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Chapter 5
Probability and Random Variables
Slide 5-2
Chapter 5 Section 1 – Probability Basics
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. There are three types: Classical, Empirical,
and Subjective.
Classical is more theoretical.
Empirical is obtained through observations.
Subjective is based on inexact methods such as
guesswork or opinion
Slide 5-3
Rounding Rule for Probabilities

Probabilities should be expressed as reduced fractions
or rounded to two or three decimal places.

When the probability of an event is an extremely
small decimal, it is permissible to round the decimal to
the first nonzero digit after the decimal point.
Slide 5-4
Experiment – is an action whose outcome cannot be
predicted with certainty
An outcome is the result of a single trial of a probability
experiment
Event – is some specified result that may or may not occur
when an experiment is performed.
Slide 5-5
Empirical Probability

Empirical probability relies on actual experience (or
observations) to determine the likelihood of outcomes.

Empirical is obtained through observations.

Given a frequency distribution, the probability of an event being
in a given class is:
Empirical Exp: The chance depends on what you observed on several
tosses of a coin. Flip coin 4 times…heads 1 out of 4 therefore ¼.
Slide 5-6
When two balanced dice are rolled, 36 equally likely
outcomes are possible:
The sum of the dice can be 11 in two ways. The probability the
sum is 11 is f/N = 2/36 or 0.056.
Doubles can be rolled in six ways. The probability of doubles is
f/N = 6/36 or 0.167.
Contains a three. The probability contains a three is f/N = 11/36
or 0.306.
Slide 5-7
2.
If an event E cannot occur (i.e., the event
contains no members in the sample space), the
probability is zero.
Example: Rolling a 7 on a die.
3.
0
P (7 )   0
6
If an event E is certain, then the probability of E is 1.
Example: rolling a number less than 7 on a die.
6
P(number  7)   1
6
4.
The sum of the probabilities of all the outcomes in the
sample space is 1.
Slide 5-8
5.8
Oklahoma State Officials:
Governor
G
Lieutenant Governor
L
Secretary of State
S
Attorney General
A
Treasurer
T
a) List the possible samples without
replacement of size 3 that can
be obtained from the population
of five officials
G,L,S
G,S,T
L,A,T
G,L,A
G,A,T
S,A,T
G,L,T
L,S,A
G,S,A
L,S,T
If a simple random sample without replacement of three
officials is taken from the five officials, determine the
probability that the Governor, Attorney General, and
Treasurer are obtained.
One sample includes the governor, attorney general, and
treasurer. Therefore the probability is f/N = 1/10 or 0.1
Slide 5-9
5.8
Oklahoma State Officials:
Governor
G
Lieutenant Governor
L
Secretary of State
S
Attorney General
A
Treasurer
T
a) List the possible samples without
replacement of size 3 that can
be obtained from the population
of five officials
G,L,S
G,S,T
L,A,T
G,L,A
G,A,T
S,A,T
G,L,T
L,S,A
G,S,A
L,S,T
If a simple random sample without replacement of three
officials is taken from the five officials, determine the
probability that the Governor, and Treasurer are included in
the sample.
Three samples included the governor and treasurer.
Therefore, the probability is f/N = 3/10 or 0.3
Slide 5-10
5.8
Oklahoma State Officials:
Governor
G
Lieutenant Governor
L
Secretary of State
S
Attorney General
A
Treasurer
T
a) List the possible samples without
replacement of size 3 that can
be obtained from the population
of five officials
G,L,S
G,S,T
L,A,T
G,L,A
G,A,T
S,A,T
G,L,T
L,S,A
G,S,A
L,S,T
If a simple random sample without replacement of three
officials is taken from the five officials, determine the
probability that the Governor, is included in the sample.
Six samples include the governor. Therefore, the probability
is f/N = 6/10 or 0.6
Slide 5-11
Property 1 - states that probabilities cannot be negative or greater than
one.
Property 2 - If an event E cannot occur (i.e., the event contains no
members in the sample space), the probability is zero.
Example: Rolling a 7 on a die. P (7)  0  0
6
Property 3 - If an event E is certain, then the probability of
E is 1.
Example: rolling a number less than 7 on a die.
P(number  7) 
6
1
6
Slide 5-12
Chapter 5 Section 2 - Events
Slide 5-13
Venn diagrams are used to represent probabilities pictorially
Figure 5.9
Slide 5-14
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.
Slide 5-15
Figure 5.14
Two mutually
exclusive events
Two non-mutually
exclusive events
Figure 5.15
Three mutually
exclusive events
Three non-mutually
exclusive events
Three non-mutually
exclusive events
Slide 5-16
Addition Rules

Addition Rule 1—When two events A and B are mutually
exclusive, the probability that A or B will occur is:

Addition Rule 2—If A and B are not mutually exclusive, then:
Slide 5-17
EXAMPLES
Determine whether these event are mutually exclusive
(cannot occur at the same time).
a)
Roll a die: Get an even number, and get a number less than 3.
NO – can occur at the same time. Can get a 2.
b)
Roll a die: Get a prime number (2,3,5) and get an odd number.
NO – can occur at the same time. Can get a 3 or 5.
c)
Roll a die: Get a number greater than 3, and get a number less than 3.
YES – cannot occur at the same time.
d)
Select a student in class: The student has blond hair, and the student has
blue eyes.
NO – can occur at the same time. A student can have blond hair and blue
eyes.
Slide 5-18
EXAMPLES
Determine whether these event are mutually exclusive
(cannot occur at the same time).
e)
Select a student at Rio: The student is a sophomore, and the student is a
business major.
NO – can occur at the same time. A student can be a Sophomore with a
business major.
f)
Select any course: It is a calculus course, and it is an English course.
YES – cannot occur at same time. A course is one or the other not both.
g)
Selected a registered voter: The voter is a Republican, and the voter is a
Democrat.
YES – cannot occur at same time. A voter is one or the other not both.
Slide 5-19
Chapter 5 Section 3 – Some Rules of Probability
Probability Notation
If E is an event, then P(E) represents the probability that
event E occurs. It is read “the probability of E.”
Slide 5-20

Addition Rule 2
If A and B are not mutually exclusive
Slide 5-21
Examples
Roll a six-sided die. What is the probability of rolling
1 = 1/6
2 = 1/6
3 = 1/6
4 = 1/6
5 = 1/6
6 = 1/6
P(roll a number > 4) = P(roll 5 or roll 6)
= P(roll 5) + P(roll 6)
= 1/6 + 1/6 = 2/6 or 1/3
P(roll a number ≤ 4) = P(roll 1 or roll 2 or roll 3 or roll 4)
= P(roll 1) + P(roll 2) + P(roll 3) + P(roll 4)
= 4/6 or 2/3
How are the statements above related?
P(roll a number ≤ 4) + P(roll a number > 4) = 1
Slide 5-22
EXAMPLES
An automobile dealer had 10 Fords, 7 Buicks, and 5 Plymouths on her used car
lot. If a person purchased a used car, find the probability that it is a Ford or
Buick.
Total = 10 + 7 + 5 = 22
P(Ford) = 10/22
P(Buick) = 7/22
P(Ford or Buick) = P(Ford) + P(Buick) =
10/22 + 7/22 = 17/22
The probability that a student owns a car is 0.65, and the probability that a student
owns a computer is 0.82. If the probability that a student owns both is 0.55, what
is the probability that a given student owns neither a car nor a computer?
P(car or computer) = 0.65 + 0.82 – 0.55 = 0.92
P(neither) = 1 – 0.92 = 0.08
Slide 5-23
EXAMPLES
A single card is drawn from a deck. Find the probability of
selecting the following:
a) A 4 or a diamond.
P(4 or diamond) = P(4) + P(diamonds) – P(4 of diamonds)
= 4/52 +13/52 – 1/52 = 16/52 or 4/13
There are four 4’s and 13 diamonds, but the 4 of
diamonds is counted twice.
b) A club or a diamond.
P(club or diamond) = 13/52 + 13/52 = 26/52 or ½
c) A jack or a black card.
P(jack or black) = 4/52 +26/52 – 2/52 = 28/52 or 7/13
Slide 5-24
EXAMPLES
In a certain geographic region, newspapers are classified as
being published daily morning, daily evening, and weekly.
Some have a comic section and some do not. The distribution is
shown here.
Have Comic Section
Morning
Evening
Weekly
TOTAL
Yes
2
3
1
6
No
3
4
2
9
Total
5
7
3
15
If a newspaper is selected at random, find these probabilities.
a) The newspaper is a weekly publication.
P(weekly) = 3/15 or 1/5
Slide 5-25
EXAMPLES
In a certain geographic region, newspapers are classified as being
published daily morning, daily evening, and weekly. Some have a comic
section and some do not. The distribution is shown here.
Have Comic Section
Morning
Evening
Weekly
TOTAL
Yes
2
3
1
6
No
3
4
2
9
Total
5
7
3
15
If a newspaper is selected at random, find these probabilities.
b) The newspaper is a daily morning publication or has comics.
P(morning or has comics) =
P(morning) + P(has comics) – P(morning with comics) =
5/15 + 6/15 - 2/15 = 9/15 or 3/5
Slide 5-26
EXAMPLES
In a certain geographic region, newspapers are classified as being published
daily morning, daily evening, and weekly. Some have a comic section and
some do not. The distribution is shown here.
Have Comic Section
Morning
Evening
Weekly
TOTAL
Yes
2
3
1
6
No
3
4
2
9
Total
5
7
3
15
If a newspaper is selected at random, find these probabilities.
c) The newspaper is published weekly or it does not have comics.
P(weekly or no comics) =
P(weekly) + P(no comics) – P(weekly with no comics) =
3/15 + 9/15 – 2/15 = 10/15 or 2/3
Slide 5-27
Module P3
The Multiplication Rule: Independence
Independent Events
Two events A and B are independent if the
fact that A occurs does not affect the
probability of B occurring.
Slide 5-28
Multiplication Rule 1

There are two multiplication rules they can be used
to find the probability of two or more events that
occur in sequence. Such as roll a die and toss a coin
at the same time.

Multiplication Rule 1—When two events are
independent, the probability of both occurring is:
Slide 5-29
Example -- Independent
A washer and dryer have been purchased; both are under
warranty.
outcome 1 -- the washer needs service
outcome 2 -- the dryer needs service
Does knowing that the washer needs service affect
the likelihood that the dryer needs service? NO
What is the probability that both the washer and the dryer
need service while under warranty?
P(washer needs service and dryer needs service) =
P(washer needs service) * P(dryer needs service)
Slide 5-30
Examples - - independent event
1.
Tossed a coin twice, the probability of getting two heads is ½ * ½ = ¼
NOTE: The sample space would be. HH, HT, TH, TT; then P(HH) = ¼
2.
Toss a coin and roll a die. What is the probability of getting heads on the
coin and a 4 on the die.
P(H and 4) = P(H) * P(4) = 1  1  1
2 6 12
NOTE: The sample space for coin is H, T and for Die is 1, 2, 3, 4, 5, 6. The
sample space for both is H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6;
then P(head and 4) = 1
12
3.
A urn contains 3 red balls, 2 blue balls, and 5 white balls.
Find probability of selecting 2 blue balls.
P(B and B) = P(B) * P(B) = 2/10 * 2/10 = 4/100 = 1/25
selecting 1 blue and 1 white.
P(B and W) = P(B) * P(W) = 2/10 * 5/10 = 10/100 = 1/10
selecting 1 red and 1 blue.
P(R and B) = P(R) * P(B) = 3/10 * 2/10 = 6/100 = 3/50
Slide 5-31
Dependent Events

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 events.
Slide 5-32
Multiplication Rule 2

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). Does NOT mean
divide it means the probability that event B occurs given
that event A has already occurred.

Multiplication Rule 2—When two events are dependent,
the probability of both occurring is:
Slide 5-33
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:
Slide 5-34
Example - Dependent
Select a card from a deck of 52 cards. What is the
likelihood of selecting an ace? P(ace) = 4/52
Do not replace the first card. Select another card. What
is the likelihood of selecting an ace? P(ace) = 4/51
The second outcome depends on the first.
What is the likelihood of selecting an ace knowing that
the first was an ace?
P(ace|ace) = 3/51
What is the likelihood of selecting two aces if the first and
second cards are not replaced? P(two aces) = 4/50
Slide 5-35
Tree Diagram

A tree diagram is a device used to list
all possibilities of a sequence of events
in a systematic way.

Can be used for independent or
dependent and can also be used for
sequences of three of more events.
Slide 5-36
Slide 5-37
EXAMPLE
If 18% of all Americans are underweight, find
the probability that if three Americans are
selected at random, all will be underweight.
P(all three underweight)
= 0.18 * 0.18 * 0.18 = (0.18)3 = 0.005832 or 0.6%
Slide 5-38
EXAMPLE
If 25% of U.S. federal prison inmates are not
U.S. citizens, find the probability that two
randomly selected federal prison inmates will
not be U.S. citizens.
P(two inmates are not citizens)
= 0.25 * 0.25 = (0.25)2 = 0.0625 or 6.3%
Slide 5-39
EXAMPLES
A Flashlight has six batteries, two of which
are defective. If two are selected at random
without replacement, find the probability that
both are defective.
2 1 2
1
or
P(both are defective) =  
6 5 30 15
Slide 5-40
EXAMPLES
In a scientific study there are 8 guinea pigs, 5 of which are
pregnant. If three are selected at random without
replacement, find the probability that all are pregnant.
5
4
3
5



P(all are pregnant) =
8 7
6
28
Find the probability that none are pregnant.
3 2 1
6
1



or
P(none are pregnant) =
8 7
6
336
56
Slide 5-41
EXAMPLES
An automobile manufacturer has three factories A, B, and C.
They produce 50%, 30%, and 20%, respectively, of a specific
model of car.
30% of the cars produced in factory A are white,
40% of those produced in factory B are white and
25% produced in factory C are white.
If an automobile produced by the company is selected at random,
find the probability that it is white.
Slide 5-42
0.3
0.5
0.3
0.2
A
B
C
0.7
Factory
W
(0.5)(0.3) = 0.15
0.4
NW
W
(0.3)(0.4) = 0.12
0.6
0.25
NW
W
(0.2)(0.25) = 0.05
0.75
NW
P(white) = 0.15 + 0.12 + 0.05 = 0.32
Slide 5-43
EXAMPLES
In a pizza restaurant, 95% of the
customers order pizza. If 65% of the
customers order pizza and a salad, find
the probability that a customer who
orders pizza will also order a salad.
0.65
P(pizza|salad) =
 0.684 or 68.4%
0.95
Slide 5-44
EXAMPLES
At a teachers’ conference, there were 4
English teachers, 3 mathematics teachers, and
5 science teachers. If 4 teachers are selected
for a committee, find the probability that at
least one is a science teacher.
4+3+5 = 12
P(at least one science) = 1 – P(no science) =
7 6 5 4
840
7 92
1     1
 1

12 11 10 9
11880
99 99
Slide 5-45
EXAMPLES
Eighty students in a school cafeteria were asked if they
favored a ban on smoking in the cafeteria. The results
are shown in the table.
Class
Favor
Oppose
No Opinion
Total
FR
15
27
8
50
SOPH
23
5
2
30
TOTAL
38
32
10
80
If a student is selected at random, find these
probabilities:
a) Given that the student is a freshman, he or she opposes
the ban.
27 50 27 80 2160 27





P(fr and oppose) =
80
80
80 50
4000
50
Slide 5-46
EXAMPLES
Eighty students in a school cafeteria were asked if they
favored a ban on smoking in the cafeteria. The results are
shown in the table.
Class
Favor
Oppose
No Opinion
Total
FR
15
27
8
50
SOPH
23
5
2
30
TOTAL
38
32
10
80
If a student is selected at random, find these probabilities:
b) Given that the student favors the ban, the student is a
sophomore.
23 38 23 80 1840 23
   

P(favor and soph) =
80 80
80 38
3040
38
Slide 5-47
Module P3 - The Counting Rule
Fundamental Counting Rule

Many times we wish to know the number of outcomes for a
sequence of events. We can use the fundamental counting rule to
determine this number.

The fundamental counting rule can be used to determine the total
number of outcomes in a sequence of events.

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:

Note: “And” in this case means to multiply.
Slide 5-48
Consider the digits 0,1,2,3, and 4. If they are used on a
four-digit ID card, How many different cards are possible?
(5 * 5 * 5 * 5) = 54 = 625
How does the fundamental counting rule apply here?
Since there are 4 spaces to fill and 5 digits for each
space:
k1  k2  k3  kn = 5 * 5 * 5 * 5
In this example, repetition is allowed. If repetition is
allowed then number stays the same from left to right. If
not allowed then the number decreases by 1 from left to
right.
If not allowed then we would have the first digit can be
chosen 5 ways, but the second digit can be chosen 4 ways,
since there are only four digits left, etc.
5 * 4 * 3 * 2 = 120
Slide 5-49
EXAMPLE
The call letters of a radio station must have 4 letters.
The first letter must be a K or a W. How many
different station call letters can be made if repetitions
are not allowed? Note: 26 letters in the alphabet.
2 * 25 * 24 * 23 = 27,600
If repetitions are allowed?
2 * 26 * 26 * 26 = 35,152
Slide 5-50
Permutations

In order to understand the permutation and combination rules, we
need a special definition of 0!
Factorial Formula for any counting n is:
n! n(n  1)(n  2)......1
0! 1

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. It is written as nPr, and the formula is:
Slide 5-51
Combinations

A selection of distinct objects without regard to
order is called a combination.

The number of combinations of r objects selected
from n objects is denoted nCr and is given by the
formula:
Slide 5-52
Permutation:
Suppose an organization has 4 members: A, B, C, D.
How many ways can a president and a VP be selected? In
other words, select 2 from 4 when order is important.
Combination:
How many ways can a committee of two be selected? In
other words, select 2 from 4 when order is not important.
Slide 5-53
EXAMPLES
In a board of directors composed of 8 people.
How many ways can 1 chief executive officer, 1
director, and 1 treasurer be selected?
Which rule will we need to use?
Permutation
8!
8  7  6  5  4  3  2  1 8  7  6  5  4  3  2 1 40320



 336
8 P3 
(8  3)!
5!
5  4  3  2 1
120
Slide 5-54
EXAMPLES
If a person can select three presents form 10
presents under a Christmas tree.
How many different combinations are there?
Which rule will we need to use?
Combination
10! 10  9  8  7! 720


 120
10 C3 
7!3!
7!3  2  1
6
Slide 5-55
EXAMPLES
In a train yard there are 4 tank cars, 12 boxcar, and 7 flatcars. How many
ways can a train be made up consisting of 2 tank cars, 5 boxcars, and 3
flatcars? (In this case order is not important.)
We need to use the
combination rule.
4
C2 12 C5 7 C3 
4!
12!
7!



(4  2)!2! (12  5)!5! (7  3)!3!
4  3  2! 12  11  10  9  8  7! 7  6  5  4!



2!2  1
7!5  4  3  2  1
4!3  2  1
12 95,040 210



2
120
6
6  792  35 
166,320
Slide 5-56
EXAMPLES
There are 7 women and 5 men in a department store.
How many way can a committee of 4 people be selected?
12!
12 11 10  9  8! 11880


 495
12 C4 
(12  4)!4!
8!4  3  2  1
24
How many ways can this committee be selected if there must be 2 men and 2 women
on the committee?
7 C2 5 C 2 
7!
5!
7! 5! 7  6  5! 5  4  3! 42 20 840





  
 210
(7  2)!2! (5  2)!2! 5!2! 3!2! 5!2 1 3!2 1 2 2
4
How many ways can this committee be selected if there must be at least 2 women on
the committee?
7
C2 5 C2  7 C3 5 C1  7 C4 
7!
5!
7!
5!
7!




(7  2)!2! (5  2)!2! (7  3)!3! (5  1)!1! (7  4)!4!
7  6  5! 5  4  3! 7  6  5  4! 5  4! 7  6  5  4  3!






5!2  1
3!2  1
4!3  2  1 4!1
3!4  3  2  1
42 20 210 5 840


 
 21  10  35  5  35  210  175  35  420
2
2
6
1
24
Slide 5-57
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.
Slide 5-58
Summary (cont’d.)

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.
Slide 5-59
Summary (cont’d.)

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.
Slide 5-60
Summary (cont’d.)
Rule
Definition
Multiplication rule
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.
Permutation rule
The arrangement of n objects in a
specific order using r objects at a time
Combination rule
The number of combinations of r
objects selected from n objects (order
is not important)
Slide 5-61
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.
Slide 5-62
Conclusions (cont’d)

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.
Slide 5-63