Homework2 - (d Note that since the intersection is...

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HOMEWORK PROBLEMS #2 SOLUTIONS TO EVEN PROBLEMS 44. (a) P (bad taxi is dispatched to airport C ) = 8 3 5 ! 9 3 5 1 ! = 8! 3!5! 9! 3!5!1! = 1 9 . 11 (b) P (every airport receives a bad taxi) = P (3 , 3) × ± 6 2 4 ² 9 3 5 1 ! = 3! × 6! 2!4! 9! 3!5!1! = 5 28 . 18 58. (a) P (3 aces and 2 kings) = 4 3 ! 4 2 ! 52 5 ! = 1 108290 9 . 2 × 10 - 6 (b) P (full house) = P (13 , 2) × 4 3 ! 4 2 ! 52 5 ! = 6 4165 . 0014 74. Note that: P ( A ) = . 61 ,P ( B ) = . 30 ,P ( C ) = . 09 ,P ( D ) = . 30. (a) Not independent, since P ( A D ) = . 20 and P ( A ) × P ( D ) = . 183 (b) Independent, since P ( B ) × P ( D ) = P ( B D ) = . 09 (c) Not independent, since P ( C D ) = . 01 and P ( C ) × P ( D ) = . 027 76. Let D = the consumer was dissatisfied, A = the job was done by plumber A , then we know that P ( D ) = . 10, P ( A | D ) = . 50, P ( A ) = . 40. (a) P ( D | A ) = P ( D A ) P ( A ) = P ( A | D ) P ( D ) P ( A ) = . 50 × . 10 . 40 = . 125 (b) P ( D | A ) = 1 - P ( D | A ) = 1 - . 125 = . 875 86. (a) No, since then P ( A B ) = P ( A ) + P ( B ) - P ( A B ) = . 8 + . 7 - . 1 = 1 . 4 > 1, which is not possible. (b) Note that, P ( A B ) 1 P ( A ) + P ( B ) - P ( A B ) 1 . 8 + . 7 - P ( A B ) 1 1 . 5 - 1 P ( A B ) So P ( A B ) has to be at least .5. (c) No, since then P ( A B ) = P ( B ) - P ( A B ) = . 7 - . 77 = - . 07 < 0, which is not possible.
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Unformatted text preview: (d) Note that, since the intersection is contained in both A and B , we must have that P ( A ∩ B ) ≤ P ( A ) = . 8 and P ( A ∩ B ) ≤ P ( B ) = . 7. Therefore, P ( A ∩ B ) can be at most .7. 94. (a) P ( A ∪ B ) = P ( A ) + P ( B )-P ( A ∩ B ) = . 95 + . 90-. 88 = . 97 (b) P ( A ∪ B ) = 1-P ( A ∪ B ) = 1-. 97 = . 03 96. (a) P ( A ∪ B ) = P ( A ) + P ( B )-P ( A ∩ B ) = P ( A ) + P ( B )-P ( A ) P ( B ) = . 5 + . 2-. 5 × . 2 = . 6 (b) P ( A ∩ B ) = P ( A ) P ( B ) = (1-P ( A ))(1-P ( B )) = (1-. 5)(1-. 2) = . 4 (c) P ( A ∪ B ) = P ( A ) + P ( B )-P ( A ∩ B ) = (1-. 5) + (1-. 2)-. 4 = . 9 1...
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