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Unformatted text preview: P ( A ∩ B ) = P ( A ) + P ( B )P ( A ∪ B ) = 0 . 4 + 0 . 5. 7 = 0 . 2 . (b) By the De Morgan theorem, P ( A ∪ B ) = P (( A ∩ B ) ) = 1P ( A ∩ B ) = 1. 2 = 0 . 8. Problem A (1) B =”no Aces in the hand”. P ( B ) = ( 48 5 ) ( 52 5 ) = 0 . 658842; P ( B ) = 1. 658842 = 0 . 341158. (2) B = B 1 ∪ B 2 ∪ B 3 ∪ B 4 where B k means that there are k aces in the hand. Of course, these events are mutually exclusive hence P ( B ) = ( 4 1 )( 48 4 ) ( 52 5 ) + ( 4 2 )( 48 3 ) ( 52 5 ) + ( 4 3 )( 48 2 ) ( 52 5 ) + ( 4 4 )( 48 1 ) ( 52 5 ) = 0 . 2994736356080894 + 0 . 03992981808107859 + 0 . 0017360790470034167 + . 00001846892603195124 = 0 . 341158. 2...
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This note was uploaded on 10/13/2010 for the course MATH 414 at Penn State.
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