y X < 5 } = P { X 5 < Y < X + 5 } = integraldisplay 15 15 integraldisplay x +5 x 5 1 30 · 60 dydx = 30 · 10 30 · 60 = 1 6 . The probability that the man arrives first is P { X < Y } = 1 2 by sym metry. Problem 14 Let X,Y be uniform random variables on (0 ,L ). Let Z =  Y X  . We want to find E [ Z ]. First, find F Z ( a ), for a ≥ 0. We have F Z ( a ) = P { Z ≤ a } = P { Y X  ≤ a } = P { a ≤ Y X ≤ a } = 2 aL a 2 L 2 . us ing geometric considerations. Hence, f Z ( x ) = 2 L 2 x L 2 if 0 ≤ a ≤ L . Hence, E [ Z ] = integraldisplay L x · 2 L 2 x L 2 dx = 2 L 2 parenleftbigg Lx 2 2 x 3 3 parenrightbigg  L = L 3 . Problem 18 Let X be uniform on (0 , L 2 ) and let Y be uniform on ( L 2 ,L ). We want 1 to find P braceleftbig Y X > L 3 bracerightbig . P braceleftbigg Y X > L 3 bracerightbigg = P braceleftbigg Y < L 2 + L 3 ,X < Y L 3 bracerightbigg + P braceleftbigg Y > L 2 + L 3 bracerightbigg = integraldisplay 5 L 6 L 2 integraldisplay y L 3 4 L 2 dxdy + integraldisplay L 2 5 L 6 2 L dy = 4 9 + 1 3 = 7 9 . Problem 20 If the joint density function of X and Y is f ( x,y ) = braceleftBigg xe ( x + y ) x > ,y > otherwise, then f ( x,y ) = f X ( x ) f Y ( y ), where f X ( x ) = xe x for x > 0, and f Y ( y ) = e y for y > 0 (0 otherwise), so that X and Y are independent. If f ( x,y ) = braceleftBigg 2 < x < y, < y < 1 otherwise, then X and Y are not independent because the nonzero values of f are located in a triangular domain....
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 probability density function, 1 K, 2 j, 2 L, 2 J, 2 5L

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