Download Independent Events

LESSON
20.2
Name
Independent Events
Class
Date
20.2 Independent Events
Essential Question: What does it mean for two events to be independent?
Common Core Math Standards
Resource
Locker
The student is expected to:
S-CP.2
Explore
Understand that two events A and B are independent if the probability
of A and B occurring together is the product of their probabilities, and use
this characterization to determine if they are independent. Also S-CP.3,
S-CP.4, S-CP.5
Suppose you flip a coin and roll a number cube. You would expect the probability of getting heads on the coin to
be __12 regardless of what number you get from rolling the number cube. Likewise, you would expect the probability
of rolling a 3 on the number cube to be __16 regardless of whether of the coin flip results in heads or tails.
When the occurrence of one event has no effect on the occurrence of another event, the two events are called
independent events.
Mathematical Practices
A
MP.7 Using Structure
A jar contains 15 red marbles and 17 yellow marbles. You randomly draw a marble from the
jar. Let R be the event that you get a red marble, and let Y be the event that you get a yellow
marble.
Language Objective
17
15
Since the jar has a total of 32 marbles, P(R) = _____ and P(Y) = _____ .
32
32
Work with a partner to brainstorm examples of independent events.
B
ENGAGE
Suppose the first marble you draw is a red marble, and you put that marble back in the
jar before randomly drawing a second marble. Find P(Y∣R), the probability that you get a
yellow marble on the second draw after getting a red marble on the first draw. Explain your
reasoning.
Essential Question: What does it mean
for two events to be independent?
PREVIEW: LESSON
PERFORMANCE TASK
View the Engage section online. Discuss the
photograph. Ask students to describe differences that
they observe in the plants pictured. Then preview the
Lesson Performance Task.
17
Since the jar still has a total of 32 marbles and 17 of them are yellow, P(Y∣R) = _____ .
32
C
© Houghton Mifflin Harcourt Publishing Company
Two events are independent provided the
occurrence of one event has no effect on the
occurrence of the other event. For independent
events A and B, P(A | B) = P(A); that is, the
probability of event A does not change even when
event B is assumed to have occurred.
Understanding the Independence of Events
Suppose you don’t put the red marble back in the jar before randomly drawing a second
marble. Find P(Y∣R), the probability that you get a yellow marble on the second draw after
getting a red marble on the first draw. Explain your reasoning.
17
Since the jar now has a total of 31 marbles and 17 of them are yellow, P(Y∣R) = _____ .
31
Reflect
1.
In one case you replaced the first marble before drawing the second, and in the other case you didn’t.
For which case was P(Y∣R) equal to P(Y)? Why?
P(Y∣R) = P(Y) when the red marble was replaced because the total number of marbles in
the jar, and therefore the proportion of marbles that are yellow, stayed the same.
2.
In which of the two cases would you say the events of getting a red marble on the first draw and getting a
yellow marble on the second draw are independent? What is true about P(Y∣R) and P(Y) in this case?
The events are independent when the red marble is returned to the jar. In this case,
P(Y∣R) = P(Y).
Module 20
be
ges must
EDIT--Chan
DO NOT Key=NL-B;CA-B
Correction
Lesson 2
1015
gh “File info”
made throu
Date
Class
nt Events
Name
epende
20.2 Ind
ion: What
Quest
Essential
does it mean
of
the full text
S-CP.2 For
S-CP.5
S-CP.3, S-CP.4,
endent?
s to be indep
for two event
rd, see the
this standa
g on page
table startin
e 1. Also
CA2 of Volum
nts
ce of Eve
Resource
Locker
penden
the coin to
heads on
of getting
probability
probability
expect the
expect the
you would
tails.
Likewise,
You would
Explore
heads or
number cube. coin flip results in
number cube.
and roll a
rolling the
are called
er of the
flip a coin
you get from
two events
less of wheth
__1
Suppose you of what number
r event, the
to be 6 regard
less
_1_
ence of anothe
be 2 regard on the number cube
the occurr
3
on
a
of rolling
has no effect
of one event
e from the
occurrence
draw a marbl a yellow
When the
mly
.
t events
You rando
you get
marbles.
independen
event that
ing the Inde
HARDCOVER PAGES 737748
Understand
A2_MNLESE385900_U8M20L2 1015

Turn to these pages to
find this lesson in the
hardcover student
edition.
17 yellow
Y be the
marbles and red marble, and let
ns 15 red
a
A jar contai
that you get
17
be the event
_____ .
jar. Let R
15
P(Y) =
marble.
= _____ and
32
32 marbles, P(R)
the
32
of
e back in
jar has a total
put that marbl that you get a
Since the
e, and you
probability
red marbl
n your
a
is
P(Y∣R), the
draw. Explai
e you draw
marble. Find
first marbl
e on the first
g a second
red marbl
Suppose the
getting a
mly drawin
17
after
rando
draw
jar before
_____ .
e on the second
, P(Y∣R) =
yellow marbl
32
17 of them are yellow
reasoning.
32 marbles and
of
g a second
has a total
still
drawin
jar
mly
rando
Since the
draw after
the jar before
e on the second
e back in
yellow marbl
a
the red marbl
get
put
17
that you
don’t
ing.
_____ .
Suppose you Y∣R), the probability Explain your reason
P(
draw.
, P(Y∣R) =
first
yellow
the
marble. Find
31
are
on
e
17 of them
red marbl
getting a
31 marbles and
a total of
jar now has
Since the
didn’t.
case you
in the other
second, and
les in
drawing the
er of marb
Reflect
marble before
numb
first
the
total
se the
replaced
.
to P(Y)? Why?
case you
ced becau
d the same
P Y∣R) equal
1. In one
le was repla
case was (
yellow, staye
the red marb
For which
ga
les that are
and gettin
P(Y) when
rtion of marb
P(Y∣R) =
the first draw
the propo
this case?
marble on
in
)
fore
red
(
Y
a
P
g
there
of gettin
P(Y∣R) and
the jar, and
say the events What is true about
In this case,
t?
would you
to the jar.
enden
cases
ned
two
indep
retur
of the
draw are
marble is
2. In which
e on the second nt when the red
ende
yellow marbl
s are indep
The event
Lesson 2
).
(
Y
P
=
P(Y∣R)

© Houghto
n Mifflin
Harcour t
Publishin
y
g Compan

1015
Module 20
SE38590
A2_MNLE
1015
Lesson 20.2
L2 1015
0_U8M20
8/26/14
11:15 AM
8/26/14 11:16 AM
Explain 1
Determining if Events are Independent
EXPLORE
To determine the independence of two events A and B, you can check to see whether P(A∣B) = P(A) since the
occurrence of event A is unaffected by the occurrence of event B if and only if the events are independent.
Example 1

Understanding the Independence
of Events
The two-way frequency table gives data about 180 randomly selected
flights that arrive at an airport. Use the table to answer the question.
Late Arrival
On Time
Total
Domestic Flights
12
108
120
International Flights
6
54
60
Total
18
162
180
INTEGRATE TECHNOLOGY
Students have the option of doing the Explore activity
either in the book or online.
Is the event that a flight is on time independent of the event that a flight is domestic?
QUESTIONING STRATEGIES
Let O be the event that a flight is on time. Let D be the event that a flight is domestic. Find P(O) and
P(O∣D). To find P(O), note that the total number of flights is 180, and of those flights, there are 162
162
on-time flights. So, P(O) = ___
= 90%.
180
How can you tell if two events A and B are
independent? If P(A) = P(A | B) or
P(B) = P(B | A), then A and B are independent
events.
To find P(O∣D), note that there are 120 domestic flights, and of those flights, there are 108 on-time flights.
108
So, P(O∣D), = ___
= 90%.
120
Since P(O∣D) = P(O), the event that a flight is on time is independent of the event that a flight
is domestic.

Is the event that a flight is international independent of the event that a flight arrives late?
EXPLAIN 1
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
©nmcandre/iStockPhoto.com
Let I be the event that a flight is international. Let L be the event that a flight arrives late. Find P(I)
and P(I∣L). To find P(I), note that the total number of flights is 180, and of those
__
60
1
= 33 3 %.
flights, there are 60 international flights. So, P(I) = _____
180
To find P(I∣L), note that there are 18 flights that arrive late, and of those flights, there are 6
__
6
1
international flights. So, P(I∣L) = _____ = 33 3 %.
18
Since P(I∣L) = P(I), the event that a flight is international [is/is not] independent of the
event that a flight arrives late.
Determining if Events are
Independent
QUESTIONING STRATEGIES
How can a two-way frequency table help you
determine whether two events are
independent? A two-way frequency table makes it
easy to find n(A) and n(A | B), which you can use to
calculate P(A) and P(A | B). If the two probabilities
are equal, the events are independent.
AVOID COMMON ERRORS
Module 20
1016
Lesson 2
PROFESSIONAL DEVELOPMENT
A2_MNLESE385900_U8M20L2 1016
Math Background
Two events are independent if the occurrence of one does not affect the probability
of the other. In other words, for independent events A and B, P(A) = P(A | B).
Note that you can use this criterion to determine if A and B are independent. If A
and B are independent, then P(A and B), also written P(A ∩ B), is P(A) ⋅ P(B).
More generally, events A1, A2, …, An are mutually independent if the occurrence
of any one does not affect the probability of any other. Then,
P(A1 ∩ A2 ∩ … ∩ An) = P(A1) ⋅ P(A2) ⋅ … ⋅ P(An).
8/28/14 8:47 PM
When given a two-way frequency table, some
students may have trouble identifying which values to
use to calculate the probability of A given B.
Encourage them to circle the column that represents
event B, and then put another circle around the cell
within the column that represents event A. The
number in the cell they have circled is n(A | B).
Independent Events 1016
Your Turn
EXPLAIN 2
The two-way frequency table gives data about 200 randomly selected apartments in a
city. Use the table to answer the question.
Finding the Probability of
Independent Events
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Point out that independent events are not the
Single Occupant
64
12
76
Multiple Occupants
26
98
124
Total
90
110
200
Total
76
64
an apartment has 1 bedroom. Then P(SO) = ___
= 38%, and P(SO∣1B) = ___
≈ 71%.
200
90
Since P(SO∣1B) ≠ P(SO), the events are not independent.
Is the event that an apartment has 2 or more bedrooms independent of the event that an apartment has
multiple occupants?
Let 2B represent that an apartment has 2 or more bedrooms, and let MO be the
4.
110
= 55%, and
event that an apartment has multiple occupants. Then P(2B) = ___
200
98
P(2B∣MO) = ___
≈ 89%. Since P(2B∣MO) ≠ P(2B), the events are not independent.
110
Explain 2
Finding the Probability of Independent Events
P(A ∩ B)
From the definition of conditional probability you know that P(A|B) = _ for any events A and B. If those
P(B)
P(A ∩ B)
events happen to be independent, you can replace P(A|B) with P(A) and get P(A) = _. Solving the last
P(B)
© Houghton Mifflin Harcourt Publishing Company
Once you know that two events are
independent, how can you find the probability
that both events take place? Multiply the individual
probabilities.
2+
Bedrooms
Is the event that an apartment has a single occupant independent of the event that an apartment has
1 bedroom?
Let SO be the event that an apartment has a single occupant, and let 1B be the event that
3.
same as mutually exclusive events, which are events
that cannot occur in the same trial. For example, a
coin cannot simultaneously show heads and tails. If A
and B are mutually exclusive, then P(A or B), also
written P(A ∪ B), is P(A) + P(B). This addition rule
is generally used when finding the probability for a
single trial, while the multiplication rule for
independent events is usually applied to two or more
separate trials, performed either simultaneously or in
succession.
QUESTIONING STRATEGIES
1 Bedroom
equation for P(A ∩ B) gives the following result.
Probability of Independent Events
Events A and B are independent if and only if P(A ∩ B) = P(A) · P(B).
Example 2

Find the specified probability.
Recall the jar with 15 red marbles and 17 yellow marbles from the Explore. Suppose you
randomly draw one marble from the jar. After you put that marble back in the jar, you
randomly draw a second marble. What is the probability that you draw a yellow marble first
and a red marble second?
Let Y be the event of drawing a yellow marble first. Let R be the event of drawing a red marble
17
second. Then P(Y) = __
and, because the first marble drawn is replaced before the second marble is
32
15
drawn, P(RǀY) = P(R) = __
. Since the events are independent, you can multiply their probabilities:
32
17 __
255
P(Y ∩ R) = P(Y) · P(R) = __
· 15 = ____
≈ 25%.
32 32
1024
Module 20
1017
Lesson 2
COLLABORATIVE LEARNING
A2_MNLESE385900_U8M20L2 1017
Small Group Activity
Have students work in groups to do a simple experiment like rolling a number
cube twice. Have one student perform the experiment, while another student
keeps track of the results both as a table and as a tree diagram. Have a third
student determine the sample space for the experiment (for example, HH, HT,
TH, TT), and a fourth student explain how to find the probability of the events in
the sample space. Ask students to discuss their results, explain why the events are
independent, and present their results to the class.
1017
Lesson 20.2
8/26/14 11:35 AM
B
You spin the spinner shown two times. What is the probability that the
spinner stops on an even number on the first spin, followed by an odd
number on the second spin?
12
1
9
3
8
5
7
6
Let E be the event of getting an even number on the first spin. Let O be the event of getting an odd
3
number on the second spin. Then P(E) = _ and, because the first spin has no effect on the second
8
5
spin, P(OǀE) = P(O) = _. Since the events are independent, you can multiply their probabilities:
8
3
5
15
P(E ∩ O) = P(E) · P(O) = _ · _ = _ ≈ 23 %.
8
8
64
Reflect
5.
In Part B, what is the probability that the spinner stops on an odd number on the first spin, followed by an
even number on the second spin? What do you observe? What does this tell you?
The probability of getting an odd number on the first spin followed by an even number
5 __
15
· 3 = ___
. This value is equal to the probability calculated in Part B.
on the second spin is __
64
8 8
Because the spins are independent of each other, changing the order of the outcomes does
not affect the overall probability.
Your Turn
6.
You spin a spinner with 4 red sections, 3 blue sections, 2 green sections, and 1 yellow section. If all the
sections are of equal size, what is the probability that the spinner stops on green first and blue second?
2
1
with the second spin. Then P(G) = ___
= __
and, because the first spin has no effect on the
5
10
3
second spin, P(BǀG) = P(B) = ___
. Since the events are independent, you can multiply their
10
3
1 ___
probabilities: P(G ∩ B) = P(G) · P(B) = __
· 3 = ___
= 6%.
5 10
50
7.
A number cube has the numbers 3, 5, 6, 8, 10, and 12 on its faces. You roll the number cube twice. What is
the probability that you roll an odd number on both rolls?
Let O1 be the event of getting an odd number on the first roll. Let O2 be the event of getting
2
1
= __
and, because the first roll has no
an odd number on the second roll. Then P(O1) = __
6
3
1
effect on the second roll, P(O2ǀO1) = P(O2) = __
. Since the events are independent, you can
3
© Houghton Mifflin Harcourt Publishing Company
Let G be the event of getting green with the first spin. Let B be the event of getting blue
1 __
1
multiply their probabilities: P(O1 ∩ O2) = P(O1) · P(O2) = __
· 1 = __
≈ 11%.
3 3
9
Module 20
1018
Lesson 2
DIFFERENTIATE INSTRUCTION
A2_MNLESE385900_U8M20L2 1018
Critical Thinking
8/27/14 5:06 PM
Distinguish between evaluating P(A | B), when it is important to find the
conditional probability that event A will occur given that event B has already
occurred, and P(A and B), when it is important to consider whether events A and
P(A ∩ B)
B are independent. The division rule P(A | B) = ______ is used to find the
P(B)
conditional probability of A given B, while the multiplication rule
P(A and B) = P(A) ⋅ P(B) is used to find the probability that event A will occur in
one trial and event B will occur in another trial, given that one trial does not affect
the other.
Independent Events 1018
Explain 3
EXPLAIN 3
So far, you have used the formula P(A ∩ B) = P(A) · P(B) when you knew that events A and B are independent. You
can also use the formula to determine whether two events are independent.
Showing that Events are Independent
Example 3

AVOID COMMON ERRORS
If students are confused about identifying
independent events, have them try communicating
their results from this lesson orally or in writing. Ask
them to check that they are using terms correctly. In
particular, be sure students understand that the word
independent has a specific meaning in probability
theory that may be different from its meaning in
everyday conversation.
The two-way frequency table shows data for 120 randomly selected patients who have the
same doctor. Determine whether a patient who takes vitamins and a patient who exercises
regularly are independent events.
Takes Vitamins
No Vitamins
Total
Regular Exercise
48
28
76
No regular Exercise
12
32
44
Total
60
60
120
Step 1 Find P(V), P(E), and P(V ∩ E). The total number of patients is 120.
60 = _
1.
There are 60 patients who take vitamins, so P(V) = _
2
120
19 .
76
There are 76 patients who exercise regularly, so P(B) = _ = _
120
30
48 = 40%.
There are 48 patients who take vitamins and exercise regularly, so P(V ∩ E) = _
120
Step 2 Compare P(V ∩ E) and P(V) ∙ P(E).
19 ≈ 32%
19 = _
1 ∙_
P(V) ∙ P(E) = _
2 30
60
Because P(V ∩ E) ≠ P(V) ∙ P(E), the events are not independent.
What are two ways of showing that two events
are independent? Show or P(A) = P(A | B) or
P(A and B) = P(A) ⋅ P(B).

© Houghton Mifflin Harcourt Publishing Company
events are independent. Have them identify the total
number in the sample space. Then ask them to find P
(A), P(B), and P(A and B) and compare the values.
P(A and B) = P(A) ⋅ P(B) if and only if A and B are
independent events.
Determine if the events are independent.
Let V be the event that a patient takes vitamins. Let E be the event that a patient exercises regularly.
QUESTIONING STRATEGIES
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Have students follow a pattern to show that
Showing That Events Are Independent
The two-way frequency table shows data for 60 randomly selected children at an elementary
school. Determine whether a child who knows how to ride a bike and a child who knows
how to swim are independent events.
Knows how to
Ride a Bike
Doesn’t Know how
to Ride a Bike
Total
Knows how to Swim
30
10
40
Doesn’t Know how to Swim
15
5
20
Total
45
15
60
Let B be the event a child knows how to ride a bike. Let S be the event that a child knows how to swim.
Step 1 Find P(B), P(S), and P(B ∩ S). The total number of children is 60.
There are
45
3
45 children who know how to ride a bike, so P(B) = _
= _.
4
60
40
2
There are 40 children who know how to swim, so P(S) = _ = _.
3
60
1
30
There are 30 children who know how to ride a bike and swim, so P(B ∩ S) = _ = _.
2
60
Module 20
1019
Lesson 2
LANGUAGE SUPPORT
A2_MNLESE385900_U8M20L2 1019
Connect Vocabulary
Have students create notecards to help them understand the mathematical
meaning of independent. Ask them to write, on the front of the card, an example
of a time in their lives when they acted independently, and then write, on the
back, an example of events that are mathematically independent. Connecting
the idea to personal experience may help some students remember its meaning
and use.
1019
Lesson 20.2
8/27/14 5:06 PM
Step 2 Compare P(B ∩ S) and P(B) ∙ P(S).
3
2
1
P(B) ∙ P(S) = _ ∙ _ = _
4
3
2
Because P(B ∩ S) = P(B) ∙ P(S), the events [are/are not] independent.
Your Turn
8.
A farmer wants to know if an insecticide is effective in preventing small
insects called aphids from damaging tomato plants. The farmer
experiments with 80 plants and records the results in the two-way
frequency table. Determine whether a plant that was sprayed with
insecticide and a plant that has aphids are independent events.
Has Aphids
No Aphids
Total
Sprayed with Insecticide
12
40
52
Not Sprayed with Insecticide
14
14
28
Total
26
54
80
Let S be the event that a tomato plant was sprayed with insecticide. Let A be the event that a
tomato plant has aphids.
13
13 13
13
26
169
52 _
_
=
, P(A) = _ = _, and P(S) · P(A) = _ · _ = _ ≈ 21%.
40
80
20
80
40 20
800
3
12
_
_
Because P(S ∩ A) =
= 15% and P(S ∩ A) ≠ P(S) · P(A), the events are not
=
P(S) =
80
20
9.
A student wants to know if right-handed people are more or less likely to play a musical instrument than
left-handed people. The student collects data from 250 people, as shown in the two-way frequency table.
Determine whether being right-handed and playing a musical instrument are independent events.
Right-Handed
Left-Handed
Total
Plays a Musical Instrument
44
6
50
Does not Play a Musical
Instrument
176
24
200
Total
220
30
250
Let R be the event that a person is right-handed. Let I be the event that a person plays a
musical instrument.
1
22
22 _
50
220 _
1 , and P(R) ⋅ P(I) = _
_
⋅ = _.
= 22 , P(I) = _ = _
25 5
125
5
250
25
250
22
44 = _
and P(R ∩ I) = P(R) ⋅ P(I), the events are independent.
Because P(R ∩ I) = _
P(R) =
250
Module 20
A2_MNLESE385900_U8M20L2 1020
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Bokstaz/
Shutterstock
independent.
125
1020
Lesson 2
8/26/14 11:42 AM
Independent Events 1020
Elaborate
ELABORATE
10. What are the ways that you can show that two events A and B are independent?
The ways that you can show that two events A and B are independent are by confirming
that P(A|B) = P(A) or that P(A and B) = P(A) ⋅ P(B).
QUESTIONING STRATEGIES
11. How can you find the probability that two independent events A and B both occur?
Multiply P(A) and P(B).
How can you find the probability of
independent events A and B? Use the
multiplication rule P(A and B) = P(A) ⋅ P(B).
12. Essential Question Check-In Give an example of two independent events and explain why they are
independent.
Sample answer: When flipping a coin twice, the events of getting heads on the first flip
and on the second flip are independent. The occurrence of getting heads on the first flip
SUMMARIZE THE LESSON
does not affect the occurrence of getting heads on the second flip. The probability of
Why is it important that the multiplication
rule, P(A and B) = P(A) ⋅ P(B), is true if and
only if the events are independent? The fact that it is
true if and only if the events are independent means
that it can be used to determine whether events are
independent, and vice versa.
1
getting heads on the second flip is _
regardless of what happens on the first flip.
2
Evaluate: Homework and Practice
1.
A bag contains 12 red and 8 blue chips. Two chips are separately drawn at random
from the bag.
a. Suppose that a single chip is drawn at random from the bag. Find the probability
that the chip is red and the probability that the chip is blue.
EVALUATE
Let R be the event that the chip is red. Let B be the event that the
chip is blue.
ASSIGNMENT GUIDE
Concepts and Skills
Practice
Explore
Understanding the Independence
of Events
Exercises 1–2
Example 1
Determining if Events are
Independent
Exercises 3–4, 20
Example 2
Finding the Probability of
Independent Events
Exercises 5–14
Example 3
Showing That Events Are
Independent
Exercises 15–19,
21–22
Lesson 20.2
© Houghton Mifflin Harcourt Publishing Company
P(R) =
1021
• Online Homework
• Hints and Help
• Extra Practice
3
8
2
12 _
_
= and P(B) = _ = _
20
5
20
5
b. Suppose that two chips are separately drawn at random from the bag and that
the first chip is returned to the bag before the second chip is drawn. Find the
probability that the second chip drawn is blue given the first chip drawn was red.
8
2
=
P(B|R) =
5
20
_ _
c.
Suppose that two chips are separately drawn at random from the bag and that
the first chip is not returned to the bag before the second chip is drawn. Find the
probability that the second chip drawn is blue given the first chip drawn was red.
8
P(B|R) =
19
_
d. In which situation—the first chip is returned to the bag or not returned to
the bag—are the events that the first chip is red and the second chip is blue
independent? Explain.
Events R and B are independent when the first chip is returned
to the bag because P(B|R) = P(B) in that case.
Module 20
Exercise
A2_MNLESE385900_U8M20L2 1021
Lesson 2
1021
Depth of Knowledge (D.O.K.)
Mathematical Practices
8/26/14 11:56 AM
1-2
1 Recall of Information
MP.6 Precision
3-14
2 Skills/Concepts
MP.5 Using Tools
15-19
2 Skills/Concepts
MP.4 Modeling
20
3 Strategic Thinking
MP.2 Reasoning
21
3 Strategic Thinking
MP.3 Logic
22
3 Strategic Thinking
MP.4 Modeling
2.
Identify whether the events are independent or not independent.
a. Flip a coin twice and get tails
both times.
Independent
Not Independent
b. Roll a number cube and get 1 on
the first roll and 6 on the second.
Independent
Not Independent
Independent
Not Independent
Independent
Not Independent
c.
Draw an ace from a shuffled deck,
put the card back and reshuffle the
deck, and then draw an 8.
d. Rotate a bingo cage and draw the
ball labeled B-4, set it aside, and
then rotate the cage again and draw
the ball labled N-38.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 To help students describe the independent
events represented in a two-way table and then
determine the numbers used to find the probabilities,
have them also draw a probability tree representing
the given information. Then ask them to show the
probability for each branch and the calculations for
each possible result in the experiment.
Answer the question using the fact that P(AǀB) = P(A) only when events A and B
are independent.
3.
The two-way frequency table shows data for 80 randomly selected people who live in a
metropolitan area. Is the event that a person prefers public transportation independent
of the event that a person lives in the city?
Prefers to Drive
Prefers Public
Transportation
Total
Lives in the City
12
24
36
Lives in the Suburbs
33
11
44
Total
45
35
80
Let T be the event that a person prefers public transportation. Let C
35
be the event that a person lives in the city. Then P(T) = ___
≈ 44% and
80
24
P(TǀC) = ___
≈ 67%. Since P(TǀC) ≠ P(T), the events are not independent.
36
The two-way frequency table shows data for 120 randomly selected people who take
vacations. Is the event that a person prefers vacationing out of state independent of the
event that a person is a woman?
Prefers Vacationing
Out of State
Prefers Vacationing
in State
Total
Men
48
32
80
Women
24
16
40
Total
72
48
120
© Houghton Mifflin Harcourt Publishing Company
4.
Let O be the event that a person prefers vacationing out of state.
72
Let W be the event that a person is a woman. Then P(O) = ___
= 60% and
120
24
P(OǀW) = ___
= 60%. Since P(OǀW) = P(O), the events are independent.
40
Module 20
A2_MNLESE385900_U8M20L2 1022
1022
Lesson 2
8/26/14 11:57 AM
Independent Events 1022
A jar contains marbles of various colors as listed in the table. Suppose you
randomly draw one marble from the jar. After you put that marble back in the jar,
you randomly draw a second marble. Use this information to answer the question,
giving a probability as a percent and rounding to the nearest tenth of percent when
necessary.
AVOID COMMON ERRORS
When students start finding probabilities for
successive events, it is easy to confuse the context.
Have students read each exercise and highlight the
key words and phrases.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Tree diagrams are useful for understanding
5.
Color of
Marble
Number of
Marbles
Red
20
Yellow
18
Green
12
Blue
10
What is the probability that you draw a blue marble first and a red marble second?
Let B be the event of drawing a blue marble first. Let R be the event of drawing a red marble
the formula for the probability of independent events.
Ask groups of students to practice constructing tree
diagrams from two-way tables, and vice versa.
10
1
second. Then P(B) = __
=_
and, because the first marble drawn is replaced before the
6
60
20
1
second marble is drawn, P(RǀB) = P(R) = __
=_
. Since the events are independent, you can
60
3
1
1 _
multiply their probabilities: P(B ∩ R) = P(B) ∙ P(R) = __
∙ 1 = __
≈ 5.6%.
18
6 3
6.
What is the probability that you draw a yellow marble first and a green
marble second?
Let Y be the event of drawing a yellow marble first. Let G be the event of drawing a green
18
3
marble second. Then P(Y) = __
= __
and, because the first marble drawn is replaced before
60
10
12
1
the second marble is drawn, P(GǀY) = P(G) = __
=_
. Since the events are independent, you
5
60
3
3 _
can multiply their probabilities: P(Y ∩ G) = P(Y) ∙ P(G) = ___
∙ 1 = __
= 6%.
50
10 5
© Houghton Mifflin Harcourt Publishing Company
7.
What is the probability that you draw a yellow marble both times?
Let Y1 be the event of drawing a yellow marble first. Let Y2 be the event of drawing a yellow
18 __
marble second. Then P(Y1) = __
= 3 and, because the first marble drawn is replaced before
60 10
18 __
the second marble is drawn, P(Y2ǀY1) = P(Y2) = __
= 3 . Since the events are independent,
60 10
3 __
9
you can multiply their probabilities: P(Y1 ∩ Y2) = P(Y1) ∙ P(Y2) = __
∙ 3 = ___
= 9%.
10 10
100
8.
What color marble for the first draw and what color marble for the second draw have
the greatest probability of occurring together? What is that probability?
The color marble with the greatest number in the jar has the highest probability of being
drawn, so draw a red marble on each draw. Let R1 be the event of drawing a red marble first.
20
1
= __
and, because
Let R2 be the event of drawing a red marble second. Then P(R1) = __
60
3
the first marble drawn is replaced before the second marble is drawn,
20
1
P(R2ǀR1) = P(R2) = __
=_
. Since the events are independent, you can multiply their
60
3
1 _
1
probabilities: P(R1 ∩ R2) = P(R1) ∙ P(R2) = __
∙1=_
≈ 11.1%.
9
3 3
Module 20
A2_MNLESE385900_U8M20L2 1023
1023
Lesson 20.2
1023
Lesson 2
8/26/14 12:00 PM
You spin the spinner shown two times. Each section of the spinner
is the same size. Use this information to answer the question,
giving a probability as a percent and rounding to the nearest tenth
of a percent when necessary.
9.
What is the probability that the spinner stops on 1 first and 2 second?
4
1
P(1) = ___
= __
and, because the first spin has no effect on the
12
3
3
1
second spin, P(2ǀ1) = P(2) = ___
= __
. Since the events are
12
1
4
2
4
1
1
2
2
3
1
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 If A and B are independent events, the
3
5
probability that both A and B occur is given by the
multiplication rule P(A and B) = P(A) ⋅ P(B). More
generally, if n independent events occur, then the
probability is the product of the n probabilities of the
independent events.
4
independent, multiply their probabilities: P(1 ∩ 2) =
1 __
1
P(1) ∙ P(2) = __
∙ 1 = ___
≈ 8.3%.
3 4
12
10. What is the probability that the spinner stops on 4 first and 3 second?
2
1
= __
and, because the first spin has no effect on the second spin,
P(3) = ___
6
12
2
1
P(4ǀ3) = P(4) = ___
= __
. Since the events are independent, multiply their
6
12
1 __
1
probabilities: P(4 ∩ 3) = P(4) ∙ P(3) = __
∙ 1 = ___
≈ 2.8%.
6 6
36
11. What is the probability that the spinner stops on an odd number first and an even
number second?
Let O be the event of getting an odd number on the first spin. Let E
be the event of getting an even number on the second spin. Then
7
P(O) = ___
and, because the first spin has no effect on the second spin,
12
5
P(EǀO) = P(E) = ___
. Since the events are independent, multiply their
12
35
7 ___
probabilities: P(O ∩ E) = P(O) ∙ P(E) = ___
∙ 5 = ____
≈ 24.3%.
144
12 12
12. What first number and what second number have the least probability of occurring
together? What is that probability?
The number that appears least often on the spinner has the
lowest probability of occurring, so you want to get 5 on each spin.
© Houghton Mifflin Harcourt Publishing Company
1
and, because the first spin has no effect on the second spin,
P(5) = ___
12
1
P(5ǀ5) = P(5) = ___
. Since the events are independent, you can multiply
12
1 ___
1
their probabilities: P(5 ∩ 5) = P(5) ∙ P(5) = ___
∙ 1 = ____
≈ 0.7%.
144
12 12
13. Find the probability of getting heads on every toss of a coin when the coin is tossed
3 times.
Let H1, H2, and H3 represent the events of getting heads on the first, second,
and third coin tosses, respectively. Because the events are independent,
1 __
1
1
P(H1 ∩ H2 ∩ H3) = P(H1) ∙ P(H2) ∙ P(H3) = __
∙ 1 ∙ __
= __
.
2 2 2
8
Module 20
A2_MNLESE385900_U8M20L2 1024
1024
Lesson 2
8/26/14 12:02 PM
Independent Events 1024
14. You are randomly choosing cards, one at a time and with replacement, from a
standard deck of cards. Find the probability that you choose an ace, then a red card,
and then a face card. (Remember that face cards are jacks, queens, and kings.)
Let A be the event of choosing an ace as the first card. Let R be the event of choosing a
red card as the second card. Let F be the event of choosing a face card as the third card.
Because you replace each card you choose, the events are independent, and
3
1 1 3
4 26 12
P(A ∩ R ∩ F) = P(A) · P(R) · P(F) =
·
·
=
· ·
=
≈ 0.9%.
52 52 52
13 2 13
338
_ _ _ _ _ _ _
Determine whether the given events are independent using the fact that
P(A ⋂ B) = P(A) · P(B) only when events A and B are independent.
15. The manager of a produce stand wants to find out whether there is a connection
between people who buy fresh vegetables and people who buy fresh fruit. The
manager collects data on 200 randomly chosen shoppers, as shown in the two-way
frequency table. Determine whether buying fresh vegetables and buying fresh fruit are
independent events.
Bought Vegetables
No Vegetables
Total
Bought Fruit
56
20
76
No Fruit
49
75
124
Total
105
95
200
Let V be the event that a shopper bought fresh vegetables. Let F be the event
that a shopper bought fresh fruit.
399
105 _
76
19
21 ( ) _
21 19
_
=
,P F =
= _, and P(V) · P(F) = _ · _ = _ ≈ 20%.
200
40
200
50
40 50
2000
56
Because P(V ∩ F) = _ = 28% and P(V ∩ F) ≠ P(V) · P(F), the events are not
P(V) =
200
© Houghton Mifflin Harcourt Publishing Company
independent.
16. The owner of a bookstore collects data about the reading preferences of 60 randomly
chosen customers, as shown in the two-way frequency table. Determine whether
being a female and preferring fiction are independent events.
Prefers Fiction
Prefers
Nonfiction
Total
Female
15
10
25
Male
21
14
35
Total
36
24
60
Let Fe be the event that a reader is female. Let Fi be the event that a
reader prefers fiction.
5 ( ) _
3
25 _
36 _
5 3
1
_
=
, P Fi =
= , and P(Fe) · P(Fi) = _ · _ = _.
5
4
60
12
60
12 5
15
1
_
_
Because P(Fe ∩ Fi) =
= and P(Fe ∩ Fi) = P(Fe) · P(Fi), the events
P(Fe) =
60
4
are independent.
Module 20
A2_MNLESE385900_U8M20L2 1025
1025
Lesson 20.2
1025
Lesson 2
8/26/14 12:04 PM
17. The psychology department at a college collects data about whether there is a
relationship between a student’s intended career and the student’s like or dislike
for solving puzzles. The two-way frequency table shows the collected data for 80
randomly chosen students. Determine whether planning for a career in a field
involving math or science and a like for solving puzzles are independent events.
Plans a Career in a
Math/Science Field
Plans a Career in a
Non-Math/Science
Field
Total
35
15
50
Dislikes Solving Puzzles
9
21
30
Total
44
36
80
Likes Solving Puzzles
Let MS be the event that a student plans a career in a math/science field. Let L be
the event that a student likes solving puzzles.
5
50 _
11
11 ( ) _
44 _
11 5
_
=
,P L =
= , and P(MS) · P(L) = _ · _ = _ ≈ 34%.
80
20
80
8
20 8
32
35
Because P(MS ∩ L) = _ ≈ 44% and P(MS ∩ L) ≠ P(MS) · P(L), the events are not
P(MS) =
80
independent.
18. A local television station surveys some of its viewers to determine the primary
reason they watch the station. The two-way frequency table gives the survey data.
Determine whether a viewer is a man and a viewer primarily watches the station for
entertainment are independent events.
Primarily Watches
for Entertainment
(Comedies, Dramas)
Total
Men
28
12
40
Women
35
15
50
Total
63
27
90
© Houghton Mifflin Harcourt Publishing Company
Primarily Watches for
Information (News,
Weather, Sports)
Let M be the event that a viewer is a man. Let E be the event that a viewer
primarily watches the station for entertainment.
3
40 _
4
2
27
4 3
_
= , P(E) = _ = _ and P(M) · P(E) = _ · _ = _.
15
90
9
90
10
9 10
2
12
Because P(M ∩ E) = _ = _ and P(M ∩ E) = P(M) · P(E), the events
P(M) =
90
15
are independent.
Module 20
A2_MNLESE385900_U8M20L2 1026
1026
Lesson 2
8/26/14 12:09 PM
Independent Events 1026
19. Using what you know about independent events, complete the two-way frequency
table in such a way that any event from a column will be independent of any event
from a row. Give an example using the table to demonstrate the independence of
two events.
Women
Men
Total
Prefers Writing
with a Pen
40
60
100
Prefers Writing
with a Pencil
20
30
50
Total
60
90
150
The ratio of women to men is 60:90, or 2:3. For independence of events,
the 100 people who prefer writing with a pen must also be divided into a
ratio of 2:3. Let m represent the common multiplier of 2 and 3, and solve
2m + 3m = 100 to get m = 20. So, there must be 2(20) = 40 women who
prefer writing with a pen and 3(20) = 60 men who prefer writing with a
pen. To find the number of women who prefer writing with a pencil, subtract
40 from 60 to get 20. To find the number of men who prefer writing with a
pencil subtract 60 from 90 to get 30.
Examples showing independence of events will vary. Sample answer:
Let W be the event that a person is a woman. Let Pe be the event that a
60
2 ( ) ___
2
=_
, P Pe = 100
=_
, and
person prefers writing with a pen. P(W) = ___
5
150
150
3
40
2 _
4
4
P(W) · P(Pe) = _
· 2 = __
. Since P(W ∩ Pe) = ___
= __
and P(W ∩ Pe) = P(W) · P(Pe),
5 3
15
15
150
the events are independent.
© Houghton Mifflin Harcourt Publishing Company
H.O.T. Focus on Higher Order Thinking
20. Make a Prediction A box contains 100 balloons. The balloons come in two colors:
80 are yellow and 20 are green. The balloons are also either marked or unmarked:
50 are marked “Happy Birthday!” and 50 are not. A balloon is randomly chosen
from the box. How many yellow “Happy Birthday!” balloons must be in the box if
the event that a balloon is yellow and the event that a balloon is marked “Happy
Birthday!” are independent? Explain.
Because half of the balloons are marked “Happy Birthday!”, the probability
1
of drawing a “Happy Birthday!” balloon is . To be independent of the
2
event that a balloon is yellow, the conditional probability of drawing a
_
balloon marked “Happy Birthday!” given that the balloon is yellow also
1
1
needs to have a probability of . So, (80) = 40 yellow “Happy Birthday!”
2
2
balloons would need to be in the box.
_
Module 20
A2_MNLESE385900_U8M20L2 1027
1027
Lesson 20.2
_
1027
Lesson 2
8/26/14 12:12 PM
21. Construct Arguments Given that events A and B are independent, prove that the
complement of event A, A c, is also independent of event B.
P(A c∣ B) = 1 - P(A∣B)
PEERTOPEER DISCUSSION
Have students discuss various pairs of independent
events that they can create from the spinner below.
Definition of complementary events
= 1 - P(A)
Definition of independent events
= P(A c)
Definition of complementary events
c
So, events A and B are also independent.
8
22. Multi-Step The two-way frequency table shows two events, A and B, and their
complements, A c and B c. Let P(A) = a and P(B) = b. Using a, b, and the grand
total T, form the products listed in the table to find the number of elements in A ∩ B,
A ∩ B c, A c ∩ B, and A c ∩ B c.
A
B
abT
(1 - a)bT
Bc
a(1 - b)T
(1 - a)(1 - b)T
Total
Row total for event B c: a(1 - b)T + (1 - a)(1 - b)T = ⎡⎣a + (1 - a)⎤⎦ (1 - b)T = (1 - b)T
Column total for event A c: (1 - a)bT + (1 - a)(1 - b)T = (1 - a)⎡⎣b + (1 - b)⎤⎦T = (1 - a)T
bT
Show that events A and B c are independent.
n(A ∩ B ) _
a(1 - b)T
_
= a = P(A)
=
c
(1 - b)T
n(B )
c
d. Show that events A c and B are independent.
P(A c∣B) =
(1 - a)bT
n(A ∩ B) _
_
= 1 - a = P(A )
=
c
c
n(B)
4
© Houghton Mifflin Harcourt Publishing Company
n(A ∩ B) _
abT
_
= a = P(A)
=
P(A∣B c) =
3
Have students explain what is meant by independent
events and give an example of how to determine
whether two events are independent.
b. Show that events A and B are independent using the fact that P(A∣ B) = P(A)
only when events A and B are independent.
c.
6
JOURNAL
Column total for event A: abT + a(1 - b)T = a⎡⎣b + (1 - b)⎤⎦T = aT
n(B)
2
Have them describe in their own words the process
they would use to find the probabilities for pairs of
independent events.
T
a. Find the table’s missing row and column totals in simplest form.
Row total for event B: abT + (1 - a)bT = ⎣⎡a + (1 - a)⎦⎤ bT = bT
P(A∣B) =
7
5
Total
Ac
1
bT
e. Show that events A c and B c are independent.
P(A c∣B c) =
Module 20
A2_MNLESE385900_U8M20L2 1028
(1 - a)(1 - b)T
n(A ∩ B ) __
_
=
= 1 - a = P(A )
c
c
n(B )
c
c
(1 - b)T
1028
Lesson 2
4/2/14 2:31 PM
Independent Events 1028
Lesson Performance Task
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Patterns
MP.8 Megan drew a Punnett Square to find the
Before the mid-1800s, little was known about the way that plants pass along characteristics
such as color and height to their offspring. From painstaking observations of garden peas,
the Austrian monk Gregor Mendel discovered the basic laws of heredity. The table shows
the results of three of Mendel’s experiments. In each experiment, he looked at a particular
characteristic of garden peas by planting seeds exclusively of one type.
results of crossing two violet-round seed-yellow seed
parents. Let V, R, and Y represent dominant violet,
round, and yellow. Let v, r, and y represent recessive
white, wrinkled, and green.
How many rows and columns will Megan’s Punnett
Square have? 8 rows, 8 columns
Characteristic
Type Planted
Results in Second Generation
Flower color
100% violet
705 violet, 224 white
Seed texture
100% round
5474 round, 1850 wrinkled
Seed color
100% yellow
6022 yellow, 2011 green
1. Suppose you plant garden peas with violet flowers and round, yellow seeds.
Estimate the probability of obtaining second-generation plants with violet flowers,
the probability of obtaining second-generation plants with round seeds, and the
probability of obtaining second-generation plants with yellow seeds. Explain how
you made your estimates.
List the letter groups that will appear atop the
columns and beside the rows in the square. VRY, VRy,
VrY, Vry, vRY, vRy, vrY, vry
Mendel saw that certain traits, such as violet flowers and round seeds, seemed stronger
than others, such white flowers and wrinkled seeds. He called the stronger traits
“dominant” and the weaker traits “recessive.” Both traits can be carried in the genes of a
plant, because a gene consists of two alleles, one received from the mother and one from
the father. (For plants, the “father” is the plant from which the pollen comes, and the
“mother” is the plant whose pistil receives the pollen.) When at least one of the alleles has
the dominant trait, the plant exhibits the dominant trait. Only when both alleles have the
recessive trait does the plant exhibit the recessive trait.
What is the probability that a second-generation
flower will be white and wrinkled, and have green
1
seeds? ___
64
© Houghton Mifflin Harcourt Publishing Company
You can use a 2 × 2 Punnett square, like the one shown, to see the results of crossing the
genes of two parent plants. In this Punnett square, V represents the dominant flower color
violet and v represents the recessive flower color white. If each parent’s genes contain both
V and v alleles, the offspring may receive, independently and with equal probability, either
a V allele or a v allele from each parent.
V
v
V
VV
Vv
v
vV
vv
2. After planting a first generation of plants exhibiting only dominant traits, Mendel
observed that the second generation consisted of plants with a ratio of about 3:1
dominant-to-recessive traits. Does the Punnett square support or refute Mendel’s
observation? Explain.
Module 20
A2_MNLESE385900_U8M20L2 1029
1029
Lesson 20.2
1029
Lesson 2
8/26/14 12:13 PM
3. Draw a 4 × 4 Punnett square for finding the results of crossing two
violet-flower-and-round-seed parent plants. Let V and R represent
the dominant traits violet flowers and round seeds, respectively. Let v
and r represent the recessive traits white flowers and wrinkled seeds,
respectively. Each column heading and row heading of your Punnett
square should contain a two-letter combination of V or v and R or r. Each
cell of your Punnett square will then contain four letters. Use the Punnett
square to find the probability that a second-generation plant will have
white flowers and round seeds. Explain your reasoning.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Critical Thinking
MP.3 A gardener claimed to have planted 10
violet-round seed-yellow garden pea seeds and to
have obtained 10 white-wrinkled seed-green plants.
Was the gardener telling the truth? Explain. Sample
answer: The gardener’s results were possible but
highly unlikely. The probability that a single secondgeneration white-wrinkled seed-green plant will be
1. Estimate the probabilities using Mendel’s data: If V represents violet
705
705
flowers, then P(V) = _______
= ___
≈ 76%. If R represents round
705 + 224
929
5474
5474
= ____
≈ 75%. If Y represents yellow
seeds, then P(R) = _________
5474 + 1850
7324
6022
6022
= ____
≈ 75%.
seeds, then P(Y) = _________
6022 + 2011
8033
_1 . The probability that 10 will be
64
1
produced is (_) , which is very small.
produced is
2. The Punnett square supports Mendel’s observation. The combination
1 _
1
·1=_
. Since
of alleles in each table cell occurs with a probability of _
4
2 2
three of the four table cells contain a dominant V allele, you can expect
three-fourths of the offspring to have violet flowers. Since one table
cell contains two recessive v alleles, you can expect one-fourth of the
offspring to have white flowers. So, the ratio of offspring with violet
10
64
_3
3
.
flowers to offspring with white flowers is __41 = _
1
4
3.
VR
VR
Vr
vR
vr
VVRR
VVRr
VvRR
VvRr
Vr
VVRr
VVrr
VvRr
Vvrr
vR
VvRR
VvRr
vvRR
vvRr
vr
VvRr
Vvrr
vvRr
vvrr
© Houghton Mifflin Harcourt Publishing Company
The combination of alleles in each table cell occurs with a probability
1 _
1 _
1
of _
·1·_
· 1 = __
. For offspring to have white flowers, which is
16
2 2 2 2
recessive, a table cell must contain two v alleles. For offspring to have
round seeds, which is dominant, a cell must contain two R alleles or a
combination of an R allele and an r allele. Three of 16 table cells satisfy
3
those requirements (vvRR, vvRr, and vvRr), giving a probability of __
.
16
Module 20
1030
Lesson 2
EXTENSION ACTIVITY
A2_MNLESE385900_U8M20L2 1030
Have students work in pairs. Each pair should have two coins and a sheet of paper
on which to record their results. Ask students to conduct 50 trials, each time
flipping two coins and recording the results (2 heads, a heads and a tails, or 2
tails). Ask them to interpret their results in terms of the laws of heredity
discovered by Mendel. Sample answer: We let H represent a dominant trait and
T a recessive trait. We interpreted every HH, HT, and TH occurrence as an
offspring displaying the dominant trait and every TT occurrence as an offspring
displaying the recessive trait. About 75% of our trials resulted in HH, HT, or TH
(offspring displaying the dominant trait), closely matching Mendel’s results.
8/26/14 12:14 PM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Independent Events 1030