Chapter 10
c. There are 10 possible results.
Chapter 10 Opener
Out of 20 spins, you think you will spin a 4 two times.
Check students’ work with tallies in a table and
closeness of their guess.
Try It Yourself (p. 399)
1.
Baseballs
2
1
=
=
Footballs
6
3
So, the ratio of baseballs to footballs is
2.
2. a. an even number; Spinning an even number has 5
possible results: 2, 4, 6, 8, or 10. Spinning a multiple
of 4 has only 2 possible results: 4 or 8.
1
.
3
b. neither; Spinning an even number has 5 possible
results: 2, 4, 6, 8, or 10. Spinning an odd number has 5
possible results: 1, 3, 5, 7, or 9.
Footballs
6
3
=
=
Total pieces of equipment
8
4
So, the ratio of footballs to total pieces of equipment
3
is .
4
3. a. Answer should include, but is not limited to:
Students will work in pairs to play Rock Paper
Scissors 30 times and record the results in a table.
b. There are 9 possible outcomes because there are
Sneakers
2
1
3.
=
=
Ballet slippers
4
2
9 results possible in the table.
Player A
c.
Player B
1
So, the ratio of sneakers to ballet slippers is .
2
Sneakers
2
1
4.
=
=
Total number of shoes
6
3
So, the ratio of sneakers to total number of shoes is
5.
1
.
3
green beads
3
1
=
=
blue beads
6
2
1
The ratio of green beads to blue beads is .
2
6.
red beads
4
=
green beads
3
4
The ratio of red beads to green beads is .
3
7.
Section 10.1
Out of 20 spins, you think you will spin orange
4 times.
Check students’ work with tallies in a table and
closeness of their guess.
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Rock
Tie
A wins
B wins
Paper
B wins
Tie
A wins
Scissors
A wins
B wins
Tie
because there are three ways for each player to win.
4. Make a list or table that shows all possible results.
10.1 On Your Own (p. 403)
1. a. The possible outcomes are choosing A, B, C, D, E, F,
G, H, I, J, and K.
b.
consonants
vowels
B, C, D, F, G, H, J, K
A, E, I
The favorable outcomes of choosing a vowel are
choosing A, E, and I.
2. a. There are 8 marbles. So, there are 8 possible
outcomes.
b.
1. a. There are 2 possible results.
b. There are 5 possible results.
Scissors
d. no; Each player has an equal chance of winning
10.1 Activity (pp. 400 –401)
Out of 20 flips, you think you will flip heads 10 times.
Check students’ work with tallies in a table and
closeness of their guess.
Paper
There are three ways Player A can win, three ways
Player B can win, and three ways the players can tie.
green beads
3
1
=
=
total beads
15
5
The ratio of green beads to the total number of beads
1
is .
5
Rock
blue
not blue
blue, blue
green, red, purple,
yellow, yellow, yellow
There are 2 blue marbles. So, choosing blue can occur
in 2 ways.
c.
yellow
not yellow
yellow, yellow, yellow
blue, blue, green,
red, purple
There are 5 marbles that are not yellow. So, choosing
not yellow can occur in 5 ways. The favorable
outcomes of choosing not yellow are blue, blue, green,
red, and purple.
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Chapter 10
11.
10.1 Exercises (pp. 404 –405)
divisible by 3
not divisible by 3
3, 6, 9
1, 2, 4, 5, 7, 8
Vocabulary and Concept Check
1. Rolling an even number on a number cube is an event
The favorable outcomes of choosing a number divisible
by 3 are choosing 3, 6, and 9.
because it is a collection of several outcomes.
2. Sample answer: An outcome is any of the possible results
of an experiment, while a favorable outcome is an
outcome of a specific event of the experiment.
12. a. There are 2 blue marbles. So, choosing blue can occur
in 2 ways.
b. The favorable outcomes of the event are blue and blue.
Practice and Problem Solving
3. Because the spinner has 8 different numbers, there are
13. a. There is 1 green marble. So, choosing green can occur
in 1 way.
8 possible outcomes.
b. The favorable outcome of the event is green.
4.
even
odd
2, 4, 6, 8
1, 3, 5, 7
14. a. There are 2 purple marbles. So, choosing purple can
occur in 2 ways.
b. The favorable outcomes of the event are purple and
There are 4 ways to spin an even number and 4 ways
to spin an odd number.
purple.
15. a. There is 1 yellow marble. So, choosing yellow can
occur in 1 way.
5. The possible outcomes are choosing 1, 2, 3, 4, 5, 6, 7, 8,
and 9.
6.
b. The favorable outcome of the event is yellow.
16. a.
6
not 6
6
1, 2, 3, 4, 5, 7, 8, 9
The favorable outcome is choosing a 6.
7.
odd
not odd
1, 3, 5, 7, 9
2, 4, 6, 8
not red
red, red, red
green, blue, blue,
purple, purple, yellow
There are 6 marbles that are not red. So, choosing not
red can occur in 6 ways.
b. The favorable outcomes of the event are green, blue,
blue, purple, purple, and yellow.
The favorable outcomes of choosing an odd number are
choosing 1, 3, 5, 7, and 9.
8.
red
greater than 5
not greater than 5
6, 7, 8, 9
1, 2, 3, 4, 5
17. a.
blue
not blue
blue, blue
green, red, red, red,
purple, purple, yellow
There are 7 marbles that are not blue. So, choosing not
blue can occur in 7 ways.
The favorable outcomes of choosing a number greater
than 5 are choosing 6, 7, 8, and 9.
b. The favorable outcomes of the event are green, red,
red, red, purple, purple, and yellow.
9.
odd number less than 5
not an odd
number less than 5
1, 3
2, 4, 5, 6, 7, 8, 9
The favorable outcomes of choosing an odd number less
than 5 are choosing 1 and 3.
10.
less than 3
not less than 3
1, 2
3, 4, 5, 6, 7, 8, 9
The favorable outcomes of choosing a number less than 3
are choosing 1 and 2.
18. There are 7 marbles that are not purple, even though there
are only 4 colors. Choosing not purple could be red, red,
red, blue, blue, green, or yellow.
19.
Presidential
not Presidential
Presidential,
Presidential,
Presidential
Susan B. Anthony, Susan B. Anthony,
Susan B Anthony, Susan B. Anthony,
Susan B. Anthony, Kennedy, Kennedy
There are 7 coins that are not Presidential Dollars. So,
choosing not a Presidential Dollar can occur in 7 ways.
20. false; Spinning blue and spinning red have the same
number of favorable outcomes on Spinner A.
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Chapter 10
21. true; There are three blue sections and two green sections.
Section 10.2
22. false; There are five possible outcomes when spinning
10.2 Activity (pp. 406 –407)
Spinner A.
1. a. Answer should include, but is not limited to: Students
will write rules for a game that uses the given spinner.
The rules should be written so players can clearly
understand the object of the game.
23. true; There are four red sections.
24. false; Spinning not green can occur in eight ways on
Spinner B.
b. Answer should include, but is not limited to: After
playing the game and analyzing the outcome, students
will determine if they should revise the rules that they
wrote for the game in part (a).
25. Because the events “choosing a rock CD” and “not
choosing a rock CD” have the same number of favorable
outcomes, half of the CDs must be rock CDs. So, there
are 30 rock CDs.
c. yes; Each section measures 60°. When each section is
the same size, it is equally likely to spin a given
number. When you increase or decrease the angle of a
section, you increase or decrease the likelihood of
landing on that section because its area is changing.
26. There are 5 cards, so there are 5 possible outcomes. After
choosing a card, there are 4 cards left. So, the number of
possible outcomes changes to 4.
d. Sample answer: Each section of the spinner is the
same size. So, your friend has an equal chance of
landing on any of the numbers.
Fair Game Review
27.
28.
29.
30.
x
1
=
10
5
x
1
• 10 = • 10
10
5
x = 2
2. a.
Angle
1
2
3
4
5
6
Measure
60°
60°
90°
45°
45°
60°
Section 3 measures 90° and makes up one-quarter of
the spinner. Your chance of landing on 3 is greater
than the other numbered sections.
60
20
=
n
7
60 • 7 = 20 • n
420 = 20n
b.
21 = n
Angle
1
2
3
4
5
6
Measure
60°
120°
90°
45°
30°
15°
1
w
=
3
36
1
w
• 36 =
• 36
3
36
12 = w
25
100
=
b
17
25 • b = 100 • 17
25b = 1700
b = 68
31. C; S = 2lw + 2lh + 2 wh
= 2(12)(6) + 2(12)(5) + 2(6)(5)
= 144 + 120 + 60
= 324
The surface area of the rectangular prism is 324 in.2
Section 2 measures 120° and makes up one-third of
the spinner. Your chance of landing on 2 is greater
than the other numbered sections.
Answer should include, but is not limited to: Students
should explain that the rules still make sense, but the
game is not fair because of the unequal sections.
3. The rules of the game are fair for Activities 1 and 2b. The
sum of the angle measures for the odd numbered pie
pieces is equal to the sum of the angle measures for the
even numbered pie pieces, so there is an equal likelihood
of the events happening. The rules of the game are not
fair for Activity 2a. The sum of the angle measures for
the odd numbered pie pieces is 195°, and the sum of the
angle measures for the even numbered pie pieces is 165°.
So, Player 1 has a better chance of winning.
4. Sample answer: You can describe an event as impossible,
unlikely to occur, likely to occur, certain, or equally
likely to happen or not happen.
5. It is impossible to spin an 8 in Activity 1 because the
numbers on the spinner are only 1 through 6.
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Chapter 10
6. Sample answers: medical fields, weather forecasters,
insurance policy writers, various sports managers.
10.2 On Your Own (pp. 408 –409)
1. Because the probability of landing a jump on a
snowboard is
1
, it is equally likely to happen or not
2
happen.
2. Because the probability is 100%, it is certain that the
temperature will be less than 120°F tomorrow.
4
6
2
=
3
3. P(greater than 2) =
The probability of rolling a number greater than 2 is
2
or 66.7%.
3
7. The probability that you will grow 12 more feet is 0, so
growing 12 more feet is impossible.
8. The probability of the sun rising tomorrow is 1, so it is
certain that the sun will rise tomorrow.
1
of the days in July, the probability
5
1
1
of it raining on a day in July is . Because is close to
5
5
1
, it is unlikely that it will rain on a day in July.
4
9. Because it rains on
10. Your probability of playing the correct note on a violin is
50%, so it is equally likely to play the correct note or not
to play the correct note.
number of red shirts
total number of shirts
1
=
10
11. P( red ) =
4. The number cube has only the numbers 1 through 6. So,
rolling a number 7 is impossible, and its probability is
0 or 0%.
number of short straws
total number of straws
1
n
=
15
75
75 = 15n
5. P(short ) =
The probability of choosing a red shirt is
number of green shirts
total number of shirts
2
=
10
1
=
5
12. P(green ) =
5 = n
There are 5 short straws.
The probability of choosing a green shirt is
10.2 Exercises (pp. 410 –411)
Vocabulary and Concept Check
number of shirts that are not white
total number of shirts
9
=
10
of the number of favorable outcomes to the number of
possible outcomes.
than 1, because there can never be more favorable
outcomes than total possible outcomes.
3. Sample answer: An impossible event is a lake in Florida
freezing during the summer. A certain event is the sun
setting tonight.
The probability of not choosing a white shirt is
number of shirts that are not black
total number of shirts
8
=
10
4
=
5
14. P( not black ) =
4. You should spin Spinner B when you want to move
5. either; Both spinners have the same number of sections
9
,
10
or 90%.
Practice and Problem Solving
down, because Spinner B has more sections labeled
“Down.”
1
, or 20%.
5
13. P( not white) =
1. To find the probability of an event, you find the ratio
2. no; The probability of an event can never be greater
1
, or 10%.
10
The probability of not choosing a black shirt is
4
,
5
or 80%.
labeled “Forward.”
3
of the time, so it is likely that
4
your soccer team will win.
6. Your soccer team wins
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Chapter 10
c. There are six numbers greater than 6 (7, 8, 9, 10, 11,
number of orange shirts
total number of shirts
0
=
10
= 0
15. P(orange) =
and 12).
P(greater than 6) =
The probability of rolling a number greater than 6
1
is . So, it is equally likely that you will roll a
2
number greater than 6 or not greater than 6.
The probability of choosing an orange shirt is 0 or 0%.
16. The numerator should be the number of shirts that are not
blue, instead of the number of shirts that are blue.
20.
6
3
=
10
5
Father’s Genes
P( not blue) =
number of prize winners
17. P( winning a prize) =
total number of people
n
0.05 =
400
20 = n
Mother’s Genes
X
X
X
XX
XX
Y
XY
XY
21. The probability of two parents having a boy or having a
There will be 20 people who win a prize in the contest.
number of winning ducks
total number of ducks
n
0.24 =
25
6 = n
18. a. P( winning ) =
girl is equally likely because there are two ways the genes
can combine to create a girl and two ways the genes can
combine to create a boy.
22. a.
Parent 1
C
s
Parent 2
The number of winning ducks is 6, so there are
25 − 6 = 19 not winning ducks.
C
CC
Cs
s
Cs
ss
P(CC ) =
number of not winning ducks
b. P( not winning) =
total number of ducks
19
=
25
= 0.76
outcome of CC
1
=
total number of outcomes
4
The probability of a child having the gene
1
combination is , or 25%.
4
Because 0.76 is close to 0.75, it is likely that you will
not choose a winning duck.
19. a. There are eight numbers less than 9 (1, 2, 3, 4, 5, 6, 7,
and 8).
P(less than 9) =
6
1
=
12
2
b. P(curly) =
number of outcomes with a C
3
=
total numbers of outcomes
4
The probability of a child having curly hair is
3
, or
4
75%.
8
2
=
12
3
The probability of rolling a number less than 9 is
Fair Game Review
2
.
3
2
3
is close to , it is likely that you will roll
3
4
a number less than 9.
Because
23. x + 5 < 9
x < 4
1
3
4
5
6
7
−5
−4
−3
−2
24. b − 2 ≥ −7
b. There are four multiples of 3 (3, 6, 9, and 12).
4
1
P( multiple of 3) =
=
12
3
2
b ≥ −5
−8
−7
−6
1
The probability of rolling a multiple of 3 is .
3
1
1
Because is close to , it is unlikely that you will
3
4
roll a multiple of 3.
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Chapter 10
25.
6. Sample answer: Most likely this is true because there are
w
3
− 3 < w, or
1 > −
only 20 chips in the bag and you did not select an orange
chip in 50 tries. However, you cannot say this for certain
because there may be 1 orange chip in the bag and you
never selected it.
w > −3
−6
26.
−5
−4
−3
−2
−1
7. Sample answer:
0
6 ≤ −2 g
−3 ≥ g , or
Green Orange
g ≤ −3
Yellow
Red
Blue
−6
−5
−4
−3
−2
−1
0
8. a. yes; Because the sizes of the sections are equal, each
27. C; 46 + 135 + 86 + x = 360
outcome is equally likely to occur.
267 + x = 360
b. no; Because the sizes of the sections are not equal,
x = 93
each outcome is not equally likely to occur.
c. no; Because the sizes of the sections are not equal,
Section 10.3
each outcome is not equally likely to occur.
10.3 Activity (pp. 412–413)
10.3 On Your Own (pp. 415 –416)
1. a. Check students’ work.
1. The bar graphs shows 4 twos, 11 fours, and 6 sixes. So,
b. Check students’ work.
an even number was rolled 4 + 11 + 6 = 21 times in a
total of 50 rolls.
c. Check students’ work. Relative frequencies should get
closer to the probabilities of
1
.
2
number of times the event occurs
total number of trials
21
P(even ) =
50
P(event ) =
d. Check students’ work. Relative frequencies should get
closer to the probabilities of
1
.
2
The experimental probability is
2. a. no; The relative frequencies can help you estimate the
number of each type of chip in the bag, but you cannot
be sure of the exact numbers.
b. Sample answer: 100 times; The greater the number of
times you perform the experiment, the more accurate
your approximations will be.
3. a. no; Just because there are two outcomes does not
mean that each outcome is equally likely.
b. no; Each outcome is not equally likely.
Check students’ work.
4. Sample answer: You can find the probability of an event
occurring based on a relative frequency. You randomly
draw a marble from a bag and replace it. You do this 50
times. You draw a red marble 25 times. So, the
1
probability of drawing a red marble is , or 50%.
2
1
, or 50%.
2
The relative frequency of rolling an odd number should
1
be close to . So, your friend should roll an odd number
2
about 250 times.
5. The probability of rolling an odd number is
2.
21
, 0.42, or 42%.
50
number of times the event occurs
total number of trials
5
P(defective jeans) =
200
P(event ) =
To make a prediction, multiply the probability of
defective jeans by the number of jeans shipped.
5
• 5000 = 125.
200
So, you can expect that there will be 125 defective pairs
of jeans when 5000 are shipped.
3. P( X) =
number of Xs
1
=
total number of letters
7
The probability of choosing an X is
1
, or about 14.3%.
7
number of odd sections
total number of sections
n
0.6 =
10
0.6(10) = n
4. P(odd ) =
6 = n
There are 6 sections that have odd numbers.
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5.
7. An even number was spun 6 + 11 + 7 = 24 times.
1
• 540 = 90
6
You would expect 90 bobbleheads to be won.
relative frequency =
24
50
12
=
25
6. The bar graphs shows 53 twos, 50 threes, 52 fours, 49
=
fives, and 48 sixes. So, a number greater than 1 was
rolled 53 + 50 + 52 + 49 + 48 = 252 times in a total
of 300 rolls.
number of times the event occurs
total number of trials
252
P(greater than 1) =
300
21
=
25
P(event ) =
The experimental probability is
21
, or 84%.
25
Because there are 5 possible outcomes for numbers
greater than 1 on a number cube (2, 3, 4, 5, and 6), the
theoretical probability of rolling a number greater than 1
5
1
is
= 83 %, which is close to the experimental
6
3
probability.
10.3 Exercises (pp. 417–419)
an experiment and find the ratio of the number of times
the event occurs to the total number of trials in the
experiment.
2. yes; You could flip tails 7 out of 10 times, but with more
trials the probability of flipping tails should get closer
to 0.5.
3. An event that has a theoretical probability of 0.5 means
there is a 50% chance you will get a favorable outcome.
4. Sample answer: The theoretical probability of spinning
red on a spinner with four equal sections colored red,
1
blue, green, and yellow is .
4
5. Because the pollster surveys only a subset of the
population, experimental probability should be used.
The pollster cannot find theoretical probability without
surveying every person who will participate in the
election.
Practice and Problem Solving
6. The total number of times the spinner was spun is
8 + 6 + 9 + 11 + 9 + 7 = 50.
number of times 6 spun
7
=
total number of spins
50
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8. A number less than 3 was spun 8 + 6 = 14 times.
P(less than 3) =
number of times less than 3 spun
total number of spins
14
50
7
=
25
=
The experimental probability of spinning a number less
7
than 3 is
, or 28%.
25
spun is 6 + 9 + 11 + 9 + 7 = 42.
1. To find the experimental probability of an event, perform
The relative frequency of spinning a 6 is
The relative frequency of spinning an even number is
12
, or 48%.
25
9. The total number of times a number that is not 1 was
Vocabulary and Concept Check
relative frequency =
number of times an even number spun
total number of spins
P ( not 1) =
number of times not 1 spun
42
21
=
=
total number of spins
50
25
The experimental probability of not spinning a 1 is
21
,
25
or 84%.
10. A 1 or a 3 was spun 8 + 9 = 17 times.
P (1 or 3) =
number of times a 1 or a 3 spun
17
=
total number of spins
50
The experimental probability of spinning a 1 or a 3 is
17
, or 34%.
50
11. P (7 ) =
number of times 7 spun
0
=
= 0
total number of spins
50
The experimental probability of spinning a 7 is 0, or 0%.
12. P(at least 1 cracked egg )
number of cartons with at least 1 cracked egg
total number of cartons checked
3
=
20
=
The experimental probability that a carton of eggs has at
3
least one cracked egg is
, or 15%.
20
7
, or 14%.
50
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13. P( vowels) =
number of vowels chosen
3
=
total number of letters chosen
7
3
• 105 = 45
7
1
• 20 = 5
4
You can expect 5 cards to have flowers on them.
number of strikes
total number of chips
3
9
=
10
n
3 • n = 9 • 10
3n = 90
There are 30 chips in the bag.
1
1
, or about 33 %.
3
3
number of sections with 1
1
=
total number of sections
6
The probability of spinning a 1 is
22. P(strike) =
n = 30
number of red sections
2
1
=
=
total number of sections
6
3
The probability of spinning red is
25
, or
26
about 96.2%.
number of cards chosen with flowers 1
14. P(flowers) =
=
total number of cards chosen
4
16. P(1) =
number of letters not Z
25
=
total number of letters
26
The probability of choosing any letter except Z is
You can expect 45 tiles to be vowels.
15. P( red ) =
21. P(letter except Z) =
23. P( pop song ) =
pop songs played
total songs played
n
80
0.45(80) = n
0.45 =
2
1
, or 16 %.
6
3
36 = n
There are 36 pop songs on your MP3 player.
17. P(odd ) =
number of odd sections
3
1
=
=
total number of section
6
2
The probability of spinning an odd number is
1
, or 50%.
2
18. P( multiple of 2)
number of sections with a multiple of 2
total number of sections
3
=
6
1
=
2
=
1
The probability of spinning a multiple of 2 is , or 50%.
2
number of sections less than 7
total number of sections
6
=
6
=1
19. P(less than 7) =
The probability of spinning a number less than 7 is 1, or
100%.
20. P(9) =
number of females
total number in class
16
=
16 + 20
16
=
36
4
=
9
24. a. P(female) =
The probability of randomly choosing a female is
4
,
9
or about 44.4%.
number of females
total number in class
4
f
=
9
45
4 • 45 = 9 • f
b. P(female) =
180 = 9 f
20 = f
Number of males = 45 − 20 = 25
So, 25 − 20 = 5 males joined the class.
number of sections with 9
0
=
= 0
total number of sections
6
The probability of spinning a 9 is 0, or 0%.
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25. P ( 4) =
number of “4” spins
37
=
total number of spins
200
29. a. P(TT) =
The experimental probability of spinning a 4 is
37
,
200
or 18.5%.
P( 4) =
number of “4” sections
1
=
total number of sections
5
The theoretical probability of spinning a 4 is
1
, or 20%.
5
number of “3” spins
39
=
total number of spins
200
39
,
200
number of “3” sections
1
=
total number of sections
5
number of TT’s
22
11
=
=
number of trials
100
50
The experimental probability of flipping two tails is
11
, or 22%.
50
expected number of TT’s
total number of trials
n
11
=
50
600
n
11
• 600 =
• 600
50
600
132 = n
P(TT ) =
1
, or 20%.
5
The experimental probability is very close to the
theoretical probability.
27. P(greater than 4) =
50 = n
b. P(TT) =
or 19.5%.
The theoretical probability of spinning a 3 is
expected number of TT’s
total number of trials
1
n
=
12
600
600 = 12n
You can expect to flip two tails 50 times in 600 trials.
The experimental probability of spinning a 3 is
P(3) =
The experimental probability of flipping two tails is
1
1
, or 8 %.
12
3
P(TT) =
The experimental probability is close to the theoretical
probability.
26. P (3) =
number of TT’s
1
=
number of trials
12
number of spins greater than 4
total number of spins
40
200
1
=
5
=
You can expect to flip two tails 132 times in 600 trials.
c. Sample answer: A larger number of trials should result
in a more accurate probability, which gives a more
accurate prediction.
The experimental probability of spinning a number
1
greater than 4 is , or 20%.
5
number of sections greater than 4
total number of sections
1
=
5
P(greater than 4) =
The theoretical probability of spinning a number greater
1
than 4 is , or 20%.
5
The experimental and theoretical probabilities are equal.
28. theoretical; Spinning the spinner 10,000 times is very
time consuming if using experimental probability.
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Chapter 10
30. P( 2) =
number of times sum of 2 rolled
2
1
=
=
total number of rolls
60
30
31. P ( 2) =
P(3) =
number of times sum of 3 rolled
4
1
=
=
total number of rolls
60
15
P (3) =
number of favorable outcomes
2
1
=
=
number of possible outcomes
36
18
P( 4) =
number of times sum of 4 rolled
5
1
=
=
total number of rolls
60
12
P ( 4) =
number of favorable outcomes
3
1
=
=
number of possible outcomes
36
12
P(5) =
number of times sum of 5 rolled
6
1
=
=
total number of rolls
60
10
P (5) =
number of favorable outcomes
4
1
=
=
number of possible outcomes
36
9
P(6) =
number of times sum of 6 rolled
13
=
total number of rolls
60
P ( 6) =
number of favorable outcomes
5
=
number of possible outcomes
36
P(7) =
number of times sum of 7 rolled
10
1
=
=
total number of rolls
60
6
P (7 ) =
number of favorable outcomes
6
1
=
=
number of possible outcomes
36
6
P(8) =
number of times sum of 8 rolled
6
1
=
=
total number of rolls
60
10
P (8) =
number of favorable outcomes
5
=
number of possible outcomes
36
P(9) =
number of times sum of 9 rolled
8
2
=
=
total number of rolls
60
15
P (9 ) =
number of favorable outcomes
4
1
=
=
number of possible outcomes
36
9
number of favorable outcomes
1
=
number of possible outcomes
36
P(10) =
number of times sum of 10 rolled
2
1
=
=
total number of rolls
60
30
P (10) =
number of favorable outcomes
3
1
=
=
number of possible outcomes
36
12
P(11) =
number of times sum of 11 rolled
3
1
=
=
total number of rolls
60
20
P (11) =
number of favorable outcomes
2
1
=
=
number of possible outcomes
36
18
P(12) =
number of times sum of 12 rolled
1
=
total number of rolls
60
P (12) =
number of favorable outcomes
1
=
number of possible outcomes
36
Each sum is not equally likely because they do not have
the same outcomes. The event with the greatest
experimental probability is rolling a sum of 6. So, a sum
of 6 is most likely.
Each sum is not equally likely because they do not have
the same outcomes. The event with the greatest
probability is rolling a sum of 7. So, a sum of 7 is most
likely.
32. The experimental and theoretical probabilities from
Exercises 30 and 31 are similar.
33. a. Sample answer: After 500 trials, you would expect a
sum of 6, 7, or 8 to be most likely. After 1000 or
10,000 trials, a sum of 7 would probably be most
likely.
b. As the number of trials increases, the experimental
probability approaches the theoretical probability.
34. a. Check students’ work. The cup should land on its side
most of the time.
b. Check students’ work.
c. Check students’ work.
d. more likely; Due to the added weight, the cup will be
more likely to hit open-end up and thus more likely to
land open-end up. Some students may justify by
performing multiple trials with a quarter taped to the
bottom of the cup.
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Chapter 10
Fair Game Review
35.
4. a. 10,000 passswords ×
I = Prt
16 = 200( r )( 2)
16 = 400r
0.04 = r
The annual interest rate is 0.04, or 4%.
36.
≈ 16.7 minutes
It could take up to 16.7 minutes, or 16 minutes
40 seconds to guess the password.
b. 100,000 passswords ×
I = Prt
 18 
26.25 = 500( r ) 
 12 
26.25 = 750r
It could take up to 2.8 hours to guess the password.
c. 308,915,776 passswords ×
1 h
1 day
×
60 min
24 h
≈ 357.5 days
3
Section 10.4
1 min
600 passwords
×
The annual interest rate is 0.035, or 3.5%.
 3 ft 
27 ft 3
3
= 243 ft 3
 = 9 yd ×
1 yd3
 1 yd 
1 min
1h
×
60 min
600 passwords
≈ 2.8 hours
0.035 = r
37. D; 9 yd3 × 
1 min
600 passwords
It could take up to 357.5 days to guess the password.
d. 2,821,109,907,456 passswords ×
10.4 Activity (pp. 420 –421)
1 day
1 h
×
60 min
24 h
1 yr
×
365 days
1. a. 000; 999; 1000
×
b. 10; 10; 10; Sample answer: Multiply to find the total
number of possible combinations:
10 • 10 • 10 = 1000 combinations.
≈ 8945.7 years
c. 40 • 40 • 40 = 64,000 combinations
d. 10 • 10 • 10 • 10 = 10,000 combinations
e. The lock in part (c) is most difficult to guess because
it has the greatest number of possible combinations
compared to the other two locks.
2. a. There are 10 choices for each digit.
10 • 10 • 10 • 10 = 10,000
It could take up to 8945.7 years to guess the password.
10.4 On Your Own (pp. 422–424)
1. Crust
Style
Outcome
Thin
Hawaiian
Mexican
Pepperoni
Veggie
Thin Crust Hawaiian
Thin Crust Mexican
Thin Crust Pepperoni
Thin Crust Veggie
Stuffed
Hawaiian
Mexican
Pepperoni
Veggie
Stuffed Crust Hawaiian
Stuffed Crust Mexican
Stuffed Crust Pepperoni
Stuffed Crust Veggie
Deep
Dish
Hawaiian
Mexican
Pepperoni
Veggie
Deep Dish Crust Hawaiian
Deep Dish Crust Mexican
Deep Dish Crust Pepperoni
Deep Dish Crust Veggie
There are 10,000 possible passwords.
b. There are 10 choices for each digit.
10 • 10 • 10 • 10 • 10 = 100,000
There are 100,000 possible passwords.
c. There are 26 choices for each letter.
26 • 26 • 26 • 26 • 26 • 26 = 308,915,776
There are 308,915,776 possible passwords.
d. There are 26 + 10 = 36 choices for each character.
36 • 36 • 36 • 36 • 36 • 36 • 36 • 36
= 2,821,109,907,456
There are 2,821,109,907,456 possible passwords.
The password requirement in part (d) in most secure
because it has the greatest number of possible passwords.
3. Sample answer: You can use an organized list, a table, a
tree diagram, or multiplication.
1 min
600 passwords
There are 12 different outcomes. So, there are
12 different pizzas possible.
2. Event 1: Spinning the spinner (4 outcomes)
Event 2: Choosing a number from 1 to 5 (5 outcomes)
4 × 5 = 20
There are 20 possible outcomes.
3. Event 1: Choosing a T-shirt (4 outcomes)
Event 2: Choosing a pair of jeans (5 outcomes)
Event 3: Choosing a pair of shoes (5 outcomes)
4 × 5 × 5 = 100
You can make 100 different outfits.
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4. There are four favorable outcomes in the sample space
for rolling at most 4 and flipping heads: 1H, 2H, 3H and
4H.
P(event ) =
number of favorable outcomes
number of possible outcomes
4
12
1
=
3
P(at most 4 and heads) =
The probability is
1
1
, or 33 %.
3
3
5. There are four favorable outcomes in the sample space
for flipping at least two tails: HTT, THT, TTH, and TTT.
P(event ) =
number of favorable outcomes
number of possible outcomes
4
8
1
=
2
P(at least 2 tails) =
1
The probability is , or 50%.
2
6. Using the table from Exercise #31 of Section 10.3, there
is one favorable outcome in the sample space for rolling
double threes.
P(event ) =
P(double threes) =
The probability is
number of favorable outcomes
number of possible outcomes
1
36
1
7
, or 2 %.
36
9
7. There is one favorable outcome in the sample space for
choosing a stuffed crust Hawaiian pizza.
P(event ) =
number of favorable outcomes
number of possible outcomes
1
P(double threes) =
8
1
1
The probability is , or 12 %.
8
2
10.4 Exercises (pp. 425 –427)
Vocabulary and Concept Check
1. A sample space is the set of all possible outcomes of an
event. Use a table or tree diagram to list all the possible
outcomes.
2. To use the Fundamental Counting Principle, first identify
the number of possible outcomes for each event. Then the
total number of possible outcomes is the product of these
numbers.
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3. One way is to use a tree diagram to find the total number
of possible outcomes. Another way is to use the
Fundamental Counting Principle by finding the number
of possible outcomes of spinning the spinner and
multiplying it by the number of possible outcomes of
rolling the number cube. The total number of possible
outcomes is 30.
4. Sample answer: An example of a compound event is
drawing two names out of a hat.
Practice and Problem Solving
5. Event 1: Choosing a number from 0 to 49 (50 outcomes)
Event 2: Choosing a number from 0 to 49 (50 outcomes)
Event 3: Choosing a number from 0 to 49 (50 outcomes)
50 × 50 × 50 = 125,000
There are 125,000 possible combinations for the lock.
6. Event
Time
Outcome
Miniature
Golf
1 P.M. – 3 P.M. Miniature Golf, 1 P.M. – 3 P.M.
6 P.M. – 8 P.M. Miniature Golf, 6 P.M. – 8 P.M.
Laser
Tag
1 P.M. – 3 P.M. Laser Tag, 1 P.M. – 3 P.M.
6 P.M. – 8 P.M. Laser Tag, 6 P.M. – 8 P.M.
Roller
Skating
1 P.M. – 3 P.M. Roller Skating, 1 P.M. – 3 P.M.
6 P.M. – 8 P.M. Roller Skating, 6 P.M. – 8 P.M.
There are 6 possible outcomes for the birthday party.
7. Type
Style
Outcome
Lion
Realistic Realistic lion
Cartoon Cartoon lion
Bear
Realistic Realistic bear
Cartoon Cartoon bear
Hawk
Realistic Realistic hawk
Cartoon Cartoon hawk
Dragon
Realistic Realistic dragon
Cartoon Cartoon dragon
There are 8 possible outcomes for the new school mascot.
8. Event 1: There are 3 possible sizes.
Event 2: There are 7 possible flavors.
3 × 7 = 21
There are 21 possible beverages.
9. Event 1: There are 4 possible amounts of memory.
Event 2: There are 5 possible colors.
4 × 5 = 20
There are 20 possible types of MP3 players.
10. Event 1: There are 3 possible suits.
Event 2: There are 2 possible wigs.
Event 3: There are 4 possible talents.
3 × 2 × 4 = 24
There are 24 possible types of clowns.
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Chapter 10
11. Event 1: There are 5 possible appetizers.
15–20. Spinner
Event 2: There are 4 possible entrées.
Coin
Outcome
1
H
T
1H
1T
2
H
T
2H
2T
3
H
T
3H
3T
4
H
T
4H
4T
5
H
T
5H
5T
Event 3: There are 3 possible desserts.
5 × 4 × 3 = 60
There are 60 possible different meals.
12. Method 1:
Type
Type 1
Type 2
Type 3
Size
Outcome
Size 1
Type 1, Size 1
Size 2
Type 1, Size 2
Size 3
Type 1, Size 3
Size 1
Type 2, Size 1
Size 2
Type 2, Size 2
Size 3
Type 2, Size 3
Size 1
Type 3, Size 1
Size 2
Type 3, Size 2
Size 3
Type 3, Size 3
15. There is one favorable outcome in the sample space for
spinning a 1 and flipping heads.
number of favorable outcomes
number of possible outcomes
1
P(1 and H) =
10
P(event ) =
The store sells a total of 9 different note cards.
The probability is
Method 2:
Event 1: There are 3 possible types.
1
, or 10%.
10
16. There are two favorable outcomes in the sample space for
Event 2: There are 3 possible sizes.
spinning an even number and flipping heads: 2H and 4H.
3×3 = 9
number of favorable outcomes
number of possible outcomes
2
P(even and H) =
10
1
=
5
P(event ) =
The store sells a total of 9 different note cards.
13. The total number of outcomes is the product of the
possible outcomes of each event, not the sum.
Event 1: Question 1 (2 possible answers)
Event 2: Question 2 (2 possible answers)
The probability is
Event 3: Question 3 (2 possible answers)
Event 4: Question 4 (2 possible answers)
17. There are two favorable outcomes in the sample space for
Event 5: Question 5 (2 possible answers)
spinning a number less than 3 and flipping tails: 1T and
2T.
2 × 2 × 2 × 2 × 2 = 32
The quiz can be answered 32 different ways.
Principle.
b. Let G = green, B = blue, R = red, and Y = yellow.
Marble 2
Outcome
G
B
R
R
B
Y
G
R
Y
G
B
Y
Y
number of favorable outcomes
number of possible outcomes
2
P(less than 3 and T) =
10
1
=
5
P(event ) =
14. a. Use a tree diagram or the Fundamental Counting
Marble 1
1
, or 20%.
5
G
B
R
GB GR GY BG BR BY RG RB RY YG YB YR
There are 12 possible outcomes.
Using the Fundamental Counting Principle, there are 4
possible outcomes for choosing the first marble, and 3
possible outcomes for choosing the second marble. So,
there are 4 × 3 = 12 possible outcomes.
The probability is
1
, or 20%.
5
18. There are no favorable outcomes in the sample space for
spinning a 6 and flipping tails.
number of favorable outcomes
number of possible outcomes
0
P(6 and T) =
10
= 0
P(event ) =
The probability is 0, or 0%.
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19. There are four favorable outcomes in the sample space
for not spinning a 5 and flipping heads: 1H, 2H, 3H,
and 4H.
P(event ) =
4
10
2
=
5
for spinning a prime number and not flipping heads: 2T,
3T, and 5T.
The probability is
Coin
H
1 (red)
T
H
2 (blue)
T
H
3 (yellow)
T
spinning an even number, flipping tails, then spinning an
odd number: 2T1 and 2T3.
P(event ) =
number of favorable outcomes
number of possible outcomes
2
18
1
=
9
1
1
The probability is , or 11 %.
9
9
3
10
P(odd then H then yellow ) =
3
, or 30%.
10
Spinner
Outcome
1 (red)
1 (red), H, 1 (red)
2 (blue)
1 (red), H, 2 (blue)
3 (yellow)
1 (red), H, 3 (yellow)
1 (red)
1 (red), T, 1 (red)
2 (blue)
1 (red), T, 2 (blue)
3 (yellow)
1 (red), T, 3 (yellow)
1 (red)
2 (blue), H, 1 (red)
2 (blue)
2 (blue), H, 2 (blue)
3 (yellow)
2 (blue), H, 3 (yellow)
1 (red)
2 (blue), T, 1 (red)
2 (blue)
2 (blue), T, 2 (blue)
3 (yellow)
2 (blue), T, 3 (yellow)
25. a. Event 1: There are 3 possible choices.
1 (red)
3 (yellow), H, 1 (red)
Event 2: There are 3 possible choices.
2 (blue)
3 (yellow), H, 2 (blue)
3 (yellow)
3 (yellow), H, 3 (yellow)
1 (red)
3 (yellow), T, 1 (red)
2 (blue)
3 (yellow), T, 2 (blue)
3 (yellow)
3 (yellow), T, 3 (yellow)
21. There is one favorable outcome in the sample space for
spinning blue, flipping heads, then spinning a 1.
P(event ) =
P( blue then H then 1) =
The probability is
23. There are two favorable outcomes in the sample space for
number of favorable outcomes
number of possible outcomes
21–24.
Spinner
number of favorable outcomes
number of possible outcomes
2
18
1
=
9
1
1
The probability is , or 11 %.
9
9
20. There are three favorable outcomes in the sample space
P( prime and not H ) =
P(event ) =
P(odd then H then yellow ) =
2
, or 40%.
5
P(event ) =
spinning an odd number, flipping heads, then spinning
yellow: 1H-yellow, and 3H-yellow.
number of favorable outcomes
number of possible outcomes
P( not spinning 5 and H) =
The probability is
22. There are two favorable outcomes in the sample space for
number of favorable outcomes
number of possible outcomes
1
18
24. There are four favorable outcomes in the sample space
for not spinning red, flipping tails, then not spinning an
even number: blue-T1, blue-T3, yellow-T1, and yellowT3.
P(event ) =
number of favorable outcomes
number of possible outcomes
4
18
2
=
9
2
2
The probability is , or 22 %.
9
9
P( not red then T then not even ) =
3×3 = 9
There are 9 possible outcomes, but only one way to
guess both answers correctly.
P(event ) =
P( both correct ) =
number of favorable outcomes
number of possible outcomes
1
9
The probability that you guess the correct answers to
1
both questions is , or about 11.1%.
9
1
5
, or 5 %.
18
9
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b. If you can eliminate one of the choices for each
question, then you have two choices left to choose
from for each question.
Event 1: There are 2 possible choices.
2×2 = 4
There are 4 possible outcomes, but only one way to
guess both answers correctly.
P( both correct ) =
P(event ) =
number of favorable outcomes
number of possible outcomes
1
4
The probability that your choice is correct is
1
, or 25%.
4
1
1000
1
,
1000
or 0.1%.
b. There are 1000 possible combinations. With 5 tries,
you would guess 5 out of the 1000 possibilities. So,
the probability of getting the correct combination is
5/1000, or 0.5%.
Event 2: There are 10 possible choices.
10 × 10 = 100
There are 100 possible outcomes, but only one way to
guess both digits correctly.
28. Event 1: 1 possible engine in the first position
Event 2: 8 possible train cars in the second position
number of favorable outcomes
number of possible outcomes
Event 3: 7 possible train cars in the third position
1
100
Event 5: 5 possible train cars in the fifth position
Event 4: 6 possible train cars in the fourth position
Event 6: 4 possible train cars in the sixth position
The probability that your choice is correct is
1
,
100
or 1%.
Event 7: 3 possible train cars in the seventh position
Event 8: 2 possible train cars in the eighth position
Event 9: 1 possible train car in the ninth position
b. Because you know the digits are even, you have 5
choices for each digit.
Event 1: There are 5 possible choices.
Event 2: There are 5 possible choices.
5 × 5 = 25
There are 25 possible outcomes, but only one way to
guess both digits correctly.
P(event ) =
There are 1000 possible outcomes, but only one way
to guess all three digits correctly.
P(correctly guessed) =
26. a. Event 1: There are 10 possible choices.
P( both correct ) =
Event 3: There are 10 possible choices.
number of favorable outcomes
number of possible outcomes
The probability increases to
P(event ) =
Event 2: There are 10 possible choices.
10 × 10 × 10 = 1000
Event 2: There are 2 possible choices.
P(event ) =
27. a. Event 1: There are 10 possible choices.
number of favorable outcomes
number of possible outcomes
1
P( both digits) =
25
The probability increases to
1
, or 4%.
25
1 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40,320
The train can be arranged 40,320 ways.
29. a. There are 9 events, and each event has 10 possible
outcomes, so a tree diagram would be too large.
b. 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 1,000,000,000
There are 1 billion identification numbers possible.
c. Answer should include, but is not limited to: Students
will explain that certain number combinations are not
valid for Social Security numbers. For example, the
first three digits cannot be a number in the 800’s or
900’s. Also, the first three digits cannot all be zeros,
the fourth and fifth digits cannot both be zeros, and
the sixth through the ninth digits cannot all be zeros.
30. 10; Let the 5 candidates be represented by A, B, C, D,
and E. Then, a committee of 3 people could be one of the
10 following ways: ABC, ABD, ABE, ACD, ACE, ADE,
BCD, BCE, BDE, and CDE. Order does not matter.
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b. yes; The probability of getting a green marble on the
Fair Game Review
second draw depends on the color of the first marble
because if a green marble is drawn first, then there is a
50% chance of drawing a green marble on the second
draw. If a purple marble is drawn first, then there is a
100% chance of drawing a green marble on the second
draw.
31. Sample answers:
adjacent: ∠XWY and ∠ZWY , ∠XWY and ∠XWV
vertical: ∠VWX and ∠YWZ , ∠YWX and ∠VWZ
32. Sample answers:
adjacent: ∠LJM and ∠LJK , ∠LJM and ∠NJM
vertical: ∠KJL and ∠PJN , ∠PJQ and ∠MJL
3. a. Answer should include, but is not limited to: Students
will perform an experiment by drawing a marble from
a bag, putting it back, and drawing a second marble.
The experiment should be performed 36 times.
Students will record their results in a table and then
draw a bar graph representing the results. Bar graphs
should be neatly drawn and clearly labeled.
1 cm
1 cm
33. B; 1 cm : 1 m =
= 1 : 100
=
1m
100 cm
Section 10.5
10.5 Activity (pp. 428 –429)
b. Answer should include, but is not limited to: Students
will perform a second experiment by drawing 2
marbles from a bag 36 times. Students will record
their results in a table and then draw a bar graph
representing the results. Bar graphs should be neatly
drawn and clearly labeled.
1. a.
First draw:
Second draw:
GG
GG
GP
GG
GG
GP
PG
PG
PP
will use the results of the experiments in parts (a) and
(b) to determine the experimental probability of
drawing two green marbles.
number of favorable outcomes
P(green, green ) =
number of possible outcomes
=
c. Answer should include, but is not limited to: Students
d. The second experiment represents dependent events
4
9
The probability that both marbles are green is
4
.
9
b. no; The probability of getting a green marble on the
second draw does not depend on the color of the first
marble because the probability of getting a green
marble is the same for both draws.
2. a.
because the probability of the second marble being
green or purple depends on the color of the first
marble. The first experiment represents independent
events because the color of the second marble does not
depend on the color of the first marble.
4. Two events are dependent if the occurrence of one event
does affect the likelihood that the other event will occur.
Two events are independent if the occurrence of one
event does not affect the likelihood that the other event
will occur.
Sample answer: An example of dependent events is
drawing two names out of a basket without replacing the
first name. An example of independent events is flipping
a coin twice.
First draw:
Second draw:
GG
GP
GG
GP
PG
PG
number of favorable outcomes
P(green, green ) =
number of possible outcomes
2
6
1
=
3
This event is not more likely than the event in
1
Activity 1 because the probability of is not
3
4
greater than the probability of .
9
300 Big Ideas Math Red Accelerated
Worked-Out Solutions
outcome of spinning a spinner. So, the events are
independent.
6. The outcome of selecting the first group leader does
=
The probability that both marbles are green is
5. The outcome of rolling a number cube does not affect the
1
.
3
affect the outcome of selecting the second group leader
because there are fewer students to choose from when
picking the second group leader. So, the events are
dependent.
7. The outcome of spinning red on one spinner does not
affect the outcome of spinning green on the other spinner.
So, the events are independent.
Copyright © Big Ideas Learning, LLC
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Chapter 10
2 2
; ; Find the product of the
3 3
2 2
4
probabilities, which is •
= .
3 3
9
8. For Activity 1:
So, the probability of drawing two green marbles in
4
Activity 1 is .
9
2 1
For Activity 2: ; ; Sample answer: Find the product
3 2
2 1
2
1
of the probabilities, which is •
=
= .
3 2
6
3
So, the probability of drawing two green marbles
1
in Activity 2 is .
3
10.5 On Your Own (pp. 430 –432)
2
1. P( multiple of 2) =
5
1
P( heads) =
2
P( A and B) = P( A) • P( B)
P(multiple of 2 and heads) = P(multiple of 2) • P(heads)
2 1
= •
5 2
1
=
5
The probability of spinning a multiple of 2 and flipping
1
heads is , or 20%.
5
2. There are 88 other audience members who are not you,
your relatives, or your friends. Choosing an audience
member changes the number of audience members left.
So, the events are dependent.
88
22
=
100
25
87
29
P(second other audience member ) =
=
99
33
P(first other audience member ) =
P( A and B ) = P( A) • P( B after A)
P( 2 other audience members)
= P(1st other ) • P( 2nd other after 1st )
22 29
•
25 33
58
=
75
=
3. P( #1 correct and #2 correct and #3 correct )
= P( #1 correct ) • P( #2 correct ) • P( #3 correct )
1 1 1
• •
4 4 4
1
=
64
=
The probability of answering all three questions correctly
1
, or about 1.56%. This is greater than the probability
is
64
in Example 3. So, the probability of answering all three
questions correctly increases when one choice is
eliminated.
10.5 Exercises (pp. 433 –435)
1. All four questions have first events that are the same,
because “choosing a 1” is the same as “choosing a green
chip” and “choosing a number less than 2.” However,
the second event of question #1 does not match the
others. Of the chips remaining, the even numbers are
also the chips that are not red, and these chips are blue
and yellow, not just blue.
P(1 and blue) = P(1) • P( blue after 1)
1 2
•
6 5
1
=
15
=
P(1 and even ) = P(1) • P(even after 1)
1 3
•
6 5
1
=
10
=
2. When events A and B are independent, the probability
of both events is the probability of event A times the
probability of event B. When events A and B are
dependent, the probability of both events is the
probability of event A times the probability of event B
after event A occurs.
Practice and Problem Solving
3. The outcome of the first roll does not affect the outcome
of the second roll. So, the events are independent.
4. After you draw your lane number, there is one less lane
number available for your friend. So, the events are
dependent.
The probability that you, your relatives, and your friends
are not chosen to be either of the first two contestants
58
, or about 77.3%.
is
75
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Big Ideas Math Red Accelerated
Worked-Out Solutions
301
Chapter 10
5.
P( A and B) = P( A) • P( B )
P(3 and heads) = P(3) • P( heads)
1 1
•
4 2
1
=
8
=
9. P(first is 5) =
P( A and B) = P( A) • P( B after A)
P(5 and 6) = P(5) • P(6 after 5)
2 1
•
4 2
1
=
4
The probability of choosing a 5 and then a 6 is
1
, or
42
about 2.4%.
=
10. P(first is odd ) =
The probability of spinning an even number and flipping
1
tails is , or 25%.
4
P( A and B) = P( A) • P( B )
P(greater than 1 and tails) = P(greater than 1) • P( tails)
3 1
•
4 2
3
=
8
=
The probability of spinning a number greater than 1 and
3
flipping tails is , or 37.5%.
8
8.
1 1
•
7 6
1
=
42
=
P( A and B) = P( A) • P( B )
P(even and tails) = P(even ) • P( tails)
7.
1
6
P(second is 6) =
The probability of spinning a 3 and flipping heads is
1
, or 12.5%.
8
6.
1
7
P( A and B) = P( A) • P( B)
P( not 2 and heads) = P( not 2) • P( heads)
3 1
•
4 2
3
=
8
=
The probability of not spinning a 2 and flipping heads is
3
, or 37.5%.
8
3
7
P(second is 20) =
1
6
P( A and B) = P( A) • P( B after A)
P(odd and 20) = P(odd ) • P( 20 after odd )
3 1
•
7 6
1
=
14
=
The probability of choosing an odd number and then 20 is
1
, or about 7.1%.
14
11. P(first is less than 7) =
2
7
P(second is multiple of 4) =
2
1
=
6
3
P( A and B) = P( A) • P( B after A)
P(less than 7 and multiple of 4) = P(less than 7)
 multiple of 4 
• P

 after less than 7 
2 1
= •
7 3
2
=
21
The probability of choosing a number less than 7 and
2
then a multiple of 4 is
, or about 9.5%.
21
302 Big Ideas Math Red Accelerated
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Chapter 10
4
7
12. P(first is even ) =
P(second is even ) =
16. P(choose correct path at first fork ) =
3
1
=
6
2
P(choose correct path at second fork ) =
P( both even ) = P(even ) • P(even after even )
P( both correct paths) = P(correct 1st path ) • P(correct 2nd path )
4 1
•
7 2
2
=
7
1 1
•
2 2
1
=
4
=
=
The probability of choosing two even numbers is
2
, or
7
The probability that you are still on the correct path is
13. The events are dependent, not independent.
17. P(first is purple) =
1
4
3
14
P( A and B ) = P( A) • P( B after A)
P( A and B) = P( A) • P( B after A)
P( both purple) = P( purple) • P( purple after purple)
P( red and green ) = P(red ) • P(green after red )
4
3
•
15 14
2
=
35
=
1 1
•
4 3
1
=
12
=
The probability that both balloons are purple is
14. The tree diagram shows that three marbles are available
for the first draw and only two marbles are available for
the second draw. So, the events are dependent.
2
, or
35
about 5.7%.
18.
P( A and B and C ) = P( A) • P( B ) • P(C )
P( 4 and heads and 7) = P( 4) • P( heads) • P(7)
2
1
=
4
2
1
P(second is silver ) =
3
15. P(first is silver ) =
1 1 1
• •
9 2 9
1
=
162
=
P( A and B) = P( A) • P( B after A)
1 1
1
• =
2 3
6
The probability that both are silver hoop earnings is
or about 16.7%.
4
15
P(second is purple) =
1
3
P( both silver ) = P(silver ) • P(silver ) =
1
,
4
or 25%.
about 28.6%.
P(second is green ) =
1
2
P( A and B ) = P( A) • P( B after A)
P( A and B) = P( A) • P( B after A)
P(first is red) =
1
2
1
,
6
The probability is
19.
1
, or about 0.6%.
162
P( A and B and C ) = P( A) • P( B) • P(C )
P(odd and heads and 3) = P(odd ) • P( heads) • P(3)
5 1 1
• •
9 2 9
5
=
162
=
The probability is
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5
, or about 3.1%.
162
Big Ideas Math Red Accelerated
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303
Chapter 10
20. P( A and B and C ) = P( A) • P( B ) • P(C )
 even and tails 
P
 = P(even ) • P( tails) • P(odd )
 and odd

4 1 5
= • •
9 2 9
10
=
81
The probability is
10
, or about 12.3%.
81
P( A and B and C ) = P( A) • P( B) • P(C )
21.
P( not 5 and heads and 1) = P( not 5) • P( heads) • P(1)
8 1 1
• •
9 2 9
4
=
81
=
4
The probability is , or about 4.9%.
81
22.
24. If 20% of the shoes are black, then 80% are not black.
P( A and B and C ) = P( A) • P( B) • P(C )
P( none black ) = P( not black ) • P( not black )
• P( not black )
= 0.80 • 0.80 • 0.80
= 0.512
The probability that none of the shoes are black is 0.512,
or 51.2%.
25. a. no; If you and your best friend were in the same
group, then the probability that you both are chosen
would be 0 because only one leader is chosen from
each group. Because the probability that both you and
1
your best friend are chosen is
, you and your best
132
friend are not in the same group.
b. P( both are chosen ) = P( you are chosen )
• P(friend is chosen )
1
1
=
• x
132
12
12
= x
132
1
= x
11
P( A and B and C ) = P( A) • P( B) • P(C )
 odd and not heads 
 not 
P
 = P(odd) • P heads  • P( not 6)
 and not 6



5 1 8
= • •
9 2 9
20
=
81
The probability is
20
, or about 24.7%.
81
15
23. P( not chosen first ) =
16
14
15
13
P( not chosen third) =
14
12
P( not chosen fourth ) =
13
P( not chosen second) =
P( not one of first four students chosen )
= P( not 1st ) • P( not 2nd ) • P(not 3rd ) • P(not 4th )
15 14 13 12
•
•
•
16 15 14 13
12
=
16
3
=
4
The probability that your best friend is chosen as a
1
group leader is , or about 9.1%.
11
c. The probability that you are chosen to be a leader
1
, so there are 12 students in your group. The
12
probability that your best friend is chosen to be a leader
1
is , so there are 11 students in that group. The total
11
number of students in the class is 12 + 11 = 23.
is
26. 25% =
25
1
=
100
4
Eliminate all but 2 of the choices in each question.
a. P( both correct ) = P(one question correct )
• P( the other question correct )
=
The probability that you are not one of the first four
3
students chosen is , or 75%.
4
1
1 1
=
•
4
2 2
1
3
b. 8 % =
25
25
1
1
•
=
% =
3
3
100
12
Eliminate one of the choices in one of the questions
and eliminate two of the choices in the other question.
P( both correct ) = P(one question correct )
• P( the other question correct )
1
1 1
=
•
12
4 3
304 Big Ideas Math Red Accelerated
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Chapter 10
2. Sample answer: Place 7 green and 3 red marbles in a bag.
Fair Game Review
Let the green marbles represent a win and the red marbles
represent a loss. Randomly pick one marble to simulate
the first game. Replace the marble and repeat two more
times. This is one trial. Run 30 trials. Check students’
work. The probability should be close to 34.3%
(depending on the number of trials, because that is the
theoretical probability).
27. Sample answer:
608
308
908
The triangle has no congruent sides and one right angle.
So, it is a right scalene triangle.
28. Sample answer:
508
1108
208
The triangle has no congruent sides and one obtuse angle.
So, it is an obtuse scalene triangle.
29. Sample answer:
508
508
808
3.
1
2
3
4
5
6
7
8
9
10
11
A
7584
3762
3024
4547
1719
7938
6951
4714
0797
3300
B
3974
3805
1554
6220
0662
9551
0578
4511
3022
5454
C
8614
2725
2708
9497
1814
8552
5560
5115
9067
5351
D
2500
7320
1126
7530
6218
4321
0740
6952
2193
6319
E
4629
6487
9395
3036
2766
8043
4479
5609
6553
0387
F
Sample answer: Using the spreadsheet in Example 3 and
using digits 1 – 4 as successes, the experimental
8
probability is
, or 0.16, or 16%.
50
The triangle has two congruent angles, and all angles are
acute. So, it is an acute isosceles triangle.
30. C;
For A:
2
= 0.66...
3
0.6 = 0.60
67% = 0.67
For B: 44.5% = 0.445
4
= 0.444...
9
0.46 = 0.466...
For C: 0.269 = 0.269
27% = 0.270
3
= 0.2727...
11
1
≈ 2.1429
7
214% = 2.1400
For D: 2
2.14 = 2.1414...
10.5 Extension (p. 437)
Practice
1. a. Sample answer: Roll four number cubes. Let an odd
number represent a correct answer and an even
number represent an incorrect answer. Run 40 trials.
b. Check students’ work. The probability should be
“close” to 6.25% (depending on the number of trials,
because that is the theoretical probability).
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Worked-Out Solutions
305
Chapter 10
4. Example 1: The events are independent, and
1
P( boy) = .
2
P( three boys) = P( boy) • P( boy) • P( boy)
1 1 1
• •
2 2 2
1
=
8
=
The theoretical probability of having three boys is
1
, or 12.5%.
8
Example 2: The events are independent.
P( rain on Monday) = 0.6
P( rain both days) = P( rain on Monday) • P( rain on Tuesday)
= (0.6)(0.2)
= 0.12
The theoretical probability of it raining on both Monday
and Tuesday is 0.12, or 12%.
Example 3: Each year, there are two possible equal
outcomes: cancellation due to weather or no cancellation.
So, over four years, there are 2 × 2 × 2 × 2 = 16
possible outcomes.
There are five favorable outcomes: Cancellations in years
1, 2, 3, and 4, years 1, 2, and 3, years 1, 2, and 4, years 1,
3, and 4, or years 2, 3, and 4.
number of favorable outcomes
P(event ) =
number of possible outcomes
5
16
= 31.25%
P(at least 3 years of cancellations out of 4) =
So, the theoretical probability is of having at least three
5
years of four with cancellations is , or 31.25%.
16
Alternatively, Example 3 could have been solved using
the following method.
P (1, 2, 3, 4) = P(1) • P( 2) • P(3) • P( 4)
1 1 1 1
1
=
• • •
=
2 2 2 2
16
P (1, 2, 3, not 4) = P(1) • P( 2) • P(3) • P( not 4)
1 1 1 1
1
• • •
=
2 2 2 2
16
P (1, 2, not 3, 4) = P(1) • P( 2) • P( not 3) • P( 4)
=
1 1 1 1
1
• • •
=
2 2 2 2
16
P (1, not 2, 3, 4) = P(1) • P( not 2) • P(3) • P( 4)
=
1 1 1 1
1
• • •
=
2 2 2 2
16
306 Big Ideas Math Red Accelerated
Worked-Out Solutions
=
1 1 1 1
1
• • •
=
2 2 2 2
16
5
1
So, the theoretical probability is 5  =
, or 31.25%.
 16  16
Exercise 1: The events are independent, and
1
P(correct guess) = .
2
P( 4 correct guesses) = P(correct guess) • P(correct guess)
• P(correct guess) • P(correct guess)
1 1 1 1
• • •
2 2 2 2
1
=
16
1
The theoretical probability is , or 6.25%.
16
=
P( rain on Tuesday) = 0.2
=
P (not 1, 2, 3, 4) = P( not 1) • P( 2) • P(3) • P( 4)
Exercise 2: The events are independent, and
P( winning ) = 0.7.
P( three wins) = P( winning ) • P( winning ) • P( winning)
= (0.7)(0.7)(0.7)
= 0.343
The theoretical probability is 0.343 or 34.3%.
Exercise 3: The events are independent.
P(cancellation ) = 0.4 =
2
5
P( no cancellation ) = 0.6 =
3
5
P(1, 2, 3, 4) = P(1) • P( 2) • P(3) • P( 4)
2 2 2 2
16
• • •
=
5 5 5 5
625
P(1, 2, 3, not 4) = P(1) • P( 2) • P(3) • P( not 4)
=
2 2 2 3
24
• • • =
5 5 5 5
625
P(1, 2, not 3, 4) = P(1) • P( 2) • P( not 3) • P( 4)
=
2 2 3 2
24
• • •
=
5 5 5 5
625
P(1, not 2, 3, 4) = P(1) • P( not 2) • P(3) • P( 4)
=
2 3 2 2
24
• • •
=
5 5 5 5
625
P (not 1, 2, 3, 4) = P( not 1) • P( 2) • P(3) • P( 4)
=
=
3 2 2 2
24
• • •
=
5 5 5 5
625
So, the theoretical probability is
16
24
24
24
24
112
+
+
+
+
=
= 17.92%.
625 625 625 625 625
625
When you increase the number of trials in a simulation,
the experimental probability approaches the theoretical
probability of the event that you are simulating.
Copyright © Big Ideas Learning, LLC
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Chapter 10
Study Help
9. A multiple of 3 was rolled 21 + 23 = 44 times.
Available at BigIdeasMath.com.
Quiz 10.1–10.5
1. There are 2 red butterflies. So, choosing a red butterfly
can occur in 2 ways.
The probability of rolling a multiple of 3 is
2. There are no brown butterflies. So, choosing a brown
butterfly can occur in 0 ways.
3. There are 4 butterflies that are not blue. So, choosing a
not blue butterfly can occur in 4 ways.
4. P(green ) =
=
number of green paper clips
total number of paper clips
3
,
10
number of yellow paper clips
total number of paper clips
5
5
1
=
=
6+3+ 4+ 2+5
20
4
The probability of choosing a yellow paper clip is
1
,
4
or 25%.
6. P( not yellow ) =
=
number of not yellow paper clips
total number of paper clips
6+3+ 4+ 2
15
3
=
=
6+3+ 4+ 2+5
20
4
The probability of choosing a paper clip that is not yellow
3
is , or 75%.
4
7. P( purple) =
10. A 2 or a 3 was rolled 22 + 21 = 43 times.
P( 2 or 3) =
number of times a 2 or a 3 rolled
43
=
total number of rolls
120
43
, or about
120
35.8%.
or 30%.
=
11
, or about
30
36.7%.
The probability of rolling a 2 or a 3 is
6
6
3
=
=
6+3+ 4+ 2+5
20
10
The probability of choosing a green paper clip is
5. P( yellow ) =
number of times
multiple of 3 rolled
44
11
P( multiple of 3) =
=
=
total number of rolls
120
30
number of purple paper clips
total number of paper clips
0
=
= 0
6+3+ 4+ 2+5
11. The number of times a number less than 7 was rolled
is 120.
number of times less than 7 rolled
total number of rolls
120
=
120
=1
P(less than 7) =
The probability of rolling a number less than 7 is 1,
or 100%.
12. Event 1: There are 4 possible types.
Event 2: There are 3 possible colors.
4 × 3 = 12
There are 12 possible kinds of calculators.
13. Event 1: There are 4 possible destinations.
Event 2: There are 2 possible lengths.
4×2 = 8
There are 8 possible vacations.
14. P( black ink ) =
number of black ink pens
2
=
total number of pens
5
The probability of randomly choosing a black pen is
2
, or 40%.
5
The probability of choosing a purple paper clip is 0,
or 0%.
8. The total number of rolls is
18 + 22 + 21 + 16 + 20 + 23 = 120.
P( 4) =
number of times 4 rolled
16
2
=
=
total number of rolls
120
15
The probability of rolling a 4 is
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2
, or about 13.3%.
15
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307
Chapter 10
15. P(first is blue) =
6. no; You can have a large sample that is not representative
2
5
P(second is blue) =
1
4
P( A and B ) = P( A) • P( B after A)
P( both blue) = P( blue) • P( blue after blue)
2 1
=
•
5 4
1
=
10
The probability that you and your friend both choose a
1
blue pen is , or 10%.
10
Section 10.6
10.6 Activity (pp. 440 –441)
1. a. Population: The students in a school; Sample: The
students in a math class
b. Population: The grizzly bears in a park; Sample: The
grizzly bears with GPS collars in a park
c. Population: All quarters in circulation; Sample: 150
quarters
d. Population: All books in a library; Sample: 10 fiction
books in a library
2. a. no; Not every student in your school is equally likely
of being selected. Most of the responses may be
playing or listening to music.
b. no; Not every student in your school is equally likely
of being selected. Not all students may read the school
newspaper.
c. yes; Every student in your school is equally likely of
being selected.
d. no; Not every student in your school is equally likely
of being selected. You are only surveying students
from a certain grade.
3. a. not valid; Residents that call into the radio show most
likely have very strong opinions and not everyone may
listen to the radio.
b. not valid; The reporter only talked to 2 residents. The
size of the sample is not large enough to draw any
valid conclusions.
c. valid; You talked to a good amount of residents that
are randomly selected and your conclusion is
reasonable based on your results.
4. Sample answer: A sample accurately represents a
population when the sample is large enough, random, and
not just a certain portion of the population.
5. Check students’ work.
308 Big Ideas Math Red Accelerated
Worked-Out Solutions
of a population. Sample answer: You survey residents
about a new power plant being built but you only survey
residents who will live close to the power plant. This
could result in a large sample that is not representative of
residents in a town.
10.6 On Your Own (pp. 442–443)
1. C; The sample is representative of the population.
2. The sample that surveys every 5th student on an
alphabetical list of eighth graders is unbiased because
every eighth grader is equally likely of being selected.
Sampling 15 band members is biased because it does not
represent the entire eighth grade, and band members are
more likely to listen to music.
3. No, firefighters are more likely to support the new sign.
students in survey
students in school
(two or more movies)
(two or more movies)
4.
=
total surveyed
students in school
24
n
=
75
1200
384 = n
So, about 384 students in the school watch two or more
movies each week.
10.6 Exercises (pp. 444 –445)
Vocabulary and Concept Check
1. You would survey a sample instead of a population
because samples are easier to obtain.
2. You should make sure the people surveyed are selected at
random and are representative of the population, as well
as making sure your sample is large enough.
Practice and Problem Solving
3. The population is the residents of New Jersey. The
sample consists of the residents of Ocean County.
4. The population is all the cards in a deck. The sample
consists of the 4 cards drawn from the deck.
5. biased; The sample is not selected at random and is not
representative of the population because students in a
band class play a musical instrument.
6. unbiased; The sample is representative of the population,
selected at random, and large enough to provide accurate
data.
7. biased; The sample is not representative of the population
because people who go to the park are more likely to
think that the park needs to be remodeled.
8. The sample is representative of the population, selected at
random, and large enough to provide accurate data. So,
the sample is unbiased and the conclusion is valid.
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Chapter 10
9. The sample is not representative of the population
because people going to the baseball stadium are more
likely to support building a new baseball stadium. So, the
sample is biased and the conclusion is not valid.
Fair Game Review
20.
10. Sample A is not large enough to provide accurate data.
Sample B is a large sample, so Sample B is better for
making a prediction.
11. Sample B is not representative of the population because
the population consists of pencils from all the machines.
Sample A is representative of the population, so
Sample A is better for making a prediction.
p
100
p
=
100
p
=
100
=
= 100 •
p
100
= p
So, 30% of 60 is 18.
12.
students in school ( pizza )
students in survey (pizza)
=
number of students in survey
number of students in school
a
w
18
60
3
10
3
100 •
10
30
21.
n
58
=
125
1500
696 = n
So, there are about 696 students in the school whose
favorite food is pizza.
a
w
98
w
98
w
980
p
100
70
=
100
7
=
10
= 7w
=
140 = w
13. Because the population size is very large, you would
So, 70% of 140 is 98.
survey a sample.
14. Because the population size is very small, you would
22.
survey the population.
15. Because the population size is very large, you would
survey a sample.
Number surveyed
who said yes
Student tickets bought
16.
=
Total surveyed
Students in school
12
210
=
72
n
n = 1260
There are about 1260 students in the school.
17. Not everyone has an email address, so the sample may
not be representative of the entire population. Sample
answer: When the survey question is about technology or
which email service you use, the sample may be
representative of the entire population.
18. a. Sample answer: The person could ask “Do you agree
with the town’s unfair ban on skateboarding on public
property?”
b. Sample answer: The person could ask “Do you agree
that the town’s ban on skateboarding on public
property has made the town quieter and safer?”
19. Sample answer: no; 75% of the students in the sample
said that they plan to attend college. Because 75% of 900
is 675, the counselor’s prediction was too high. The
counselor included students that replied “maybe,”
increasing the number to 80% of 900, or 720 students.
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a
w
30
w
30
w
600
p
100
15
=
100
3
=
20
= 3w
=
200 = w
So, 30 is 15% of 200.
23.
a
w
a
500
a
500
1000a
p
100
0.6
=
100
6
=
1000
= 3000
=
a = 3
So, 3 is 0.6% of 500.
1
Bh
3
1
= ( 4)(6)(5)
3
= 40
24. A; V =
So, the volume of the pyramid is 40 cm3.
Big Ideas Math Red Accelerated
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309
Chapter 10
10.6 Extension (pp. 446 –447)
Activity 1
Step 5: Number of students who prefer pop music:
10 + 13 + 8 + 10 + 9 = 50
Number of students surveyed:
20 + 20 + 20 + 20 + 20 = 100
Step 1: Number you surveyed Students in school
who prefer pop
who prefer pop
=
Total surveyed
Students in school
Number surveyed Students in school
who prefer pop
who prefer pop
=
Total surveyed
Students in school
10
n
=
20 840
420 = n
50
n
=
100 840
420 = n
So, there are 420 students who prefer pop music.
Step 2: Number Kevin surveyed Students in school
who prefer pop
who prefer pop
=
Total surveyed
Students in school
13
n
=
20 840
546 = n
So, there are 546 students who prefer pop music.
Using Kevin’s results, this inference is more than
the inference in Step 1.
Step 3: Number Steve surveyed Students in school
who prefer pop
who prefer pop
=
Total surveyed
Students in school
So, there are 420 students who prefer pop music
when all five samples are combined.
Practice
1. a. Check students’ work.
make a more accurate prediction by increasing the
number of random samples.
Activity 2
Step 1:
6+8+6+6+7+
4 + 10 + 8 + 7 + 8
70
Mean 1:
=
=7
10
10
10 + 4 + 4 + 6 + 8
+6 + 7 + 12 + 8 + 8 73
Mean 2:
=
= 7.3
10
10
8
n
=
20 840
336 = n
10 + 9 + 8 + 6 + 5
+8 + 6 + 6 + 9 + 10 77
Mean 3:
=
= 7.7
10
10
Number Laura surveyed Students in school
who prefer pop
who prefer pop
=
Total surveyed
Students in school
4+8+4+4+5
+4 + 4 + 6 + 5 + 6 50
Mean 4:
=
=5
10
10
10
n
=
20 840
420 = n
6 + 8 + 8 + 6 + 12
+4 + 10 + 8 + 6 + 12 80
Mean 5:
=
=8
10
10
Number Ming surveyed Students in school
who prefer pop
who prefer pop
=
Total surveyed
Students in school
9
n
=
20 840
378 = n
Mean 6:
10 + 10 + 8 + 9 + 16
+8 + 7 + 12 + 16 + 14 110
=
= 11
10
10
4+5+6+6+4
+5 + 6 + 6 + 4 + 4 50
Mean 7:
=
=5
10
10
Steve: 336 students prefer pop music.
Laura: 420 students prefer pop music.
16 + 20 + 8 + 12 + 10
+8 + 8 + 14 + 16 + 8
120
Mean 8:
=
= 12
10
10
Ming: 378 students prefer pop music.
Step 4: Sample answer: The greatest is 546 students.
The least is 336 students. 420 is the median and
the mode of the data. So, use the inference of
420 students.
b. Check students’ work.
c. Check students’ work. Sample answer: Yes, you can
Step 2:
3
4
5
6
7
8
9
10 11 12 13
Mean hours
worked each
week
Step 3: Sample answers: The actual mean number of hours
probably lies within the interval 6 to 9.5 hours
(the box). So, about 7.5 hours is a good estimate.
The mean of the entire data set is 7.875. So, the
estimate is close.
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Chapter 10
Activity 3
b. Both distributions are approximately symmetric.
Step 1: Check students’ work.
c. For Male Students in Grade 8:
Step 2: Check students’ work. Sample answer: Because the
actual percent of students is 70%, students can say
that they expect the number of red peanuts in their
samples to be between 60% –80%, or 55% –85%, or
65% –75%, etc.
Step 3: Check students’ work.
Practice
7+9+8+7
1
1
+8
2
2
1
+10 + 6 + 6 + 8 + 8
2
1
1
1
+8 + 9 + 11 + 7 + 8
2
2
2 = 123 = 8.2
Mean:
15
15
From part (a):
2. a.
Median: 8
0
10 20 30 40 50 60 70 80 90 100
Percent that
prefer water
Mode: 8, 8
b. Sample answer: The actual percent of student-athletes
that prefer water over a sports drink probably lies
within the interval 50% to 70% (the box). So, about
60% is a good estimate.
60 + 70 + 60 + 50 + 80
+70 + 30 + 70 + 80 + 40 610
Mean:
=
= 61
10
10
The mean of the data is 61%. So, the estimate is close.
3. First order all the data to find the median of each sample.
Step 1: Medians are: 7, 7.5, 8, 4.5, 8, 10, 5, and 11.
Step 2:
3
4
5
6
7
8
9
10 11 12 13
Median hours
worked each
week
Step 3: Sample answers: The actual median number of
hours probably lies within the interval 6 to 9 hours
(the box). So, about 7.5 is a good estimate. The
median of the data is 8. So, the estimate is close.
1
2
Range: 11 − 6 = 5
Interquartile range (IQR): 9 − 7.5 = 1.5
Mean Absolute Deviation (MAD): Find the mean of
how far each data point deviates from the mean of 8.2.
1.2 + 0.8 + 0.2 + 0.7 + 0.3
+1.8 + 2.2 + 1.7 + 0.2 + 0.2
+0.3 + 0.8 + 2.8 + 0.7 + 0.3 14.2
MAD:
=
≈ 0.95
15
15
For Male Students in Grade 6:
1
1
1
6+5 +6+6 +7
2
2
2
1
1
1
+8 + 7 + 5 + 5 + 5
2
2
2
1
1
+6 + 7 + 4 + 6 + 6
93
2
2
=
= 6.2
Mean:
15
15
From part (a):
Median: 6
4. Check students’ work.
Mode: 6
5. The more samples you have, the more accurate your
inferences will be. By taking multiple random samples,
you can find an interval where the actual measurement of
a population may lie.
1
1
Range: 8 − 4 = 4
2
2
Interquartile range (IQR): 7 − 5.5 = 1.5
Mean Absolute Deviation (MAD): Find the mean of
how far each data point deviates from the mean of 6.2.
Section 10.7
10.7 Activity (pp. 448 –449)
0.2 + 0.7 + 0.2 + 0.3 + 1.3
+2.3 + 0.8 + 0.7 + 1.2 + 0.7
+0.3 + 0.8 + 1.7 + 0.2 + 0.2 11.6
MAD:
=
≈ 0.77
15
15
1. a. Sample answer:
Male students (8th Grade)
Male students (6th Grade)
Mean Median Mode
Shoe sizes
4
5
6
7
8
9
5
6
7
8
9
10 11 12 13
4
5
6
7
8
9
10 11 12 13
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Interquartile
Range (IQR)
Mean
Absolute
Deviation
(MAD)
10 11 12
Shoe Size
4
Range
Male Students in
Eighth-Grade Class
8.2
8
8, 8 2
5
1.5
0.95
Male Students in
Eighth-Grade Class
6.2
6
6
4
1.5
0.77
1
Male students
(8th grade)
Male students
(6th grade)
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Chapter 10
d. The mean, median, and one of the modes of the shoe
sizes for male students in the eighth-grade class are
each 2 more than the mean, median, and mode of the
shoe sizes for male students in the sixth-grade class.
e. yes; The IQRs are the same. The range and MAD are
slightly greater for the male students in the eighthgrade class. So, the shoe sizes for the male students in
the eighth-grade class are slightly more spread out
than the shoe sizes for the male students in the sixthgrade class.
2. a. male students: symmetric; female students: skewed
left; yes; Sample answer: The data set for the female
students completely overlaps the data set for the male
students. The overlaps between the centers and
between the extreme values are shown.
Male students
Female students
f. yes; Sample answer: The smallest shoe size of the
male students in the eighth-grade class is about the
same as the mean shoe size of the male students in the
sixth-grade class. The largest shoe size of the male
students in the sixth-grade class is about the same as
the mean shoe size of the male students in the
eighth-grade class.
5
6
7
8
9
10 11 12 13
5
6
7
8
9
10 11 12
4
5
6
7
8
9
10 11 12
Hours
of sleep
Male students
Female students
Hours
of sleep
b. male students: symmetric; female students:
Shoe Size
4
4
Male students
(8th grade)
symmetric; yes; Sample answer: The overlaps
between the centers and between the extreme values
are shown.
Heights (inches)
4
5
6
7
8
9
10 11 12 13
Male students
(6th grade)
Male students
g. no; Sample answer: You could have the following data
56 57 58 59 60 61 62 63 64 65
sets.
1 1 1 1 1
Tigers: 8 , 8 , 9 , 9 , 9 , 10, 10, 11, 12
2 2 2 2 2
1
Bobcats: 6, 7, 7, 7, 7 , 8, 8, 8, 9
2
yes; Sample answer: From the double box-andwhisker plot, you know that at least one girl on the
Bobcats has a shoe size of 9. You also know that at
1
least one girl on the Tigers has a shoe size of 8 .
2
Female students
56 57 58 59 60 61 62 63 64 65
Heights (inches)
Male students
56 57 58 59 60 61 62 63 64 65
Female students
56 57 58 59 60 61 62 63 64 65
c. 8:00 P.M. Class: symmetric; 10:00 A.M. Class:
skewed right; no; The oldest person in the 8:00 P.M.
class is 40. The youngest person in the 10:00 A.M.
class is 42.
3. Sample answer: You can compare the measures of center,
the measures of variation, the shapes of the distributions,
and the overlap of the two distributions.
10.7 On Your Own (pp. 450 –451)
1. In part (a), Candidate A’s mean increases by 30 to 114,
but the MAD is unchanged. In part (b), the quotient is
56
= 3.5. So, the difference in the means is now
now
16
3.5 times the MAD. The number is greater, indicating
less overlap in the data.
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Chapter 10
2. No, the sample size is too small to make a conclusion
b. The water snakes have greater measures of center
about the population.
because the mean, median, and mode are greater. The
water snakes also have greater measures of variation
because the interquartile range and mean absolute
deviation are greater.
10.7 Exercises (pp. 452–453)
Vocabulary and Concept Check
1. When comparing two populations, use the mean and the
MAD when each distribution is symmetric. Use the
median and the IQR when either one or both distributions
are skewed.
4. a. Team A: median = 3, IQR = 4 − 2 = 2
Team B: median = 7, IQR = 8 − 6 = 2
The variation in the goals scored is the same, but
Team B usually scores about 4 more goals per game.
2. There will probably be little or no visual overlap of the
data. The core (center) portions of the data are too far
apart.
b.
Practice and Problem Solving
3. a. Garter snake lengths ordered:
15, 18, 21, 22, 24, 24, 25, 26, 28, 30, 32, 35
Measures of center:
5. a. Class A: median = 90,
IQR = Q3 − Q1 = 95 − 82.5 = 12.5
15 + 18 + 21 + 22
+24 + 24 + 25 + 26
300
+28 + 30 + 32 + 35
Mean =
=
= 25 in.
12
12
Class B: median = 80,
IQR = Q3 − Q1 = 85 − 75 = 10
The variation in the test scores is about the same, but
Class A has greater test scores.
Median = 24.5 in.
Mode = 24 in.
Measures of variation:
Range = 35 − 15 = 20 in.
IQR = Q3 − Q1 = 29 − 21.5 = 7.5 in.
10 + 7 + 4 + 3
+1 + 1 + 0 + 1
52
+3 + 5 + 7 + 10
MAD =
=
≈ 4.33 in.
12
12
Water snake lengths ordered:
21, 24, 25, 27, 30, 32, 32, 34, 35, 37, 40, 41
Measures of center:
21 + 24 + 25 + 27
+30 + 32 + 32 + 34
378
+35 + 37 + 40 + 41
Mean =
=
= 31.5 in.
12
12
median for Team B − median for Team A
4
=
= 2
IQR for Team A
2
median for Team B − median for Team A
4
=
= 2
IQR for Team B
2
The difference in the medians is 2 times the IQR.
b.
median for Class A − median for Class B
10
=
= 0.8
IQR for Class A
12.5
median for Class A − median for Class B
10
=
=1
IQR for Class B
10
The difference in the medians is 0.8 to 1 times the
IQR.
6. a. Volleyball Game Attendance:
112 + 75 + 49 + 95 + 88
+54 + 84 + 93 + 85 + 106
+127 + 74 + 62 + 98 + 88
1720
+68 + 117 + 132 + 53 + 60
Mean =
=
= 86
20
20
Mode = 32 in.
26 + 11 + 37 + 9 + 2
+32 + 2 + 7 + 1 + 20
+41 + 12 + 24 + 12 + 2
392
+18 + 31 + 46 + 33 + 26
MAD =
=
= 19.6
20
20
Measures of variation:
Basketball Game Attendance:
Median = 32 in.
Range = 41 − 21 = 20 in.
IQR = Q3 − Q1 = 36 − 26 = 10 in.
10.5 + 7.5 + 6.5 + 4.5
+1.5 + 0.5 + 0.5 + 2.5
61
+3.5 + 5.5 + 8.5 + 9.5
MAD =
=
≈ 5.08 in.
12
12
202 + 176 + 163 + 190 + 141
+186 + 173 + 152 + 184 + 155
+181 + 207 + 169 + 198 + 219
3700
+188 + 214 + 228 + 195 + 179
Mean =
=
= 185
20
20
17 + 9 + 22 + 5 + 44
+1 + 12 + 33 + 1 + 30
+4 + 22 + 16 + 13 + 34
354
+3 + 29 + 43 + 10 + 6
MAD =
=
= 17.7
20
20
The variation in the attendances is about the same, but
basketball has a greater attendance.
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313
Chapter 10
b. The difference in the means is 99. This is about 5.1
14. B;
times the MAD for the volleyball game attendance and
5.6 times the MAD for the basketball attendance.
percent of increase =
125 − 100
100
25
=
100
= 25%
7. The value for Exercise 6(b) is the greatest. This means
=
that the attendances have the least overlap of the data sets
in Exercises 4–6.
8. a. Sports magazine:
9 + 21 + 15 + 14 + 25
+26 + 9 + 19 + 22 + 30
190
Mean =
=
= 19
10
10
10 + 2 + 4 + 5 + 6
58
+7 + 10 + 0 + 3 + 11
MAD =
=
= 5.8
10
10
The number of students in the marching band increased
by 25%.
Quiz 10.6–10.7
1. Sample A is not large enough to make a prediction.
Sample B is large, so Sample B is better for making
a prediction.
Political magazine:
31 + 22 + 17 + 5 + 23
+15 + 10 + 20 + 20 + 17
180
Mean =
=
= 18
10
10
13 + 4 + 1 + 13 + 5
+3 + 8 + 2 + 2 + 1
52
MAD =
=
= 5.2
10
10
The mean and MAD for the sports magazine are close
to the mean and the MAD for the political magazine.
However, the sample size is small and the variability
is too great to conclude that the number of words per
sentence is about the same.
b. The sample means vary much less than the sample
2. biased; The sample is not selected at random and is not
representative of the population because students on the
basketball team use the gymnasium regularly when
practicing.
3. The sample is representative of the population, selected at
random, and large enough to provide accurate data. So,
the sample is unbiased and the conclusion is valid.
4.
students in survey (aquarium)
students in school (aquarium)
=
total surveyed
students in school
16
n
=
60
720
192 = n
numbers of words per sentence.
c. The number of words per sentence is generally greater
in the political magazine than in the sports magazine.
9. a. Check students’ work. Experiments should include
taking many samples of a manageable size from each
grade level. This will be more doable if the work of
sampling is divided amongst the whole class, and the
results are pooled together.
About 192 students would choose the aquarium.
5. a. Team A: median = 14,
IQR = Q3 − Q1 = 22 − 6 = 16
Team B: median = 32,
b. Check students’ work. The data may or may not
IQR = Q3 − Q1 = 36 − 20 = 16
support a conclusion.
The variation in the points is the same, but Team B
has greater scores.
Fair Game Review
10.
11.
new amount − original amount
original amount
b.
2
3
4
5
6
7
8
−6
−5
−4
−3
−2
−1
0
−1
0
1
4
5
6
−1.6
12.
−5
−4
−3
13.
−2
median for Team B − median for Team A
18
=
= 1.125
IQR for Team A
16
median for Team B − median for Team A
18
=
= 1.125
IQR for Team B
16
The difference in the medians is 1.125 times the IQR.
2.5
0
1
2
3
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Chapter 10
6. a. Camp A: mean = 15,
5. a. There are 8 sections with a number greater than 0. So,
spinning a number greater than 0 can occur in 8 ways.
20
MAD =
=1
20
b.
Camp B: mean = 13,
MAD =
20
=1
20
The variation in the ages is the same, but Camp A has
a greater age.
b.
mean for Camp A − mean for Camp B
2
=
= 2
MAD for Camp A
1
mean for Camp A − mean for Camp B
2
=
= 2
MAD for Camp B
1
The difference in the means is 2 times the MAD.
Chapter 10 Review
1. a. There are 2 sections with a 1. So, spinning a 1 can
occur in 2 ways.
b.
greater than 0
not greater than 0
1 green, 1 purple, 2 blue,
2 orange, 2 green, 3 orange,
3 blue, 3 purple
none
The favorable outcomes of the event are spinning
1 green, 1 purple, 2 blue, 2 orange, 2 green, 3 orange,
3 blue, and 3 purple.
6. a. There are 5 sections with a number less than 3. So,
spinning a number less than 3 can occur in 5 ways.
b.
less than 3
not less than 3
1 green, 1 purple, 2 blue,
2 orange, 2 green
3 orange, 3 blue,
3 purple
The favorable outcomes of spinning a number less
than 3 are 1 green, 1 purple, 2 blue, 2 orange, and
2 green.
1
not 1
1 green,
1 purple
2 blue, 2 orange, 2 green,
3 purple, 3 blue, 3 orange
number of even numbers
total number of numbers
3
=
6
1
=
2
7. P(even ) =
The favorable outcomes of the event are spinning
1 green and 1 purple.
2. a. There are 3 sections with a 3. So, spinning a 3 can
occur in 3 ways.
b.
3
not 3
3 purple, 3 blue,
3 orange
1 green, 1 purple, 2 blue,
2 orange, 2 green
The favorable outcomes of the event are spinning
3 purple, 3 blue, and 3 orange.
3. a. There are 5 sections with an odd number. So, spinning
an odd number can occur in 5 ways.
b.
odd
not odd
1 green, 1 purple, 3 orange,
3 blue, 3 purple
2 blue, 2 orange,
2 green
The favorable outcomes of the event are spinning
1 green, 1 purple, 3 orange, 3 blue, and 3 purple.
4. a. There are 3 sections with an even number. So,
spinning an even number can occur in 3 ways.
b.
even
not even
2 blue, 2 orange,
2 green
1 green, 1 purple, 3 orange,
3 blue, 3 purple
The favorable outcomes of the event are spinning
2 blue, 2 orange, and 2 green.
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The experimental probability of rolling an even number is
1
, or 50%.
2
8. P(3) =
number of times 3 spun
16
8
=
=
total number of spins
70
35
The experimental probability of spinning a 3 is
8
, or
35
about 22.9%.
9. The total number of times an odd number was spun is
14 + 16 + 13 = 43.
number of times
an odd number spun
43
P(odd ) =
=
total number of spins
70
The experimental probability of spinning an odd number
43
is
, or about 61.4%.
70
10. The number of times not 5 was spun is
14 + 12 + 16 + 15 = 57.
P( not 5) =
number of times not 5 spun
57
=
total number of spins
70
The experimental probability of not spinning a 5 is
57
,
70
or about 81.4%.
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315
Chapter 10
11. The number of times a number greater than 3 was spun is
15 + 13 = 28.
18. Coin
number of times
greater than 3 spun
28
2
P(greater than 3) =
=
=
total number of spins
70
5
Outcome
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
1
2
3
4
5
6
HH1
HH2
HH3
HH4
HH5
HH6
HT1
HT2
HT3
HT4
HT5
HT6
TH1
TH2
TH3
TH4
TH5
TH6
TT1
TT2
TT3
TT4
TT5
TT6
H
T
number of blue sections
2
1
=
=
total number of sections
8
4
The theoretical probability of spinning blue is
13. P(1) =
Cube
H
The experimental probability of spinning a number
2
greater than 3 is , or 40%.
5
12. P( blue) =
Coin
1
, or 25%.
4
H
T
number of "1" sections
3
=
total number of sections
8
The theoretical probability of spinning a 1 is
T
3
, or
8
37.5%.
14. P(even ) =
number of even sections
5
=
total number of sections
8
P(event ) =
The theoretical probability of spinning an even number
5
is , or 62.5%.
8
15. P( 4) =
P(T, T, even ) =
1
, or
8
19.
P( A and B) = P( A) • P( B)
P( blue and tails) = P( blue) • P( tails)
12.5%.
4 1
•
7 2
2
=
7
=
number of even sections
total number of sections
2
8
=
n
3
n = 12
16. P (even ) =
The probability of choosing a blue tile and flipping tails
2
is , or about 28.6%.
7
There are 12 sections on the spinner.
17. Event 1: Choosing a bracelet (6 possible)
Event 2: Choosing a necklace (15 possible)
6 × 15 = 90
There are 90 ways you can wear one bracelet and one
necklace.
3
1
=
24
8
So, the probability of flipping two tails and rolling an
1
even number is , or 12.5%.
8
number of "4" sections
1
=
total number of sections
8
The theoretical probability of spinning a 4 is
number of favorable outcomes
number of possible outcomes
20.
P( A and B) = P( A) • P( B)
P(G and tails) = P(G ) • P( tails)
1 1
•
7 2
1
=
14
=
The probability of choosing the letter G and flipping tails
1
is , or about 7.1%.
14
316 Big Ideas Math Red Accelerated
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Chapter 10
P( A and B) = P( A) • P( B after A)
21.
P(green and blue) = P(green ) • P( blue after green )
2 4
•
7 6
4
=
21
=
The probability of choosing a green tile and then a blue
4
tile is
, or about 19.0%.
21
P( A and B) = P( A) • P( B after A)
22.
P( red and vowel) = P( red) • P( vowel after red )
1 2
=
•
7 6
1
=
21
The probability of choosing a red tile and then a vowel is
1
, or about 4.8%.
21
23. biased; The sample is not selected at random and is not
representative of the population because students in the
biology club like biology.
24. a. Class A: median = 88,
IQR = Q3 − Q1 = 91 − 85 = 6
b.
yellow
not yellow
yellow, yellow,
yellow, yellow, yellow
red, blue, red,
green, blue
The favorable outcomes of choosing not yellow are
red, blue, red, green, and blue.
3. Event 1: 5 possible SPF’s
Event 2: 3 possible types
5 × 3 = 15
There are 15 possible sunscreens.
4. The total number of rolls is
12 + 18 + 14 + 17 + 16 + 13 = 90.
The total number of times a 1 or a 2 was rolled is
12 + 18 = 30.
number of times a
1 or 2 was rolled
30
1
P(1 or 2) =
=
=
total number of rolls
90
3
The experimental probability of rolling a 1 or a 2 is
5. The total number of times an odd number was rolled is
IQR = Q3 − Q1 = 94 − 85 = 9
number of times odd
number was rolled
42
7
P(odd ) =
=
=
total number of rolls
90
15
median for Class B − median for Class A
3
=
= 0.5
IQR for Class A
6
median for Class B − median for Class A
3
=
≈ 0.3
IQR for Class B
9
The difference in the medians is about 0.3 to 0.5 times
the IQR.
1. a. There is 1 green game piece. So, choosing green can
occur 1 way.
green
not green
green
yellow, red, blue, yellow, red,
yellow, yellow, blue, yellow
The favorable outcome of choosing green is green.
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1
, or
3
about 33.3%.
12 + 14 + 16 = 42.
Chapter 10 Test
b.
choosing not yellow can occur in 5 ways.
Class B: median = 91,
In general Class B has greater scores than Class A.
Class A has less variation than Class B.
b.
2. a. There are 5 game pieces that are not yellow. So,
The experimental probability of rolling an odd number is
7
, or about 46.7%.
15
6. The total number of times a number not 5 was rolled is
12 + 18 + 14 + 17 + 13 = 74.
number of times a number not 5 was rolled
total number of rolls
74
=
90
37
=
45
P( not 5) =
The experimental probability of not rolling a 5 is
37
, or
45
about 82.2%.
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317
Chapter 10
number of even-numbered sections
number of sections
4
=
9
7. P(even ) =
12. a. Show A: median = 45,
IQR = Q3 − Q1 = 50 − 40 = 10
Show B: median = 35,
IQR = Q3 − Q1 = 40 − 25 = 15
4
The probability of spinning an even number is , or
9
about 44.4%.
b.
8. P( A and B ) = P( A) • P( B )
P(1 and 2) =
Show B generally has a younger audience and more
variation in ages than Show A.
1 1
1
• =
9 9
81
The probability of spinning a 1 and then a 2 is
1
,
81
or about 1.2%.
median for Show A − median for Show B
10
=
=1
IQR for Show A
10
median for Show A − median for Show B
10
=
≈ 0.7
IQR for Show B
15
The difference in the medians is about 0.7 to 1 times
the IQR.
Chapter 10 Standards Assessment
1. C; Only D.C. United with 3 votes, the Minnesota Lynx
2
1
=
16
8
1
P(second is bishop) =
15
9. P(first is bishop) =
P( A and B ) = P( A) • P( B after A)
P( both bishop) = P( bishop) • P( bishop after bishop)
1 1
•
8 15
1
=
120
=
The probability of choosing a bishop first and then
1
another bishop is
, or about 0.8%.
120
10. P(first is king) =
1
16
P(second is queen ) =
with 4 votes, the New York Knicks with 5 votes, and
possibly the other 6 votes are non-Florida teams. So, it is
unlikely, but not impossible that this team member’s
favorite professional sports team is not located in Florida.
students that
voted Sunday
6
1
1
=
=
2.
or 0.2; P(Sunday) =
total number
30
5
5
of students
2
1
=
of the way
12
6
around the circle (the circumference).
3. G; In 2 hours, the hour hand travels
The circumference of the circle is
 22 
C = 2π r ≈ 2 (84) = 528 mm.
7
So, the hour hand travels
1
15
P( A and B) = P( A) • P( B after A)
P( king and queen ) = P( king) • P(queen after king)
1
1
•
16 15
1
=
240
=
The probability of choosing a king first and then a queen
1
is
, or about 0.4%.
240
11. biased; The sample size is too small and students
standing in line are more likely to say they prefer to buy
their lunches at school.
318 Big Ideas Math Red Accelerated
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4. C;
1
(528) = 88 mm.
6
16
p
=
40
27
16 • 27 = 40 • p
432 = 40 p
432
40 p
=
40
40
10.8 = p
5. H; The area of one face-off circle is
A = π r 2 ≈ (3.14)(15) = 706.5 ft 2 .
2
So, the total area of the 5 face-off circles is
5(706.5) = 3532.5 ft 2 .
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Chapter 10
6.
1
2
1
, or 0.0625, or 6.25%; P( yellow ) =
=
16
8
4
P( both yellow ) = P( yellow ) • P( yellow )
1 1
•
4 4
1
=
16
=
The probability of spinning two yellows is
1
, or
16
0.0625, or 6.25%.
7. C;
Area of base = 6 • 6 = 36 in.2
Area of lateral face =
1
• 6 • 8 = 24 in.2
2
S = area of base + areas of lateral faces
= 36 + 24 + 24 + 24 + 24
= 132 in.2
8. H;
percent of increase =
new amount − original amount
original amount
15 − 6
6
9
=
6
= 1.5
=
= 150%
9. Part A:
The events are independent because the outcome of the
first roll does not affect the second roll.
Part B:
even
not even
2, 4, 6
1, 3, 5
Each roll has 3 favorable outcomes and 6 possible
outcomes.
Part C:
P( A and B) = P( A) • P( B)
3 3
•
6 6
1 1
=
•
2 2
1
=
4
P( both even ) =
The probability of rolling two even numbers is
1
, or
4
0.25, or 25%.
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319