Chapter 7 • Answers
33
Chapter 7
Probability
Activity 1
Box 1
1. likely
4. no chance
2. some chance
5. some chance
Box 2
Answers will vary. Samples are shown.
1.
2.
3.
R
R
R
R
B
W
W
3. nearly impossible (or some chance)
6. nearly certain
R
4.
R
W
5.
R
R
B
6.
W
B
R
W
B
Activity 2
Suggestions: This activity provides an opportunity for students to play a game and formulate
probability concepts on their own before the concepts are formalized in later activities. Discussion
of the activity should focus on students’ intuitive concepts of chance and fairness, not on a formal
development of these ideas, which will be done in Activities 3 and 4. In Activity 4, students will
calculate experimental and theoretical probabilities for the game and relate them to the idea of a
fair game.
Box 1
Answers will vary. Although the probability that Spinner B will win is 5/9, the difference between
the number of times Spinner B and Spinner A win in 20 trials is likely to be small. Some students
will interpret this as meaning the game is fair (since they win about as often as they lose) while
others will interpret any difference as meaning the game is unfair.
Box 2
Answers will vary. The class totals probably will show that Spinner B wins more often and
therefore the game is not fair.
EXTENSION
Answers will vary. See EXTENSION in Activity 4 for an analysis of the game.
Activity 3
EXPERIMENTAL PROBABILITY
1. Answers will vary.
2. Answers will vary.
3. Since there is no 1 on the spinner, you can never get an outcome of 1.
0
Thus, P(1) =
= 0.
total number of trials
4. Since all of the numbers on the spinner are less than 10, the outcome of every spin is less
total number of trials
than 10. Thus, P(a number less than 10) =
= 1.
total number of trials
(Continued on next page.)
34
Chapter 7 • Probability
Activity 3 Continued
5. a. The event can never occur.
b. The event must occur.
6. To have a probability greater than 1, an event would have to occur more often than the
experiment was conducted, and to have a probability less than 0, it would have to occur a
negative number of times, neither of which is possible.
COMPLEMENTARY EVENTS
1. P(even number) + P(odd number) = 1
2. If A and B are complementary events, then P(A) + P(B) = 1.
MUTUALLY EXCLUSIVE EVENTS
Since Exercises 1 and 3 involve experimental probabilities, the answers will vary. The responses
shown are approximations based on theoretical probabilities.
1. a. P(2 or 4) = 2/3
b. P(2 and 4) = 0
c. P(2) + P(4) = (1/3) + (1/3) = 2/3
d. P(A or B) = P(A) + P(B)
2. If A and B are complementary events, then A ∩ B = φ and A ∪ B = {all possible outcomes}.
If A and B are mutually exclusive events, then A ∩ B = φ but A ∪ B may not be equal to
{all possible outcomes}.
3. a. P(an odd number or a square number) = 2/3
b. P(an odd number) + P(a square number) = (1/3) + (2/3) = 1
c. P(the number is odd and square) = 1/3
d. No, 9 is odd and square.
e. Subtract P(the number is odd and square) from P(an odd number) + P(a square number).
f. P(A or B) = P(A) + P(B) – P(A and B)
Activity 4
Box 1
1. P(2) = 1/3
P(square number) = 2/3
2. Answers will vary.
P(4) = 1/3
P(odd number) = 1/3
P(9) = 1/3
P(even number) = 2/3
Box 2
1. Answers will vary. One possibility is
.
2. Answers will vary.
3. Answers will vary, but the probabilities probably did not come out exactly 1/4 and 3/4.
Probabilities tell what will happen over a large number of trials, so it is likely that more trials
are needed before the theoretical probability of RED approaches 1/4.
(Continued on next page.)
Chapter 7 • Answers
35
Activity 4 Continued
Box 3
1. Answers will vary.
3. a.
A = Spinner
A wins
3
2
B
4
A
9
A
Spinner A
B = Spinner
B wins
2. Answers will vary.
b. Since the outcomes for Spinner B are equally
likely, for each outcome of a spin of Spinner A
Spinner B
the chance of getting any of the three possible
outcomes on Spinner B is the same. But, the
5
7
outcomes for Spinner A are also equally likely,
so over a large number of trials, we would expect
B
B
each outcome on Spinner A to occur about the
same number of times. Thus, each of the 9
B
B
possible combinations of a spin of Spinner A and
a spin of Spinner B represented by cells in the
A
A
matrix are equally likely.
4. P(spinner A wins) = 4/9
P(spinner B wins) = 5/9
5. a. - b. The answers will vary. Because of the larger number of trials, the probabilities based
on the class totals should be close to the theoretical probabilities.
6. The game is not fair since P(spinner A wins) < P(spinner B wins).
EXTENSION
3
5
7
2
B
B
B
4
A
B
B
9
A
A
A
B is more likely
to win than A.
A = A wins
C = C wins
Spinner C
1
6
8
2
A
C
C
4
A
C
C
9
A
A
A
B = B wins
C = C wins
Spinner B
Spinner B
Spinner A
Spinner A
A = A wins
B = B wins
A is more likely
to win than C.
Spinner C
1
6
8
3
B
C
C
5
B
C
C
7
B
B
C
C is more likely
to win than B.
Letting the first
player choose a spinner does not make this a fair game. In fact, it gives the second player an
advantage. The second player’s strategy is:
If the first player chooses Spinner A, choose Spinner B.
If the first player chooses Spinner B, choose Spinner C.
If the first player chooses Spinner C, choose Spinner A.
In any of the three cases, the probability that the second player wins is 5/9, so it is not a fair game.
36
Chapter 7 • Probability
Activity 5
1. a. 5/8 b.
2/8 or 1/4
2. a. 0
1/8
1/4
3/8
d
c
4.
6.
7.
d. 0
5/8
a
e
0.25
0
3.
b
c. 1/8
0.5
e. 3/8
f. 7/8
7/8
f
0.75
1
b. a - likely
b - some chance
c - nearly impossible (or some chance)
d - no chance
e - some chance
f - nearly certain
4 blue squares; For the probability of drawing a blue square to equal 0.5, half the squares in
the bag must be blue. Since there are 2 blue squares and 6 non-blue squares, you would have
to add 4 blue squares to the bag.
Results will vary.
5. Answers will vary.
Answers will vary, but the probabilities should be close to those in Exercise 1.
a. - c. Answers will vary.
d. To find the number of each color square in the bag, multiply the experimental probability
of drawing that color by 9 (the number of squares in the bag) and round the product to the
nearest whole number.
Activity 6
EXPERIMENTAL PROBABILITIES USING A MATRIX
Results of the experiment and answers to questions will vary.
MORE EXPERIMENTAL PROBABILITIES
1. Answers will vary.
2. Ideally the answers to Exercise 1 Parts (c) and (d) will be about equal. However, since
experimental probabilities are being used, this may not be the case.
3. Ideally the two results will be about equal.
THEORETICAL PROBABILITIES USING A MATRIX
1.
Player B
Paper
Scissors
Rock
Player A
A = A wins
B = B wins
C = C wins
Paper
T
B
A
Scissors
A
T
B
Rock
B
A
T
2. P(A wins) = 3/9 = 1/3
P(B wins) = 3/9 = 1/3
P(Tie) = 3/9 = 1/3
3. The game is fair since the probability that a player wins is the same as the probability that
they lose.
(Continued on next page.)
Chapter 7 • Answers
37
Activity 6 Continued
4. a. 3/9 = 1/3
b. 3/9 = 1/3
c. 1/9
d. 1/9
5. They are equal.
6. Answers will vary, but ideally the experimental and theoretical results will be close.
PROBABILITIES USING A TREE DIAGRAM
1. Answers will vary.
Player
A
1/3
1/3
P
1/3
1/3
1/3
1/3
1/3
S
1/3
1/3
1/3
1/3
R
1/3
Player
B
P
Outcome
PP
S
PS
R
P
PR
SP
S
SS
R
P
SR
RP
S
RS
R
RR
2. The outcome PP means that both players showed paper.
3. P(PS) = (1/3)(1/3)
4. P(SP or SR) = P(SP) + P(SR)
= 1/9
= 1/9 + 1/9
= 2/9
3. P(A wins) = P(SP or PR or RS) 4. P(B wins) = P(SR or PS or RP)
= 1/9 + 1/9 + 1/9
= 1/9 + 1/9 + 1/9
= 1/3
= 1/3
5. P(at least one player shows scissors)= P(SS or SP or SR or PS or RS)
= 1/9 + 1/9 + 1/9 + 1/9 + 1/9
= 5/9
Activity 7
Since all of the results are based on experimental data, the answers will vary.
38
Chapter 7 • Probability
Activity 8
Box 1
1. Yes Since the probability that any one of the children is a boy is the same as the probability
it is a girl, the outcomes are equally likely.
2. 1 + 3 + 3 + 1 = 8 outcomes
3. P(0 girls) = 1/8
P(1 girl) = 3/8
P(2 girls) = 3/8
P(3 girls) = 1/8
Box 2
1.
1st
Child
2nd
Child
3rd
Child
G
G
B
G
G
B
B
G
G
B
B
G
Key: G = girl
B = boy
B
B
4th
Child
G
B
G
B
G
B
G
B
G
B
G
B
G
B
G
B
Outcome
GGGG
GGGB
GGBG
GGBB
GBGG
GBGB
GBBG
GBBB
BGGG
BGGB
BGBG
BGBB
BBGG
BBGB
BBBG
BBBB
Number
of Girls
4
3
3
2
3
2
2
1
3
2
2
1
2
1
1
0
0 girls - 1
1 girl - 4
2 girls - 6
3 girls - 4
4 girls - 1
2. a. 1
b. The numbers at the ends of each row are 1. The other numbers in a row equal the sum of
the two numbers above them in the preceding row.
c. 1 6 15 20 15 6 1
3. P(0 girls) = 1/32
P(1 girl) = 5/32
P(2 girls) = 10/32
P(3 girls) = 10/32
P(4 girls) = 5/32
P(5 girls) = 1/32
4. a. 20/64 or 5/16
b. 42/64 or 21/32
c. 6/64 or 3/32d.
42/64 or 21/32
5. P(0 girls) = 1/512
P(1 girl) = 9/512
P(2 girls) = 36/512 P(3 girls) = 84/512 P(4 girls) = 126/512
P(5 girls) = 126/512 P(6 girls) = 84/512 P(7 girls) = 36/512 P(8 girls) = 9/512 P(9 girls) = 1/512
Activity 9
Box 1
Answers will vary. The probability that Robin will win will probably be close to 5/9.
Box 2
1. Roll the die once. An outcome of 1, 2, 3, or 4 represents the number of the card that was in
the box. An outcome of 5 or 6 would be ignored, and the die would be rolled again.
2. A trial consists of rolling a die until each of the numbers 1 through 4 has been obtained.
3. Results will vary.
4. Answers will vary.
5. Let the numbers 1, 2, and 3 represent the cards 1 through 3 respectively and the outcomes 4
and 5 represent card 4. Ignore outcomes of 6.
6. A trial would consist of rolling a die until a 1, 2, and 3 and either a 4 or 5 have been obtained.
(Continued on next page.)
Chapter 7 • Answers
39
Activity 9 Continued
EXTENSION
Answers will vary. The following response is based on the assumption that the probability Robin
wins a contest is 5/9.
Contest 1
5/9
Contest 2
R
4/9
M
R
5/9
4/9
5/9
R
4/9
M
M
Contest 3
5/9
R
Winner
R
R
Probability
4/9
5/9
M
R
M
R
80/729
4/9
M
M
M
80/729
16/81
P(Marian wins) = 80/729 + 80/729 + 16/81 = 304/729
Activity 10
Box 2
1. Answers will vary.
2. Answers will vary.
3. Answers will vary. One possibility is to find the experimental probability of each of the
outcomes 1 through 6. If enough trials have been performed, the experimental probability
represents the fraction of the six faces you would expect to be labeled with that number.
Box 3
 58
2 green faces 
× 8 ≈ 2
240
 155
 27
1 blue face 
× 8 ≈ 1 5 yellow faces 
× 8 ≈ 5
240
240
EXTENSION
1. Answers will vary.
2. Answers will vary. You would expect the results for 50 catches to be the most accurate since, in
general, the more trials that are performed the closer the experimental probability approaches the
theoretical probability.
3. The area of the Earth is approximately 196,938,800 mi2 and the surface area of the water is
greater than 138,236,600 mi2 . Thus, about 70% of the earth’s surface is covered with water.
Activity 11
Box 1
1. a.
Number of Squares
Number of Arrangements
1
1
2
2
3
6
4
24
rbg brg grb
rgb bgr gbr
b. Answers will vary. One possibility is to make an organized list by first finding all the
possible arrangements that have a red square first, then all of them that have a blue
square first, and finally, all the arrangements that have a green square first.
c. The number of arrangements of three squares is 3 times the number of arrangements of
two squares.
(Continued on next page.)
40
Chapter 7 • Probability
Activity 11 Continued
2. a. Answers will vary. For each color square, since there are 6 ways to arrange the other
three squares, there are 6 arrangements that have that color in the first position. Since
there are 4 different colors, there are 4 × 6 = 24 ways to arrange the squares.
b. yrbg yrgb ybrg ybgr ygrb ygbr
rybg rygb rbyg rbgy rgyb rgby
byrg bygr bryg brgy bgyr bgry
gyrb gybr gryb grby gbyr gbry
c. The number of arrangements of four squares is 6 times the number of arrangements of
three squares.
3. Using the Fundamental Counting Principle, there are n choices for the first square, n – 1 choices
for the second square, n – 2 for the third square, and so on until there is just 1 square remaining
for the last square, so there are n! = n(n – 1)(n – 2)(n – 3) … (3)(2)(1) arrangements altogether.
ARRANGEMENTS WITH LIKE SQUARES
1. a. 3 ways
rrg
rgr
grr
b. 6
c. 1/2 The denominator of the fraction equals the number of red squares. (Actually, it is
2!, the number of red squares factorial.
2. a. 4 ways
rrrg
rrgr
rgrr
grrr
24
1/6 The denominator of the fraction equals 3!, the number of red squares factorial.
b. 5 ways
There are 5! ways to arrange five different color squares. But because the red squares
can’t be told apart, this counts each arrangement of them 4! times. Thus, the number of
distinct arrangements of the 5 squares should be 1/4! or 1/24 of the number of
1
5!
arrangements of 5 different color squares. And, × 5!=
= 5.
4!
4!
c. 5 arrangements: rrrrg
rrrgr
rrgrr
rgrrr
grrrr
3. a. 12 arrangements: rrbg
rrgb
rbrg
rgrb
rbgr
rgbr
brrg
brgr
bgrr
grrb
grbr
gbrr
b. 6 arrangements: rrgg
rgrg
rggr
ggrr
grgr
grrg
c. 10 arrangements: rrrgg
rrgrg
rrggr
rgrrg
rgrgr
rggrr
grrrg
grrgr
grgrr
ggrrr
1
1
1
1
1
1
=
(3b) =
(3c)
=
2 2! × 1! × 1!
4 2! × 2!
12 3! × 2!
b. The denominator of each fraction is the product of the factorials of the number of each
color square.
5. If there are n squares, divide n!, the total number of arrangements of n different squares, by
the factorial of the number of each color square.
4. a. (3a)
(Continued on next page.)
Chapter 7 • Answers
41
Activity 11 Continued
1. a.
R
B
RB
B
G
RG
Y
RY
R
BR
G
BG
G
Y
BY
R
GR
B
GB
Y
Y
GY
R
YR
B
YB
G
YG
b. There are 4 ways to choose the first square and 3 ways to choose the second one, so there
are 4 × 3 = 12 permutations.
c. 4 × 3 × 2 = 24 permutations
d.
G
RBG
Y
B
RBY
RGB
Y
B
RGY
RYB
G
G
RYG
BRG
Y
R
BRY
BGR
Y
R
BGY
BYR
G
B
BYG
GRB
Y
R
GRY
GBR
Y
R
GBY
GYR
B
B
GYB
YRB
G
R
YRG
YBR
G
R
YBG
YGR
B
YGB
B
R
G
Y
R
B
G
Y
R
G
B
Y
R
Y
B
G
2. a. You still end up with the same pair of squares regardless of whether the red square is
chosen first and then the blue square (RB) or whether the blue square is chosen first and
then the red square (RB).
b. Changing the order does not change the pair.
(Continued on next page.)
42
Chapter 7 • Probability
Activity 11 Continued
3. a. 6 combinations
b. 1/2
c. The denominator equals the number of squares in a pair. (Actually, it is the factorial of
the number of squares in a pair.)
4. a. See the tree diagram above.
b. 4 combinations
c. 1/6
d. The denominator is the factorial of the number of squares in a group. The numerator is 1.
5. Divide the number of permutations that can be formed by choosing m of the n items by m!.
n(n − 1)(n − 2)L(n − m + 2)(n − m + 1)
n!