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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}
127
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View Full Document 6.10 P(Contractor 1 wins) = 2/6, P(Contractor 2 wins) = 3/6, P(Contractor 3 wins) = 1/6
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.
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This homework help was uploaded on 04/18/2008 for the course STAT 212 taught by Professor Holt during the Spring '08 term at UVA.
 Spring '08
 HOLT

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