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# solpracmt1 - SOLUTIONS to Practice Midterm Exam 1 1 You are...

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SOLUTIONS to Practice Midterm Exam 1 1. You are given the following: P ( A ) = 0 . 35 , P ( B ) = 0 . 45 , P ( AB ) = 0 . 18 . (a) P ( A or B ) = 0 . 35 + 0 . 45 - 0 . 18 = 0 . 62 . (b) P ( A c ) = 1 - 0 . 35 = 0 . 65 . (c) First, B is the disjoint union of AB and A c B . Thus, 0 . 45 = P ( B ) = P ( AB )+ P ( A c B ) = 0 . 18 + + P ( A c B ) . Thus, P ( A c B ) = 0 . 45 - 0 . 18 = 0 . 27 . 2. (a) Draw the 5 × 5 array that represents all possible combinations of X 1 and X 2 . These 25 possibilities are equally likely. Five entries correspond to X 1 = X 2 ; thus its probability is 5 / 25 = 0 . 20 . (b) Even without drawing the 5 × 5 × 5 array, we know that it has 125 cells and all cells are equally likely. Ex- actly three of these cells give X 1 + X 2 + X 3 = 4 , namely (1,1,2), (1,2,1) and (2,1,1). Thus, the desired proba- bility is 3 / 125 . 3. (a) P ( SFS ) = pqp = (0 . 85) 2 (0 . 15) = 0 . 1084 . (b) The probability of exactly three suc- cesses is 4! 3!1! (0 . 85) 3 (0 . 15) = 0 . 3685 . (c) The event ( Y = 3) can occur only one way: Two out of two successes on Friday followed by one out of two successes on Saturday. These proba- bilities are multiplied b/c trials are in- dependent, giving: P ( Y = 3) = (0 . 85) 2 2(0 . 85)(0 . 15) = 0 . 1842 .

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