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Unformatted text preview: HOMEWORK PROBLEMS #3 SOLUTIONS TO EVEN PROBLEMS 114. Let P i = the detector shows a positive reading (lie) for person i , T i = person i is telling the truth. Then we have: P ( P i  T i ) = . 10 ,P ( P i  T i ) = . 95. (a) By independence we have that P (positive reading for both suspects) = P ( P 1 ∩ P 2  T 1 ∩ T 2 ) = P ( P 1  T 1 ) P ( P 2  T 2 ) = . 10 × . 95 = . 095 (b) P ( P 1 ∩ P 2  T 1 ∩ T 2 ) = P ( P 1  T 1 ) P ( P 2  T 2 ) = (1 . 10) × . 95 = . 855 (c) P ( P 1 ∩ P 2  T 1 ∩ T 2 ) = P ( P 1  T 1 ) P ( P 2  T 2 ) = . 10 × (1 . 95) = . 005 (d) P ( P 1 ∪ P 2  T 1 ∩ T 2 ) = P ( P 1  T 1 ) + P ( P 2  T 2 ) P ( P 1 ∩ P 2  T 1 ∩ T 2 ) = . 10 + . 95 . 095 = . 955 116. Let L i = line i fails. Then P (does not completely fail) = 1 P (fails) = 1 P ( L 1 ∩ L 2 ∩ L 3 ) = 1 P ( L 1 ) P ( L 2 ) P ( L 3 ) = 1 . 01 3 = . 999999 120. (a) Number of ways to test all 6 refrigerators: P (6 , 6) = 6!. Number of ways to test all 6 refrigerators with a defective one in the first three and a defective one in the fourth position:...
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This note was uploaded on 02/23/2011 for the course MATH 444 taught by Professor Any during the Fall '10 term at Roosevelt.
 Fall '10
 Any
 Statistics, Probability

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