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p chapter 1 solutions 3rd edition

# p chapter 1 solutions 3rd edition - Probability Third...

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Sheet1 Page 1 Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solution 1.1 (i) AB{3, 5,7} n= (ii) {2, 4,6,8,10} B'= . {2, 3, 4, 5,6,7,8,10} AB '. Err:520 (iii) {1,2,3,5,7, 9} AB.= . ( ) {4, 6,8,10} AB'. Err:520 Solution 1.2 1,9 3,5,7 A B 4,6,8,10 S 2 Solution 1.3 We have: (AB)=Pr A+Pr () -Pr AB)=0.4 +0.3 -0.15 =0.55 Pr . () B (n Solution 1.4 We have: Pr (A.B)=1 -Pr ( (AB )'). .1.5Pr ()A=1 -Pr()A .Pr ()A=0.4 . Solution 1.5 By considering a Venn diagram, or otherwise, we have: Pr A +Pr B nA'=Pr A .B =0.8 and since (from the question) Pr ()=Pr BA') ()( Solutions to practice questions h Chapter 1 Probability Solutions to practice questions h Chapter 1

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Sheet1 Page 2 )() A (n , we have: Pr ()=0.4 and Pr (B nA')=0.4 A The required probability is: =- B 1 (. - () A n. Pr (B ) 1 Pr ()=-.Pr AB)Pr A +Pr ( B) .. Err:509 1 =0.5 Solution 1.6 By the law of total probability (Theorem 1.5), we have: B ) 2 ) A Pr ()=Pr (B nA1 +Pr (B nA +Pr (B n 3 ) Pr (BA )=Pr B -Pr (B nA -Pr ( .n() ) B nA ) 1 23 From the question: Pr B ()=-1 Pr (B')=1 -0.3 =0.7 and: Pr (A nB)=2 Pr (A nB)=4 Pr (A nB) 321
Sheet1 Page 3 So, we have: Pr (BA1 )=Pr ()-Pr (B nA2 )-Pr (B nA3 ) n B Err:509 1 ) - 1 - (BA .Pr (BAn 1 )=0.1 Solution 1.7 Using the information in the question: (.)=0.7 Pr AB A y 0.7 z 0.3 .++= wx .= and Pr AB (.')=0.9 .++ z =0.9 y 0.1 wx .= z So: Pr ()=wx 1 yz 1 0.1 -0.3 =0.6 A +=--=- yx B Leths solve this one graphically, using a Venn diagram.

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Sheet1 Page 4 S w w Solution 1.8 Let A ={Claims on homeowners policy} and B ={Claims on automobile policy} . From the question we have: Pr AB ( ()=0.46, Pr () Err:501 Hence: (AB.) 1 Pr( .) )= - 0.52 Pr =- (AB ' 1 =0.48 The probability that a claim is made on both policies is: Pr AB =Pr A +Pr B -Pr AB =0.46 +0.32 -0.48 =0.30
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