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# Chapter 6 - Chapter 6 6.1 a Relative frequency approach b...

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Chapter 6 6.1 a Relative frequency approach b If the conditions today repeat themselves an infinite number of days rain will fall on 10% of the next days. 6.2 a Subjective approach b If all the teams in major league baseball have exactly the same players the New York Yankees will win 25% of all World Series. 6.3 a {a is correct, b is correct, c is correct, d is correct, e is correct} b P(a is correct) = P(b is correct) = P(c is correct) = P(d is correct) = P(e is correct) = .2 c Classical approach d In the long run all answers are equally likely to be correct. 6.4 a Subjective approach b The Dow Jones Industrial Index will increase on 60% of the days if economic conditions remain unchanged. 6.5 a P(even number) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2 b P(number less than or equal to 4) = P(1) + P(2) + P(3) + P(4) = 1/6 + 1/6 + 1/6 +1/6 = 4/6 = 2/3 c P(number greater than or equal to 5) = P(5) + P(6) = 1/6 + 1/6 = 2/6 = 1/3 6.6 {Adams wins. Brown wins, Collins wins, Dalton wins} 6.7a P(Adams loses) = P(Brown wins) + P(Collins wins) + P(Dalton wins) = .09 + .27 + .22 = .58 b P(either Brown or Dalton wins) = P(Brown wins) + P(Dalton wins) = .09 + .22 = .31 c P(either Adams, Brown, or Collins wins) = P(Adams wins) + P(Brown wins) + P(Collins wins) = .42 + .09 + .27 = .78 6.8 a {0, 1, 2, 3, 4, 5} b {4, 5} c P(5) = .10 d P(2, 3, or 4) = P(2) + P(3) + P(4) = .26 + .21 + .18 = .65 e P(6) = 0 6.9 {Contractor 1 wins, Contractor 2 wins, Contractor 3 wins} 6.10 P(Contractor 1 wins) = 2/6, P(Contractor 2 wins) = 3/6, P(Contractor 3 wins) = 1/6 143

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6.11 a {Shopper pays cash, shopper pays by credit card, shopper pays by debit card} b P(Shopper pays cash) = .30, P(Shopper pays by credit card) = .60, P(Shopper pays by debit card) = .10 c Relative frequency approach 6.12 a P(shopper does not use credit card) = P(shopper pays cash) + P(shopper pays by debit card) = .30 + .10 = .40 b P(shopper pays cash or uses a credit card) = P(shopper pays cash) + P(shopper pays by credit card) = .30 + .60 = .90 6.13 {single, divorced, widowed} 6.14 a P(single) = .15, P(married) = .50, P(divorced) = .25, P(widowed) = .10 b Relative frequency approach 6.15 a P(single) = .15 b P(adult is not divorced) = P(single) + P(married) + P(widowed) = .15+ .50 + .10 = .75 c P(adult is either widowed or divorced) = P(divorced) + P(widowed) = .25 + .10 = .35 6.16 P( 1 A ) = .1 + .2 = .3, P( 2 A ) = .3 + .1 = .4, P( 3 A ) = .2 + .1 = .3. P( 1 B ) = .1 + .3 + .2 = .6, P( 2 B ) = .2 + .1 + .1= .4. 6.17 P( 1 A ) = .4 + .2 = .6, P( 2 A ) = .3 + .1 = .4. P( 1 B ) = .4 + .3 = .7, P( 2 B ) = .2 + .1 = .3. 6.18 a 57 . 7 . 4 . ) B ( P ) B and A ( P ) B | A ( P 1 1 1 1 1 = = = b 43 . 7 . 3 . ) B ( P ) B and A ( P ) B | A ( P 1 1 2 1 2 = = = c Yes. It is not a coincidence. Given 1 B the events 1 A and 2 A constitute the entire sample space. 6.19 a 67 . 3 . 2 . ) B ( P ) B and A ( P ) B | A ( P 2 2 1 2 1 = = = b P( 2 B | 1 A ) 33 . 6 . 2 . ) A ( P ) B and A ( P 1 2 1 = = = c One of the conditional probabilities would be greater than 1, which is not possible. 6.20 The events are not independent because ) A ( P ) B | A ( P 1 2 1 . 144
6.21 a P( 1 A or 1 B ) = ) B and A ( P ) B ( P ) A ( P 1 1 1 1 - + = .6 + .7 - .4 = .9 b P( 1 A or 2 B ) = ) B and A ( P ) B ( P ) A ( P 2 1 2 1 - + = .6 + .3 - .2 = .7 c P( 1 A or 2 A ) = ) A ( P ) A ( P 2 1 + = .6 + .4 = 1 6.22 25 . 60 . 20 . 20 . ) B ( P ) B and A ( P ) B | A ( P 1 1 1 1 1 = + = = ; 25 . 05 . 20 . ) A ( P 1 = + = ; the events are independent. 6.23 571 . 15 . 20 . 20 .

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Chapter 6 - Chapter 6 6.1 a Relative frequency approach b...

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