math414 hw3

# math414 hw3 - P A ∩ B = P A P B-P A ∪ B = 0 4 0 5 7 = 0...

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SOLUTIONS FOR HWK # 3 PROBLEM 1.2-3 (a) There are 12 favorable outcomes with respect to A : 4 jacks, 4 queens and 4 kings. Therefore P ( A ) = 12 52 = 3 13 . (c) A B = { 2 red “9”, 2 red “10”, 4 jacks, 4 queens, 4 kings } , M ( A B ) = 16 , P ( A B ) = 16 52 . PROBLEM 1.2-4 (a) S = { HHHH,HHHT,HHTH,HTHH,THHH,HHTT,HTTH,TTHH,HTHT,THTH, THHT,HTTT,THTTTTHT,TTTH,TTTT } (b) (i)5/16,(ii)0 (iii)11/16, (iv)4/16, (v)4/16, (vi)9/16, (vii)4/16 PROBLEM 1.2-5 A 0 A 1 A 2 A 3 = S and A 0 ,A 1 ,A 2 ,A 3 are mutually exclusive. Therefore P ( A 0 ) + P ( A 1 ) + P ( A 2 ) + P ( A 3 ) = 1, 1 64 + 9 64 + 27 64 + P ( A 3 ) = 1. Thus, P ( A 3 ) = 1 - 37 64 = 27 64 . PROBLEM 1.2-6 (a) 1/6; (b) P ( B ) = 1 - P ( B 0 ) = 1 - P ( A ) = 5 / 6; (c) P ( A B ) = P ( S ) = 1. PROBLEM 1.2-8 (a) P ( A B ) = 0 . 4 + 0 . 5 - 0 . 3 = 0 . 6; (b) A = ( A B 0 ) ( A B ); P ( A ) = P ( A B 0 ) + P ( A B ); 0 . 4 = P ( A B 0 ) + 0 . 3; P ( A B 0 ) = 0 . 4 - 0 . 3 = 0 . 1 (c) P ( A 0 B 0 ) = P [( A B ) 0 ] = 1 - P ( A B ) = 1 - 0 . 3 = 0 . 7 . 1

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PROBLEM 1.2-13 (a) P ( A ) = 10 i =1 1 2 i = 1 2 - ( 1 2 ) 11 1 2 = 1 - ( 1 2 ) 10 . (b) P ( B ) = 20 i =1 1 2 i = 1 2 - ( 1 2 ) 21 1 2 = 1 - ( 1 2 ) 20 . (c) A B . Therefore A B = B , P ( A B ) = P ( B ) = 1 - ( 1 2 ) 20 . (d) A B = A . P ( A B ) = 1 - ( 1 2 ) 10 . (e) C = B \ A,A B . Thus P ( C ) = P ( B ) - P ( A ) = 1 - ( 1 2 ) 20 - (1 - ( 1 2 ) 10 ) = ( 1 2 ) 10 - ( 1 2 ) 20 . Problem A (a) We know that P ( A B ) = P ( A ) + P ( B ) - P ( A B ). Hence
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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 ) ) = 1-P ( 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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math414 hw3 - P A ∩ B = P A P B-P A ∪ B = 0 4 0 5 7 = 0...

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