Chapter 4 Solutions:
Section 4.1:
1.)
2.)
3.)
481
» 0.184 = 18.4%
2620
371
P(brown) =
» 0.142 = 14.2%
2620
483
P(green) =
» 0.184 = 18.4%
2620
544
P(orange) =
» 0.208 = 20.8%
2620
372
P(red) =
» 0.142 = 14.2%
2620
369
P(yellow) =
» 0.141 = 14.1%
2620
Probability of choosing a particular car: (total number of people = 300)
P(blue) =
P(reason)
Safety
84
84
=
300
0.28
= 28%
Reliability
Cost
62
46
62
46
»
»
300
300
0.21
0.15
= 21% = 15%
Performance
34
34
»
300
0.11
= 11%
Comfort
47
47
»
300
0.16
= 16%
Section 4.2:
483 + 372 855
=
» 0.326 = 32.6%
2620
2620
481+ 372 + 369 1222
b.) P(blue, red, or yellow) =
=
» 0.466 = 46.6%
2620
2620
371 2249
c.) P(not brown) = 1- P ( brown ) = 1=
» 0.858 = 85.8%
2620 2620
483 2137
d.) P(not green) = 1=
» 0.816 = 81.6%
2620 2620
1.)
a.) P(green or red) =
2.)
3.)
a.) SS = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
3
b.) P(2 heads) = P({HHT, HTH, THH}) = = 0.375 = 37.5%
8
4
c.) P(at least 2 heads) = P({HHH, HHT, HTH, THH}) = = 0.50 = 50%
8
Looks
27
27
=
300
0.09
= 9%
d.) P(odd number of heads) = P({HHT, HTH, THH, TTT}) =
4
= 0.50 = 50%
8
4.)
2
= 0.25 = 25%
8
f.) P(two heads or two tails) = P({HHT, HTH, THH, TTH, THT, HTT}) =
6
= 0.75 = 75%
8
g.) P(not an odd number of heads) =
4 4
1- P ( odd number of heads ) = 1- = = 0.50 = 50%
8 8
-
5.)
a.)
e.) P(all heads or all tails) = P({HHH, TTT}) =
SS = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6),
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
2
b.) P(sum of 3) = P({(1,2), (2,1)}) =
» 0.056 = 5.6%
36
c.) P(1st die a 4) = P({(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)}) =
6
» 0.167 = 16.7%
36
5
» 0.139 = 13.9%
36
e.) P(sum of 3 or sum of 8) = P({(1,2), (2,1), (6,2), (5,3), (4,4), (3,5), (2,6)})
7
=
» 0.194 = 19.4%
36
or
P ( sum of 3 or sum of 8 ) = P ( sum of 3) + P ( sum or 8 )
d.) P(sum of 8) = P({(6,2), (5,3), (4,4), (3,5), (2,6)}) =
2
5
7
+
=
» 0.194 = 19.4%
36 36 36
f.) P(sum of 3 or 1st die a 4) = P({(1,2), (2,1), (4,1), (4,2), (4,3), (4,4), (4,5),
8
(4,6)}) =
» 0.222 = 22.2%
36
or
P ( sum of 3 or 1st die a 4 ) = P ( sum of 3) + P (1st die a 4 )
=
2
6
8
+
=
» 0.222 = 22.2%
36 36 36
g.) P(sum of 8 or 1st die a 4) = P({(6,2), (5,3), (4,4), (3,5), (2,6), (4,1), (4,2), (4,3),
10
(4,5), (4,6)}) =
» 0.278 = 27.8%
36
or
=
P ( sum of 8 or 1st die a 4 )
= P ( sum of 8 ) + P (1st die a 4 ) - P ( sum of 8 or 1st die a 4 )
=
5
6
1 10
+
=
» 0.278 = 27.8%
36 36 36 36
h.) P(not getting a sum of 8) = 1- P ( sum of 8 ) = 16.)
-
7.)
a.) P(red ball) =
8.)
5 31
=
» 0.861 = 86.1%
36 36
5
= 0.625 = 62.5%
8
3
b.) P(blue ball) = = 0.375 = 37.5%
8
58 5 8 5
c) odds of red ball =
= * = = 5 to 3
38 8 3 3
38 3 8 3
d.) odds of blue ball =
= * = = 3 to 5
58 8 5 5
-
Section 4.3:
1.)
2.)
3.)
4.)
5.)
1.)
2.)
3.)
4.)
5.)
Owning a refrigerator and owning a car are independent events, since knowing
that a person owns one of these items does not change the probability of the
person owning the other item. (One could argue that world-wide that these are
dependent events. This is because in some parts of the world if a person can
afford one, then there is a different probability that they could afford to own the
other too.)
Passing your statistics class and passing your biology class are dependent events,
since if you put more work into one class to pass it then you have less time for the
other. So the probability of passing could in fact change the probability of the
other. (One could argue that these events are independent, since passing one class
does not change the probability that you will pass the other in content
knowledge.)
SS = {2S, 3S, 4S, 5S, 6S, 7S, 8S, 9S, 10S, JS, QS, KS, AS, 2C, 3C, 4C, 5C, 6C,
7C, 8C, 9C, 10C, JC, QC, KC, AC, 2D, 3D, 4D, 5D, 6D, 7D, 8D, 9D, 10D, JD,
QD, KD, AD, 2H, 3H, 4H, 5H, 6H, 7H, 8H, 9H, 10H, JH, QH, KH, AH}
a.) - Number of meals = 2*3*5 = 30 meals
Number of phone numbers = 8*10*10*10*10*10*10 = 8,000,000 phone numbers
Number of CA license plates = 10*26*26*26*10*10*10 = 175,760,000 plates
10
P6 = 151,200
6.)
7.)
8.)
9.)
10.)
11.)
20
C4 = 4845
Permutation since order matters. 25 P9 = 7.41´1011
Combination since order doesn’t matter. 20 C7 = 77,520