hw1soln

hw1soln - Homework # 1 Solutions ApSc 116 Problem Page 18 #...

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Homework # 1 Solutions ApSc 116
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Problem Page 18 # 7 We know Pr ( AB ) = Pr ( A ) + Pr ( B ) - Pr ( A B ) = 0 . 4 + 0 . 7 - Pr ( A B ) = 1 . 1 - Pr ( A B ) Since 0 . 7 Pr ( A B ) 1 (the latter is true if A B ), we can conclude 0 . 1 Pr ( AB ) 0 . 4 .
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Problem Page 27 # 8 There are 7 5 outcomes (each of the 5 persons can choose a floor). If the passenger are to get down at different floors, the first can choose from 7, the second from 6, and so forth, therefore the probability is 7 · 6 · 5 · 4 · 3 7 5 = 360 2401
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Page 34 # 13 We want four designated lightbulbs to be in the same group. The probability is ( 20 6 ) + ( 20 10 ) ( 24 10 ) = 0 . 1140 . The term ( 20 6 ) represent the ways to choose 6 remaining bulbs out of the 20 working (assuming the 4 faulty ones have been taken by the first group). Similarly ( 20 10 ) represents the ways to choose the remaining 10 bulbs out of the 20 working light-bulbs. The denominator counts the ways of splitting the 24 light-bulbs into two groups.
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hw1soln - Homework # 1 Solutions ApSc 116 Problem Page 18 #...

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