Download LeVVon 11.1 Fundamental Theorem of Counting

LESSON OBJECTIVES
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11.1 Fundamental Theorem of
Counting
• Use the Fundamental
Theorem of Counting to
find probabilities.
11.2 Compound Events
• Find the probability of
compound events.
11.3 Collecting Data
• Analyze methods of
gathering information.
• Design an appropriate
study to gather
information.
11.4 Probability Models
• Use probability to model
random events.
• Use expected value to
make decisions.
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11.5 The Normal Distribution
• Use the Empirical Rule to
find probabilities.
• Find z-scores and use
the standard normal
distribution.
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Mental Math
Skills Review
Multiply.
Simplify each expression.
1. 5 + 11 8
18 18 9
2. 1 + 1 − 1 7
4 2 20 10
3
3. • 8 6
4 25 25
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1
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4.
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4.5 • 8 • 3 • 9 1,080
506 Chapter 11 Probability and Statistics
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Look to Your Future
Actuaries collect and scrutinize data to approximate the probability of an
event. Actuaries work for the insurance industry where probabilities of events,
such as illness, disability, or loss of property are calculated as a risk. The
probability or risk of such an event determines what policies the company can
issue and at what cost in order to remain profitable. PLANNING
THE CHAPTER
Math Labs, pp. 538–541
Data Sheet (Lab Data Sheets)
Math Applications, pp. 542–549
Chapter Review, pp. 550–551
Chapter Test, p. 552
Software Generated Assessment
Standardized Test Practice, p. 553
Grid Response Form (CRB)
Chapter Resource Book (CRB)
Reteaching, pp. 393, 397,
403, 407
Extra Practice, pp. 395, 399,
405, 409
Enrichment, pp. 401, 411
Standardized Test Response
Form, pp. 413, 414
Standardized Test Answers,
p. 415
Classroom/Journal
Topics
What’s Ahead?
In this chapter, students will
learn how to calculate the
probability of compound events.
The Fundamental Theorem of
Counting is used in many real-life
situations to find probabilities.
Students should become familiar
with probability models and the
normal distribution.
Chapter 11 Probability and Statistics
507
LESSON PLANNING
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Vocabulary
Fundamental Theorem of
Counting
independent events
dependent events
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Extra Resources
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Extra Practice 11.1
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Assignment
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Homework: 6–26
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Remind students that tree
diagrams and organized lists
are good strategies for problem
solving. Ask students to make
either a tree diagram or an
organized list to represent all of
the possible orders for Example 1.
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INSTRUCTION
Use the students’ tree
diagrams and organized lists
to help students develop
the Fundamental Counting
Principal. After the students have
expressed the mechanics of the
Fundamental Counting Principal
in their own words, present the
formal notation.
Example 1 Extend the
problem by modifying it
to include three choices of
beverages. Emphasize to students
that they can use the answer to
the original problem to obtain the
answer to the modified problem.
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R.E.A.C.T. Strategy
Cooperating
Instruct students to write a Fundamental Counting Principal problem. Select a
partner for each student. Students should solve their partner’s problem. Then
have students write a second problem that is a modification of the problem
they just solved (their partner’s problem). After the second problems are
written, ask students to exchange papers and solve the problem written by
their partner.
508 Chapter 11 Probability and Statistics
INSTRUCTION
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Ask students to provide examples
of other events which would
be considered independent. In
contrast, ask students to provide
examples of events which would
not be considered independent.
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Tell students that experimental
probability is the surveyed
chance of an event occurring,
while theoretical probability is
the mathematical or calculated
chance of an event occurring.
Point out that there is no way to
calculate a theoretical probability
concerning free throw attempts.
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Chalkboard Examples
Ask students to name two events A and B, such that:
1. P(A and B) = 1 sample answer: Event A: a card is pulled from a
standard deck and the card is either red or black; Event B: a standard
six-sided die is rolled and a number less than 7 lands face up
2. P(A and B) = 0 sample answer: Event A: a card is pulled from a
standard deck and the card is either red or black; Event B: a standard
six-sided die is rolled and a number greater than 6 lands face up
3. P(A and B) = 0.25 sample answer: Event A: a card is pulled from a
standard deck and the card is red; Event B: a standard six-sided die is
rolled and an even number lands face up
11.1 Fundamental Theorem of Counting 509
INSTRUCTION
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of other events which would
be considered dependent.
To reinforce vocabulary
comprehension, students should
justify why the events are not
independent.
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the use of the Fundamental
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P(both spades) = (number of
ways two spades can be chosen
from a deck) divided by (number
of ways two cards can be chosen
from a deck).
•
P(both spades) = 13 12 = 1
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Reteaching 11.1 (CRB)
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Diversity in the Classroom
Kinesthetic Learner
Divide students into groups to play the card game in Example 3. Each group
should play the game 100 times and record the results of each game. Each
group should use these results to calculate the experimental probability that
Alan will win the game. Finally, pool the results of each group in the class in
order to calculate the experimental probability that Alan will win the game.
Ask students to explain why the pooled results are closer to the theoretical
probability than the results of each individual group.
510 Chapter 11 Probability and Statistics
Problem Solving
Have students use pencil and
paper or Geometer’s Sketchpad
to draw a scale picture of the
dartboard.
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problem would change if the
3 -inch circle were not in the
4
center of the dartboard.
Ask students to calculate the
probability of not hitting the
bull’s-eye of the dartboard.
Emphasize that there are two
different ways of performing this
calculation.
Understand the Problem
You are given the size of a
dartboard and its bull’s-eye. You
want to find the probability that
a randomly thrown dart will land
on the bull’s-eye.
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Develop a Plan
Divide the area of the bull’s-eye
by the area of the dartboard.
Carry Out the Plan
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•
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= 0.0025 = 0.25%
Check the Results
Check: the answer seems
reasonable. The dartboard is a
few hundred times the size of the
bull’s-eye.
1. Answers will vary. Sample answer: if one event can occur m ways and
another event can occur n ways, then both events can occur in m × n
ways.
2. the outcome of one event affects the outcome of the other event
3. the outcomes of the events have no effect on each other
4. multiply the probabilities
5. multiply the probability of A by the probability of B given that A has
already occurred
11.1 Fundamental Theorem of Counting 511
WRAP UP
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To ensure mastery of objectives,
students should be able to:
• Use the Fundamental Counting
Principal to find the number of
outcomes in each situation.
• Use the Fundamental
Counting Principal to calculate
probabilities.
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Assignment
In-class practice: 1–5
Homework: 6–26
Math Applications
Exercises 4, 6, and 9 from pages
542–549
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R.E.A.C.T. Strategy
Relating
Ask students to design a 15-inch dartboard such that the probability of hitting
the bull’s-eye would be equal to the probability of not hitting the bull’s-eye.
Encourage students to write an equation as part of the solution process rather
than using guess and check.
512 Chapter 11 Probability and Statistics
Mixed Review
Additional Answers
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Answers to Math
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9. a.(0.28)3 = 2.2%
b.1 – 0.28 = 0.72
c.0.28 • 0.28 • 0.72 = 5.6%
d.0.72 • 0.28 • 0.72 = 14.5%
Sample answer: the graphs
are the same because the sine
and cosine functions are π
2
radians out of phase.
Math Applications for this
chapter are on pages 542–549.
The notes and solutions shown
below accompany the suggested
applications to assign with this
lesson.
4. a. 725 = 29
3, 000 120
b. 725 − 1 = 724
3, 000 − 1 2, 999
c. 29 • 724 ≈ 0.058
120 2, 999
≈ 5.8%
6. a.Because each click does
not depend on any
other event, they are
independent events.
b.Because each click
is independent, the
probability remains the
same: 0.14.
c.Because each click
is independent, the
probability remains the
same: 0.14.
d.0.14 • 0.14 • 0.14 • 0.14
≈ 0.0004 ≈ 0.04%
11.1 Fundamental Theorem of Counting 513
LESSON PLANNING
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Vocabulary
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compound event
mutually exclusive
inclusive events
conditional probability
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Extra Practice 11.2
Enrichment 11.2
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START UP
6ROXWLRQ
Demonstrate the concepts of
mutual exclusion and inclusion in
the following way. First, have all
female students sit on the left side
of the room and all male students
sit on the right side of the room.
Second, have all students sit on
the left side of the room and all
male students sit on the right side
of the room.
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INSTRUCTION
Example 1 Ask students to
calculate the probability that the
fruit picked is not red or orange
once they know the answer to the
original problem.
Emphasize that P(A and B) = 0
when A and B are mutually
exclusive events by asking
students to calculate the
probability P(apple and orange).
Diversity in the Classroom
English Language Learner
Some students will struggle with new vocabulary in this lesson. Students
probably computed problems such as Example 1 in previous courses without
using the formula. Remind them that Example 1 could have been calculated in
the following way.
P(orange or red) = number of winners
number of possible
P(oraange or red) = 5
12
514 Chapter 11 Probability and Statistics
INSTRUCTION
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Example 2 Show students a
Venn diagram which represents
the socks in Sigmund’s drawer.
This diagram will provide students
of a visual representation of why
P(A and B) must be subtracted.
Ask students to explain why the
selection of the socks from the
drawer might not be a completely
random event.
Tell students that another term
used to describe inclusive events is
“non-mutually exclusive” events.
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Enriching the Lesson
Ask students to justify why the formula for calculating
the P(A or B) when A and B are inclusive events can also
be used when calculating the P(A or B) when A and B are
mutually exclusive events. Tell students to use examples in
their justification.
11.2 Compound Events
515
INSTRUCTION
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Explain to students that
conditional probabilities are those
in which the sample space for
an event is restricted. Remind
students that a sample space is
the set of all possible outcomes of
a probability event.
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Tell students to think of
conditional probability in terms
of deciding what kind of car you
buy. The model of car is going to
depend on the make of car that
is available at the car dealership
where you are shopping.
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Help students to understand conditional probabilities better by constructing
tree diagrams. Each first branch in the tree diagram will represent a simple
probability. Each second branch in the tree diagram will represent a conditional
probability.
516 Chapter 11 Probability and Statistics
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Discuss with students other
inherited traits and whether these
traits are dominant or recessive.
Use the Internet for research
purposes.
Calculate the experimental
probability of a student in your
class being left handed. Compare
this experimental probability
to the theoretical probability
calculated in the activity.
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Think and Discuss Answers
1. A compound event involves two or more events. Answers will vary. Sample
answer: rolling a number cube and flipping a coin.
2. The events cannot occur at the same time.
3. The events can occur at the same time.
4. P(A or B) = P(A) + P(B) – P(A and B)
5. Answers will vary. Sample answer: the probability that one event will occur
given that another event has already occurred.
11.2 Compound Events
517
WRAP UP
To ensure mastery of objectives,
students should be able to:
• Determine the probability of
compound events which are
mutually exclusive.
• Determine the probability of
compound events which are
inclusive.
• Determine the probability of
compound events which are
conditional.
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Exercises 1, 3, 7, 10, and 12 from
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
NAME
CLASS
DATE
RETEACHING
14.2 COMPOU
A compound even
ND EVENTS
choosing two peo t involves two or more even
ts such as tossing
ple to serve on
time, the events
a
three number cub
are mutually excl committee. When two events
cannot occur at es or
usive.
If A and B are
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mut
given by the form ually exclusive events, then
the
events, then P(A ula: P(A or B) = P(A) + P(B). probability that A or B will occu
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and B) = 0 sinc
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uall
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Events that are
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not
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happen at the sam mutually exclusive are inclusiv
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e events. These
are events that
If A and B are
can
incl
the formula: P(A usive events, then the probabil
ity that A or B
or B) = P(A) +
will occur is give
P(B) – P(A and
A conditional
B).
n by
pro
event has already bability is the probability that
one event will
occurred.
occur given that
The conditional
another
probability of even
t A, given even
t B has already
the formula: P(A
occurred, is give
|B) = P( A and B)
n by
, where P(B) ≠
P( B)
0.
EXAMPLE 1
Suppose Jamal
rolls a standard
an even number
number cube. Find
or a 6.
the probability
SOLUTION
of rolling
The events are
inclusive since
rolling a 6 is an
even number.
P(rolling an even
3
number) = = 1
;
P(rolling a 6) = 1
6 2
6
P(rolling an even
number and a 6) 1
=
6
P(rolling an even
number or a 6) 1 1 1
1
= + –
2 6 6 =
EXERCISES
2
Two number cub
es
mutually exclusiv are tossed. State whether or
not the two even
1. the sum rolle e.
ts are
d is odd; the num
bers are both odd
mutually excl
usive
is less than 7; both
numbers are less
than 4
not mutually excl
is even; the num
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not mutually excl
4. the sum rolle
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CLASS
NAME
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(A or B).
events. Find P P(A) = 35%, P(B) = 42%
ually exclusive
2.
A and B are mut ) = 0.15
0.7
, P(B
1. P(A) = 0.32
1 (B) = 1
,P
6
3. P(A) = 2
7
in a
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board game. Stat e. Then find P(A or B).
mutually exclusiv
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5. A: a number
6
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1
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11.2 Compound Events
519
Answers to Math
Applications
Math Applications for this
chapter are on pages 542–549.
The notes and solutions shown
below accompany the suggested
applications to assign with this
lesson.
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1. a.The options are inclusive
because he could have a
red, long-sleeved shirt.
b.There are 4 red longsleeved shirts and 23
possible shirts, so the
probability is 4 .
23
c.There are 11 long-sleeved
shirts and 5 other red
shirts, so the probability is
16 .
23
d.There are 23 shirts,
16 pants, and 25 shoes,
so there are 23 • 16 • 25
= 9,200 different ways
Marcellus can dress his
avatar.
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3. a. 1−
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c. 274 ≈ 0.010 ≈ 1.0%
2, 674
7. a.The events are inclusive,
so P(type B or positive) =
P(type B) + P(positive) –
P(type B and positive)
= 22 + 23 − 20 = 25
38 38 38 38
b.P(type A or negative) =
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10. a.P(no seatbelt or no license) =
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0.058
b.P(no seatbelt and no license)
= 0.003 • 2,564 ≈ 8
520 Chapter 11 Probability and Statistics
12. a.44 + 25 + 27 + 31
= 127; P = 127
235
b. 27
235
c. 27
127
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Critical Thinking
Answer
Sample answer: Those
participants who know that they
are receiving the treatment might
be inclined to imagine that their
symptoms are improving because
they expect them to improve.
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Ongoing Assessment
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sample is randomly selected,
there should not be any bias
because participants do not know
if they are in the treatment group
or control group.
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1. Sample answer: Correlation;
While studying more would
likely improve grades, there
could be other factors
involved. For example, those
students might have gotten
better sleep the night before,
have been better test takers,
or have been more relaxed.
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2. Sample answer: Correlation;
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help people lose weight,
there could be other factors
involved. For example, those
in the program might have
been eating healthier.
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Answers
2. Answers will vary. Sample
answer: Are you fed up
with the incompetence of
our country’s leadership?
(biased); How many times per
day do you brush your teeth?
(unbiased)
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3. Sample answer: Causation;
There are no other possible
alternatives that could cause
the lake water to freeze other
than the water temperature
reaching the freezing point.
1. Answers will vary. Sample
answer: See if there is a
relationship between the
number of years of service of
teachers and their students’
scores on standardized tests.
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half of the participants and give a placebo to the other half so that they
do not know which group they are in. Compare the results of the two
groups to see if the treatment had an effect.
4. Sample answer: A random sample is a sample of members of a population
that is representative of the entire population. No part of the population is
more represented than any other part.
5. Sample answer: Correlation does not imply causation. Two events can
show a relationship without one event having caused the other. Causation
does imply correlation because if one event causes another event, the
events are automatically related to each other.
524 Chapter 11 Probability and Statistics
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all flowering plants, and the sample is the flowering plants to be studied.
The company wants to know if using their plant food has an appreciable
effect on the growth of flowering plants. (b) An experiment would be
most appropriate because the company will treat some of the plants with
their food. (c) The company can take a sample of 100 flowering plants
and divide them into 2 groups. Treat one half of the plants with the new
plant food and treat the other half of the plants with a regular plant food.
Give each plant the same amount of water and sunlight so that growing
conditions are identical. Measure the growth of the plants in each group
after several weeks and compare the results. The experiment should be
unbiased since a plant could be assigned to either group and will receive
the exact same growing conditions.
Practice and Problem
Solving Answers
6. Observational study; no
treatment applied to either
group of parents; no attempt
to influence the results.
7. Experiment; teacher is
applying a treatment (playing
music during test taking) for
one group but not the other.
8. Survey; a question is being
asked to gather information.
9. Unbiased; does not favor a
particular response.
10. Biased; the word thrilling
could sway the respondent to
favor action movies.
11. Likely biased; employees of a
certain make of car might be
inclined to say that it is their
favorite make.
12. Likely unbiased; each city
resident has an equal chance
of being mailed the survey.
13. Answers will vary. Sample
answer: (a) The population
is all students at Karen’s
school, and the sample is
the students whose opinions
Karen will gather. She wants
to determine whether the
students favor a school dress
code. (b) A survey would be
most appropriate because
Karen wants to gather
opinions on an issue. (c)
Karen can select a random
sample of 50 students and
ask them, “Do you support or
oppose having a school dress
code?” The survey results
should be unbiased because
the sample is representative
of the population and the
survey question is unbiased.
14. Answers will vary. Sample
answer: (a) The population is
11.3 Collecting Data
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and have the students whose
birthdays are closest to those
dates come to the board.
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Answer
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prediction if more trials are
conducted.
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Think and Discuss
Answers
1. Answers will vary. Sample
answer: Random numbers
can be used to represent each
outcome of an event with no
bias toward any particular
outcome. For example, you
can use random numbers to
decide who has to wash the
dishes after dinner.
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any particular outcome. Each
outcome is equally likely to
occur.
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answer: A simulation can
allow you to see what is likely
to happen in an experiment
without actually conducting
the experiment. It may be
costly or impractical to
conduct the experiment. The
results of the simulation can
give you insight into what
will likely happen. You can
use this information to make
predictions about the event.
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the same as the actual value of an event, but it is possible that they are the
same. If the same experiment is conducted many times, the average results
will get closer and closer to the expected value.
5. Answers will vary. Sample answer: Calculating expected value allows you
to predict what is likely to happen in an event based on probability. You
can use this information to decide what to do in situations.
530 Chapter 11 Probability and Statistics
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13. Sample answer: No, the company should not invest in the skate park
because the expected value of the investment is a loss of $900.
14. Sample answer: The expected value of developing Product A is $115,000,
and the expected value of developing Product B is $88,000. The owner
should invest in Product A because it has a higher expected value of
earnings.
Practice and Problem
Solving Answers
6. Answers will vary. Sample
answer: Roll a number cube.
Let 1 or 2 represent Terrell, 3
or 4 represent Carmen, and 5
or 6 represent Alicia.
7. Answers will vary. Sample
answer: Assign a number
from 1 to 12 to each
salesperson and use a random
number generator to generate
2 integers from 1 to 12.
8. Answers will vary. Sample
answer: Assign a number
from 1 to 32 to each player
and use a random number
generator to generate 3
integers from 1 to 32.
9. Answers will vary. Sample
answer: Create a spinner with
4 equal spaces that represent
each special offer. Spin the
spinner until each special
offer has been spun at least
once, and stop as soon as
all 4 special offers have been
spun. Keep track of the results
in a frequency table. Conduct
several trials of the simulation
and find the average number
of purchases needed to get all
4 special offers.
10. Answers will vary. Sample
answer: Use a random
number generator to
generate integers from 1
through 100. Let 1 through
35 represent an ace and let
36 through 100 represent
a non-ace. Generate
numbers until there are
3 aces in a row and count
how many “serves” it took.
Conduct several trials of the
experiment and find the
average number of serves
needed to get 3 aces.
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11.5 The Normal Distribution
535
Think and Discuss
Answers
1. Sample answer: The area
under the normal curve
represents the probability
that a random variable falls
in the range of values on the
x-axis under the curve.
2. Sample answer: If a z-score
is negative, the x-value is
less than the mean. If it is
positive, the x-value is greater
than the mean.
3. Subtract the mean from the
x-value and divide the result
by the standard deviation.
4. Sample answer: The z-score
tells you how many standard
deviations above or below
the mean the x-value lies. For
example a z-score of –2.2
tells you that the x-value
lies 2.2 standard deviations
below the mean.
5. Sample answer: You can
use standard normal tables,
calculators, and spreadsheet
programs to find probabilities
for the standard normal
curve.
Practice and Problem
Solving Answers
6.81.5%
7.95%
8.49.85%
9.15.85%
10.97.35%
11.–2.5
12.1.125
13.1.8
14.–3.1
536 Chapter 11 Probability and Statistics
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Practice and Problem
Solving Answers
15. about 85.4%
16. about 60.1%
17. about 65.5%
18. Brand B; The z-scores for
each box are 1.45 for Brand
A and –1.8 for Brand B. The
Brand A box is 1.45 standard
deviations above the mean
and the Brand B box is
1.8 standard deviations
below the mean.
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11.5 The Normal Distribution
537
MATH LAB
Activity 1
0DWK/DEV
PREPARE
• Remind students of the
meaning of a simulation.
Explain that this lab is a
simulation for how influenza
may spread throughout your
school or community.
• Discuss with students that in a
real situation of an infectious
disease, even though contact
might be made with an infected
person that is not a guarantee
of getting the infection.
TEACH
• This lab is a whole class activity.
• In a typical classroom setting,
the teacher will be activity
administrator and each
student will receive one
numbered note card.
• Keep the “infected” number
secret, as well as the identity
of the students that receives it.
• If the class is too small for this
simulation as it is described,
give each student two
numbered note cards. Instruct
the students to mentally assign
one of the numbers to the right
hand and the other number
to the left hand. When two
people come in immediate
contact, before rolling the
dice, have each person in the
pair choose right or left for the
other person. Each person then
identifies the ID number that
has been associated with the
right or left hand.
• The activity administrator
should be sure to change the
infected ID number and the
condition for being contagious
with each round. These
changes are necessary to get
the simulation as random as
possible.
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FOLLOW-UP
• Discuss the conditions used for contagious and list all possible rolls that
satisfied the condition. Determine if the condition chosen resulted in a
greater probability of being infected. Using all possible rolls, describe
a condition where very few people would likely become infected and a
condition where a lot of people would likely become infected.
Math Lab Solutions and Notes
2. The probability will vary depending on the number of students in the
simulation; 1 where n is the number of students in the simulation.
n
538 Chapter 11 Probability and Statistics
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0DWK/DEV
3 infected (from Round 1) encounter no previously infected. 3 × 5 = 15
possible newly-infected (7 or 8 expected to be newly-infected)
2 of 3 infected (from Round 1) encounter each other, and third infected
encounters no previously infected. 5 + 4 + 4 = 13 potential newly-infected (6
or 7 expected to be newly-infected)
1 of 3 infected (from Round 1) encounters both other infected people and
other 2 infected people do not encounter each other. 3 + 4 + 4 = 11
potential newly-infected people (5 or 6 expected to be newly-infected)
3 infected (from Round 1) each encounter other 2 infected. 3 + 3 + 3 = 9
potential newly-infected (4 or 5 expected to be newly-infected)
Math Lab Solutions
and Notes
6. In Round 1, each person
encounters 5, so infected
person has possibility of in­
fect­ing 5. Using theoretical
probability 1 , 5 • 1 = 2.5,
2
2
2 or 3 new people are
expected to be infected.
Expected number infected
after Round 1 is 1 + (2 or 3)
= 3 or 4.
1
7. ; 36 different outcomes
2
when two number cubes
rolled. Of those, 18 rolls have
sum of 7, 8, 9, or 10.
Table at bottom of following
page shows possible
dice combinations and
probabilities used. You
can alter the lab by using
different probabilities, such
as 1 or 1 . An example of
3
4
each is given; use the table to
find other conditions that will
result in desired probability.
1 = 12 condition with a
3
36
total of 12: 3, 4, 6, or 11
1 = 9 condition with a
4
36
total of 9: 2, 3, 4, or 10
8. During Round 2, if an
infected person encounters
another infected person,
the expected number of
newly-infected people will
be affected. Described at
the right are the 4 possible
scenarios based on 3 people
infected in Round 1. Note
that between 1 and 6 people
may be infected going into
Round 2, so the example
only considers one possible
result of Round 1 going into
Round 2.
Math Labs
539
MATH LAB
$FWLYLW\5RFN3DSHU6FLVVRUV
Activity 2
PREPARE
• Introduce the game Rock,
Paper, Scissors. Explain that
rock wins over scissors because
the rock can crush the scissors.
Scissors wins over paper
because scissors can cut paper.
Paper wins over rock because
paper can cover rock.
• Invite one student to come
to front with you and play a
coupled game. This will give
students a chance to better
understand the game before
beginning the lab.
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FOLLOW-UP
540 Chapter 11 Probability and Statistics
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3URFHGXUH
3DSHU
TEACH
• Students should work in
groups of 4 students.
• Each student will take on
the roles of participate and
recorder.
• Review the difference between
experimental probability and
theoretical probability.
• If the class size does not
accommodate groups of four,
ask students to consider a
difference format for playing
the tournament.
• Discuss how the activity
changes if it is a singleelimination tournament.
Talk about advantages and
disadvantages to playing
double elimination compare to
single elimination.
• Brainstorm different types
of tournaments and classify
each as more likely to be a
single elimination or double
elimination tournament.
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Sum of
dice
2
1+1
3
4
1 + 2; 1 + 3;
2 + 1; 3 + 1;
2 + 2;
5
6
7
8
9
10
11
12
1 + 4;
4 + 1;
3 + 2;
2 +3
1 + 5;
5 + 1;
2 + 4;
4 + 2;
3+3
1 + 6;
6 + 1;
2 + 5;
5 + 2;
3 + 4;
4+3
2 + 6;
6 + 2;
3 + 5;
5 +3;
4+4
4 + 5;
5 +4;
3 + 6;
6+3
4 + 6;
6 + 4;
5+5
5 + 6;
6+5
6+6
Number of
Rolls
1
2
3
4
5
6
5
4
3
2
1
Probability
1/36
1/18
1/12
1/9
5/36
1/6
5/36
1/9
1/12
1/18
1/36
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Math Labs
541
Math Applications
Solutions and Notes
0DWK$SSOLFDWLRQV
1. a.The options are inclusive
because he could have a
red, long-sleeved shirt.
b.There are 4 red longsleeved shirts and
23 possible shirts, so the
probability is 4 .
23
c.There are 11 long-sleeved
shirts and 5 other red
shirts, so the probability
is 16 .
23
d.There are 23 shirts,
16 pants, and 25 shoes,
so there are 23 • 16 • 25
= 9,200 different ways
Marcellus can dress his
avatar.
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Math Applications
Solutions and Notes
2. a.Sample answer: The
population is all students in
her school, and the sample
is the students whose
opinions Caitlin will gather.
b.Sample answer: whether
or not there is enough
interest among the
students in her school to
start a chess club.
c.Sample answer: Caitlin
should conduct a survey
since she wants to gather
opinions.
d.Sample answer: Ask a
random selection of
5 students from each
homeroom in the school
the survey question,
“Would you be interested
in a school chess club?”
Analyze the results of the
survey to gauge student
interest. The survey would
produce unbiased results
because the sample is a
random sample and there
is no bias in the survey
question.
3,514  • 
3,513 
1−
3. a. 1−



 12,578   12,578 
≈ 0.519 ≈ 51.9%
b. 3, 278 + 3,113 • 3, 278 + 3,112
12,578
12,578
≈ 0.258 ≈ 25.8%
c. 274 ≈ 0.010 ≈ 1.0%
2, 674
Math Applications
543
Math Applications
Solutions and Notes
4. a. 725 = 29
3, 000 120
b. 725 − 1 = 724
3, 000 − 1 2, 999
c. 29 • 724 ≈ 0.058
120 2, 999
≈ 5.8%
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2, 999
5. a.149 • 148 • 147
275 274 273
≈ 0.158 ≈ 15.8%
b.100% – 15.8% = 84.1%
c.0.158 • 0.158 • 0.842
≈ 0.021 ≈ 2.1%
d.0.158 • 0.158 • 0.158
≈ 0.004 ≈ 0.4%
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Math Applications
Solutions and Notes
6. a.Because each click does
not depend on any
other event, they are
independent events.
b.Because each click
is independent, the
probability remains the
same: 0.14.
c.Because each click
is independent, the
probability remains the
same: 0.14.
d.0.14 • 0.14 • 0.14 • 0.14
≈ 0.0004 ≈ 0.04%
7. a.The events are inclusive,
so P(type B or positive)
= P(type B) + P(positive)
– P(type B and positive)
= 22 + 23 − 20 = 25
38 38 38 38
b.P(type A or negative) =
P(type A) + P(negative)
– P(type A and negative) =
16 + 15 − 13 = 18 = 9
38 38 38 38 19
38
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Math Applications
545
Math Applications
Solutions and Notes
8. a.Sample answer: No, the
sample is representative of
the population since it is a
random sample.
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sample size increases,
the margin of sampling
error decreases, and the
estimate of the population
parameter gets more
accurate. While increasing
the sample size will
produce more accurate
results, it may not be
feasible or cost effective to
sample a very large portion
of the population.
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Solutions and Notes
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10. a.P(no seatbelt or no license)
= P(no seatbelt) + P(no
license) – P(no seatbelt
and no license) = 0.005 +
0.056 – 0.003 = 0.058
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growth plan has a greater
expected value, so Molly
might consider investing
in this plan if she still
has several years before
retirement. However,
if Molly is closer to
retirement age, she might
consider investing in the
moderate growth plan
since there is less of a risk
of losing money on her
investment.
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Math Applications
547
Math Applications
Solutions and Notes
12. a.44 + 25 + 27 + 31
= 127; P = 127
235
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Math Applications
548
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Math Applications
Solutions and Notes
14. c.Answers will vary.
Sample answer: Use a
calculator and the function
randInt(1,89) 5 times to
generate 5 integers from
1 to 89. Let integers from
1 through 7 represent a
rejected assembly, from
8 through 34 represent
an assembly that needs
a slight fix, and from 35
through 89 represent
an approved assembly.
Conduct several trials of
the simulation and use
the results to predict how
many assemblies will be
approved in the next shift.
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549
Vocabulary Review
biased (11.3)
biased sample (11.3)
causation (11.3)
compound event (11.2)
conditional probability (11.2)
control group (11.3)
correlation (11.3)
dependent events (11.1)
expected value (11.4)
experiment (11.3)
fair decision (11.4)
Fundamental Theorem
of Counting (14.1)
inclusive events (11.2)
independent events (14.1)
mutually exclusive (11.2)
normal distribution (11.5)
observational group (11.3)
population (11.3)
probability distribution (11.5)
probability model (11.4)
randInt( (11.4)
random events (11.4)
random sample (11.3)
random variable (11.5)
sample (11.3)
simulation (11.4)
standard deviation (11.5)
standard normal distribution (11.5)
survey (11.3)
treatment group (11.3)
z-score (11.5)
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Practice Additional
Answers
Open Ended Response
9. The events are not mutually
exclusive because 60 is a
multiple of 3 and a multiple
of 4; P(A and B) = 1 ;
6
P(A or B) = 1
2
Extended Response
11. a.
What type of pet do you own?
Male
Female
Dog
42
35
Cat
30
24
Both
29
20
Standardized Test
Response Form (CRB)
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