Math 227 Elementary Statistics
Bluman 5th edition
CHAPTER 4
Probability and Counting Rules
2
Objectives
•
•
•
•
Determine sample spaces and find the
probability of an event using classical
probability or empirical probability.
Find the probability of compound events using
the addition rules.
Find the probability of compound events using
the multiplication rules.
Find the conditional probability of an event.
3
Objectives (cont.)
•
Determine the number of outcomes of a sequence of
events using a tree diagram.
•
Find the total number of outcomes in a sequence of
events using the fundamental counting rule.
•
Find the number of ways r objects can be selected
from n objects using the permutation rule.
•
Find the number of ways r objects can be selected
from n objects without regard to order using the
combination rule.
•
Find the probability of an event using the counting
rules.
4
Introduction
• Probability as a general concept can be defined as the
chance of an event occurring. In addition to being used in
games of chance, probability is used in the fields of
insurance, investments, and weather forecasting, and in
various areas.
• Rules such as the fundamental counting rule, combination
rule and permutation rules allow us to count the number of
ways in which events can occur.
• Counting rules and probability rules can be used together to
solve a wide variety of problems.
5
Section 4.1 Sample Spaces and
Probability
I. Basic Concepts
• A probability experiment is a chance process that leads to
well-defined results called outcomes.
• An outcome is the result of a single trial of a probability
experiment.
• A sample space is the set of all possible outcomes of a
probability experiment.
6
Sample Spaces and Events
Examples of some sample space:
Experiment
Sample Space
Toss one coin
H(head), T(tail)
Roll a die
1, 2, 3, 4, 5, 6
Answer a true/false question
True, false
Toss two coins
H-H, T-T, H-T, T-H
7
Sample Spaces and Events
• An event consists of a set of outcomes of a
probability experiment.
Simple event - an event with one outcome
Compound event - an event with two or
more outcomes.
8
Basic Concepts (cont.)
• Equally likely events are events that have
the same probability of occurring.
• A tree diagram is a device consisting of line
segments emanating from a starting point
and also from the outcome point. It is used
to determine all possible outcomes of a
probability experiment.
9
Example: A tree diagram to find the sample space
for tossing two coins.
Outcomes
H
HH
T
HT
H
TH
T
TT
H
T
10
Example 1 : A
die is tossed one time.
(a) List the elements of the sample space S.
S = {1, 2, 3, 4, 5, 6}
(b) List the elements of the event consisting of a number that is greater than 4.
Event A: x > 4
A = {5, 6}
11
Example 2 :
A coin is tossed twice. List the elements of the sample space S, and list the
elements of the event consisting of at least one head.
H
H
T
T
H
S = {HH, HT, TH, TT}
T
Event A : At least one head (H)
A = {HT, TH, HH}
12
Example 3 :
A randomly selected citizen is interviewed and the following information is
recorded: employment status and level of education. The symbols for
employment status are Y = employed and N = unemployed , and the symbols
for level of education are 1 = did not complete high school, 2 = completed
high school but did not complete college, and 3 = completed college.
List the elements of sample space, and list the elements of the following events.
1
Y
2
3
S = {Y1, Y2, Y3, N1, N2, N3}
1
N
2
3
(a) Did not complete high school
{Y1, N1}
(b) Is unemployed
{N1, N2, N3}
13
II.
Calculating Probabilities
A. Empirical Probability
(Relative Frequency Approximation of Probability)
•
Empirical probability relies on actual experience
to determine the likelihood of outcomes.
•
Given a frequency distribution, the probability of
an event being in a given class is:
frequency for the class
f
P (E ) 

total frequencies in the distribution n
14
Example 1 : The age distribution of employees for this college is shown below:
Age
Under 20
20 – 29
30 – 39
40 – 49
50 and over
# of employees
25
48
32
15
10
n = 130
If an employee is selected at random, find the probability that he or she is in the
following age groups.
(a) Between 30 and 39 years of age
(b) Under 20 or over 49 years of age
15
Example 2 :
During a sale at men’s store, 16 white sweaters, 3 red sweaters, 9 blue
sweaters, and 7 yellow sweaters were purchased. If a customer is selected at
random, find the probability that he bought a sweater that was not white.
16
B. Classical Probability
• Classical probability uses sample spaces to
determine the numerical probability that an
event will happen.
• Classical probability assumes that all outcomes
in the sample space are equally likely to occur.
n( E )
Number of outcomes in E
P( E ) 

n( S ) total number of outcomes in the sample space
17
Example 1 :
A statistics class contains 14 males and 20 females. A student is to be
selected by chance and the gender of the student recorded. (Ref: General
Statistics by Chase/Bown, 4th Ed.)
(a) Give a sample space S for the experiment.
S = {M, F}
(b) Is each outcome equally likely? Explain.
P(M) ≠ P(F) . It is not equally likely because we have more females than males.
(c) Assign probabilities to each outcome.
18
Example 2 :
Two dice are tossed. Find the probability that the sum of two dice is greater than 8?
S = { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }
19
Example set of 52 playing cards; 13 of each suit clubs,
diamonds, hearts, and spades
20
Example 3 : If one card is drawn from a deck, find the probability of getting
(a) a club
52 cards in a deck, 13 of those are clubs.
(b) a 4 and a club
21
Example 4 :
Three equally qualified runners, Mark (M), Bill (B) and Alan (A), run a 100-meter,
sprint, and the order of finish is recorded.
(a) Give a sample space S.
B
A
A
B
M
A
A
M
M
B
B
M
M
S = {MBA, MAB, BMA, BAM, AMB, ABM}
B
A
(b) What is the probability that Mark will finish last?
{BAM, ABM}
22
III. Rounding Rule for Probabilities
• Probabilities should be expressed as
reduced fractions or rounded to two or three
decimal places.
23
Probability Rules
1. The probability of an event E is a number
(either a fraction or decimal) between and
including 0 and 1. This is denoted by:
0  P (E )  1
*Rule 1 states that probabilities cannot be
negative or greater than one.
24
Probability Rules (cont.)
2. If an event E cannot occur (i.e., the event
contains no members in the sample space), the
probability is zero.
3. If an event E is certain, then the probability of E
is 1.
4. The sum of the probabilities of the outcomes in
the sample space is 1.
25
Example 1: A probability experiment is conducted. Which
of these can be considered a probability of an
outcome?
a)
2/5
Yes
b) -0.28
No
c)
No
1.09
26
Example 2 :
Given :
S = {E1, E2, E3, E4}
P(E1) = P(E2) = 0.2 and P(E3) = 0.5
Find :
P(E4)
P(E1) + P(E2) + P(E3) +P(E4) = 1
0.2 + 0.2 + 0.5 +P(E4) = 1
P(E4) = 1 – 0.9
P(E4) = 0.1
27
IV. Complementary Events
• The complement of an event E is the set of
outcomes in the sample space that are not
included in the outcomes of event E. The
complement of E is denoted by E .
• Rule for Complementary Events
P( E )  1  P( E )
P (E )  1  P (E ),
OR
P( E )  P( E )  1
28
Example 1 : The chance of raining
tomorrow is 70%. What is the probability
that it will not rain tomorrow?
P (No Rain) = 1 – 0.70
= 0.3
30% chance it will not rain tomorrow.
29
Section 4.2 The Addition Rules for
Probability
Mutually Exclusive Events
• Two events are mutually exclusive if they cannot
occur at the same time (i.e., they have no
outcomes in common).
• The probability of two or more events can be
determined by the addition rules.
30
I. Addition Rules
• Addition Rule 1—When two events A and
B are mutually exclusive, the probability
that A or B will occur is:
P ( A or B )  P ( A )  P (B )
31
Ex1) A single card is drawn from a deck. Find the
probability of selecting a club or a diamond.
P (club or diamond)  mutually exclusive
P (club or diamond) = P (club) + P (diamond)
13

52

13
52
26 1


52 2
32
Ex2) In a large department store, there are 2 managers, 4
department heads, 16 clerks, and 4 stock persons. If a
person is selected at random, find the probability that the
person is either a clerk or a manager.
P (clerk or manager)  mutually exclusive
P (clerk or manager) = P (clerk) +
16

26

P (manager)
2
26
18 9


26 13
33
I. Addition Rules (cont.)
• Addition Rule 2—If A and B are not
mutually exclusive, then:
P ( A or B )  P ( A )  P (B )  P ( A and B )
34
Example1 : A single card is drawn from a deck. Find
the probability of selecting a jack or a black card.
P (Jack or Black) = P (Jack) + P (Black) – P (Jack and Black)
35
Example 2 :
In a certain geographic region, newspapers are classified as being published
daily morning, daily evening, and weekly. Some have a comics section and
other do not. The distribution is shown here.
Having comics Section
Morning
Evening
Weekly
Yes
2
3
1
No
3
4
2
If a newspaper is selected at random, find these probabilities.
(a) The newspaper is weekly publication.
S = 2 + 3 + 1 + 3 + 4 +2 = 15
(b) The newspaper is a daily morning publication or has comics.
P (Daily morning or has comics) = P (Daily morning) + P (Has comics)
– P (Daily morning and has comics)
36
(c) The newspaper is published weekly or does not have comics.
P (Weekly or No comics) = P (Weekly) + P (No comics)
– P (Weekly and no comics)
37
Section 4.3 The Multiplication
Rules and Conditional Probability
Independent Events
Two events A and B are independent if the
fact that A occurs does not affect the
probability of B occurring.
When the outcome or occurrence of the first
event affects the outcome or occurrence of
the second event in such a way that the
probability is changed, the events are said to
be dependent.
38
I. Multiplication Rules
• The multiplication rules can be used to find
the probability of two or more events that
occur in sequence.
• Multiplication Rule 1—When two events are
independent, the probability of both
occurring is:
P ( A and B )  P ( A )  P (B )
39
Example 1 : If 36% of college students are overweight, find
the probability that if three college students are selected
at random, all will be overweight.
P (1st overweight and 2nd overweight and 3rd overweight)
= 0.36 · 0.36 · 0.36
= 0.046656
≈ 0.047
40
Example 2 :
If 25% of U.S. federal prison inmates are not U.S. citizens, find the probability that
two randomly selected federal prison inmates will be U.S. citizens.
P (1st U.S. citizen and 2nd U.S. citizen) = 0.75 · 0.75
= 0.5625
≈ .563
Example 3 :
Suppose the probability of remaining with a particular company 10 years or longer is
1/6. A man and a woman start work at the company on the same day.
(a) What is the probability that the man will work there less than 10 years?
(b) What is the probability that both the man and woman will work there less
than 10 years?
41
Example 4 :
A smoke-detector system uses two devices, A and B. If smoke is present, the
probability that it will be detected by device A is 0.95; by device B, 0.98; and by both
devices, 0.94.
(a) If smoke is present, find the probability that the smoke will be detected by
device A or device B or by both devices.
P ( A or B) = P (A ) + P (B) – P (A and B)
= 0.95 + 0.98 – 0.94
= 0.99
(b) Find the probability that the smoke will not be detected.
P (Neither ) =1 – 0.99
= 0.01
42
Example 5 :
If you make random guesses for four multiple-choice test questions
(each with five possible answers), what is the probability of getting at
least one correct?
P (All wrong) = P (1st wrong & 2nd wrong & 3rd wrong & 4th wrong )
P (At least 1 correct) = 1 – P (All wrong)
43
Example 6 :
There are 2000 voters in a town. Consider the experiment of randomly selecting a
voter to be interviewed. The event A consists of being in favor of more stringent
building codes; the event B consists of having lived in the town less than 10
years. The following table gives the number of voters in various categories. (Ref:
General statistics by Chase/Bown, 4th Ed.)
Favor more
Stringent codes
Less than 10 years
At least 10 years
Do not favor more
stringent codes
100
700
1000
200
Find each of the following:
(a)
(b)
(c)
44
Multiplication Rules (cont.)
Multiplication Rule 2—When two events are
dependent (e.g. the outcome of event A
affects the outcomes of event B), the
probability of both occurring is:
P ( A and B )  P ( A )  P (B|A )
45
Example 1 :
A box has 5 red balls and 2 white balls. If two balls are randomly selected (one
after the other), what is the probability that they both are red?
(a) With replacement → independent case
(b) Without replacement → dependent case
46
Example 2 :
Three cards are drawn from a deck without replacement. Find the probability
that all are Jacks.
47
Multiplication Rules (cont.)
• The conditional probability of an event B in
relationship to an event A is the probability
that event B occurs after event A has
already occurred. The notation for
conditional probability is P(B|A).
48
Formula for Conditional
Probability
• The probability that the second event B occurs
given that the first event A has occurred can be
found dividing the probability that both events
occurred by the probability that the first event has
occurred. The formula is:
P ( A and B )
P (B|A ) 
P(A)
49
Example 1: Two fair dice are tossed. Consider the following events.
A = sum is 7 or more,
B = sum is even,
and C = a match (both numbers are the same).
A = { (1,6) (2,5) (2,6) (3,4) (3,5) (3, 6) (4, 3)
(4,4) (4,5) (4,6) (5,2) (5,3) (5, 4) (5, 5)
(5,6) (6,1) (6,2) (6,3) (6,4) (6, 5) (6, 6) }
B = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5)
(4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6) }
C = { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) }
50
(a) P (A and B) =
A = { (1,6) (2,5) (2,6) (3,4) (3,5) (3, 6) (4, 3)
(4,4) (4,5) (4,6) (5,2) (5,3) (5, 4) (5, 5)
(5,6) (6,1) (6,2) (6,3) (6,4) (6, 5) (6, 6) }
B = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5)
(4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6) }
(b) P (A or B) = P (A) + P (B) – P (A and B)
51
(c) P (A | B) =
(d) P (B | C) =
Example 2 :
At a local Country Club, 65% of the members play bridge and swim, and 72%
play bridge. If a member is selected at random, find the probability that the
member swims, given that the member plays bridge.
52
Example 3 :
Eighty students in a school cafeteria were asked if they favored a ban on
smoking in the cafeteria. The results of the survey are shown in the table.
Class
Favor
Oppose
No opinion
Freshman
15
27
8
Sophomore
23
5
2
If a student is selected at random, find these probabilities.
(a) The student is a freshman or favors the ban.
P (Freshman or Favor ban) = P (Freshman) + P (Favor ban)
– P (Freshman and Favor ban)
(b) Given that the student favors the ban, the student is a sophomore.
53
Section 4.4 Counting Rules
I. The Fundamental Counting Rule
In a sequence of n events in which the first one has k1
possibilities and the second event has k2 and the third
has k3, and so forth, the total number of possibilities of
the sequence will be:
k1  k2  k3      kn
Note: “And” in this case means to multiply.
54
Example 1 : Two dice are tossed. How many outcomes are in S.
How many outcomes for Task 1?
6
·
How many outcome for Task 2?
6
=
36 outcomes
Example 2 :
A password consists of two letters followed by one digit. How many different
passwords can be created? (Note: Repetitions are allowed)
1st letter → 26 outcomes (A – Z)
2nd letter → 26 outcomes (A – Z)
One digit → 10 outcomes (0-9)
26
·
26
·
10
=
6,760 ways
55
Example 3 :
Suppose four digits are to be randomly selected (with repetitions
allowed). (Note: the set of digits is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.) If the
first digit must be a 2 and repetitions are permitted, how many
different 4-digit can be made?
1st digit → has to be a 2 → 1 outcome
2nd digit → any number (0 – 9) →10 outcomes
3rd digit → any number (0 – 9)→10 outcomes
4th digit → any number (0 – 9) →10 outcomes
1
·
10
·
10
·
10
=
1,000 ways
56
II. Permutations and Combinations
• Factorial Notation
5! = 5·4·3·2·1
7! = 7·6·5·4·3·2·1
In general,
n! = n · (n-1) ·(n-2) ·(n-3) ···· 1
0!=1
57
Permutation
• A permutation is an arrangement of n objects in
a specific order.
The arrangement of n objects in a specific order
using r objects at a time is called a permutation
of n objects taking r objects at a time.
Note: Order does matter.
It is written as nPr, and the formula is:
n!
n Pr 
(n  r )!
58
Combinations
• A selection of distinct objects without regard to
order is called a combination.
(Order does NOT matter!)
• The number of combinations of r objects selected
from n objects is denoted nCr and is given by the
formula:
n!
n Cr 
(n  r )!r !
Note: Combinations are always less than permutations for the
same n and r.
59
Example 1 :
(a) 5!
5 · 4 · 3 · 2 · 1 = 120
(b) 25P8
(c) 12P4
(d) 25C8
60
Example 2 :
A television news director wishes to use three news stories on an evening
show. One story will be the lead story, one will be the second story, and the
last will be a closing story. If the director has a total of eight stories to choose
from, how many possible ways can the program be set up?
Example 3 :
If a person can select 3 presents from 10 presents, how many different
combinations are there?
61
Example 4 :
Your family vacation involves a cross-country air flight, a rental car, and a hotel
stay in Boston. If you can choose from four major air carriers, five car rental
agencies, and three major hotel chains, how many options are available for your
vacation accommodations?
Multiplication Rule
4
·
5
·
3
=
60 ways
62
Section 4.5 Probability and
Counting Rules
I. Basic Concepts
By using the fundamental counting rule, the
permutation rules, and the combination rule,
the probability of outcomes of many
experiments can be computed.
63
Example 1 :
A combination lock consists of 26 letters of the alphabet. If a three-letter
combination is needed, find the probability that the combination will consist of the
first two letters AB in that order. The same letter can be used more than once.
64
Example 2 :
Five cards are selected from a 52-card deck for a poker hand. (A poker hand
consists of 5 cards dealt in any order.)
(a) How many outcomes are in the sample space?
(b) A royal flush is a hand that contains that A, K, Q, J, 10, all in the same suit.
How many ways are there to get a royal flush?
A, K, Q, J, 10 - ♠
A, K, Q, J, 10 -♥
A, K, Q, J, 10 - ♣
A, K, Q, J, 10 - ♦
n(A) = 4 combinations
(c) What is the probability of being dealt a royal flush?
65
Example 4-52: Committee Selection
A store has 6 TV Graphic magazines and 8 Newstime
magazines on the counter. If two customers purchased a
magazine, find the probability that one of each magazine
was purchased.
TV Graphic: One magazine of the 6 magazines
Newstime: One magazine of the 8 magazines
Total: Two magazines of the 14 magazines
6
Bluman, Chapter 4
C1 8 C1 6  8 48


91
91
14 C2
66
Summary
• The three types of probability are classical,
empirical, and subjective.
• Classical probability uses sample spaces.
• Empirical probability uses frequency
distributions and is based on observations.
• In subjective probability, the researcher
makes an educated guess about the chance
of an event occurring.
67
Summary (cont)
• An event consists of one or more outcomes
of a probability experiment.
• Two events are said to be mutually exclusive
if they cannot occur at the same time.
• Events can also be classified as
independent or dependent.
• If events are independent, whether or not the
first event occurs does not affect the
probability of the next event occurring.
68
Summary (cont.)
• If the probability of the second event
occurring is changed by the occurrence of
the first event, then the events are
dependent.
• The complement of an event is the set of
outcomes in the sample space that are not
included in the outcomes of the event itself.
• Complementary events are mutually
exclusive.
69
Summary
The number of ways a sequence of n events can occur; if the first event can
occur in k1 ways, the second event can occur in k2 ways, etc.
(Multiplication rule)
k1  k2  k3      kn
The arrangement of n objects in a specific order using r objects at a time
(Permutation rule)
n!
n Pr 
(n  r )!
The number of combinations of r objects selected from n objects (order is not
important)
(Combination rule)
n!
n Cr 
(n  r )!r !
70
Conclusions
• Probability can be defined as the chance of
an event occurring. It can be used to
quantify what the “odds” are that a specific
event will occur. Some examples of how
probability is used everyday would be
weather forecasting, “75% chance of snow”
or for setting insurance rates.
71
Conclusions
• A tree diagram can be used when a list of all
possible outcomes is necessary. When only
the total number of outcomes is needed, the
multiplication rule, the permutation rule, and
the combination rule can be used.
72