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Chapter 4
Probability Concepts
4.1 Events and Probability
 Three Helpful Concepts in Understanding Probability:
 Experiment
 Sample Space
 Event
 Experiment
 An activity for which the outcome is uncertain is an experiment.
 Example 4.1.1: Examples of experiments
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Flipping a coin
Rolling two dice
Taking an exam
Observing the number of arrivals at a drive-up window over a 5-minute
period
4.1 Events and Probability (cont.)
 Sample Space
 The list of all possible outcomes of an experiment is called the sample
space.
 Example 4.1.2: Example of sample space
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Flipping a coin twice results in one of four possible outcomes. These
possible outcomes are HH, HT, HT, TT. Therefore, sample space = {HH,
HT, TH, TT}.
If there are n outcomes of an experiment, sample space lists all n
outcomes.
4.1 Events and Probability (cont.)
 Event
 An event consists of one or more possible outcomes of the experiment.
 It is usually denoted by a capital letter.
 Example 4.1.3: Examples of experiments and some corresponding events
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Experiment: Rolling two dice; events: A = rolling a total of 7, B = rolling a
total greater than 8, C = rolling two 4s.
Experiment: Taking an exam; events: A = pass, B = fail.
Experiment: Observing the number of arrivals at a drive-up window over a
5-minute period; events: A0 = no arrivals, A1 = seven arrivals, etc.
4.1 Events and Probability (cont.)
 Probability
 A numerical measure of the chance OR likelihood that a particular event
will occur.
 The probability that event A will occur is written P(A).
 The probability of any event ranges from 0 to 1, inclusive.
 P(A) = 0 means event A will never occur.
 P(A) = 1 means event A must occur.
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4.1 Events and Probability (cont.)
 How to come up with probability?
 Classical Definition of Probability

If event A occurs in m of the n outcomes in an experiment, then the
probability that event A will occur is:
m
P ( A) 
n
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This assumes all n possible outcomes have an equal chance of occurring.
Example 4.1.4: Toss a nickel and a dime. The sample space (i.e., the list of the
possible outcomes) is {HH, HT, TH, TT}. If event A is observing one head
and one tail, then m = 2 and n = 4. So according to classical definition of
probability, P(A) = m/n = 2/4 = 0.5.
4.1 Events and Probability (cont.)
 Relative Frequency Approach
 Observe an experiment n times and count the number of times event A occurs, m.
P ( A) 

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m
n
Example 4.1.5: A production process has been in operation in for 250 days and has
been accident-free for 220 days. If event A is a randomly chosen accident-free day
in the future, then, according to relative frequency approach, P(A) = 220/250 =
0.88.
 Subjective Probability
 A measure (between 0 and 1) of your belief that a particular event will occur.
 Example 4.1.6: Example of subjective probability: The probability that it will rain
today is 50%.
4.2 Basic Concepts
 Contingency Table (also called Cross-Tab Table)
 Contingency tables are used to record and analyze the relationship between two
variables.
 Example 4.2.1: Datacomp Survey: Datacomp recently conducted a survey of 200
selected purchasers of their newly introduced laptop computer to obtain a genderand-age profile of its new customers. The data are summarized in the following
contingency table.
Age (Years)
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< 30
30 - 45
> 45
Sex
(U)
(B)
(O)
Total
Male (M)
60
20
40
120
Female (F)
40
30
10
80
Total
100
50
50
200
Some Events:
M = a male is selected
F = a female is selected
U = the person selected is under 30
B = the person selected is between 30 & 45
O = the person selected is over 45
4.2 Basic Concepts (cont.)
 Contingency Table (also called Cross-Tab Table)
 Example 4.2.2: At a local University 75% of the Business faculty are
professors and 70% of the faculty are full time. 80% of the professors
are full time. Suppose the faculties are randomly assigned to courses.
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If you take a course in the Business School, what is the probability that you
will get a Professor for the course?
What is the probability that a teacher selected at random is a Professor AND
is Full Time?
What is the probability that a teacher chosen at random is Not a Professor
OR is Not Full Time?
4.2 Basic Concepts (cont.)
 Marginal Probability
 Marginal probability is the probability of one event, regardless of the
other events.
 Example 4.2.2: In Datacomp Survey, the marginal probabilities are:
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P(M) = 120/200 = 0.6
P(F) = 80/200 = 0.4
P(U) = 0.5
P(B) = 0.25
P(O) = 0.25
4.2 Basic Concepts (cont.)
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 Complement of an event
 The complement of an event A is the event that A does not occur.
 This event is denoted by A.
 For example, A = it rains tomorrow, A = it does not rain tomorrow.
 Example 4.2.3: In Datacomp Survey:
M = a male is selected. M = a male is not selected = a female is selected.
P(M) = 0.6, and so P(M) = P(F) = 0.4.
 P(A) + P(A) = 1
 P(A) = 1 – P(A)
 P(A) = 1 – P(A)
4.2 Basic Concepts (cont.)
 Joint Probability
 The probability of the occurrence of two events at the same time
 Example 4.2.4: In Datacomp Survey, what proportions are males
between 30 and 45? That is, find the probability of selecting a person
who is a male and between 30 and 45.
P(M and B) = 20/200 = 0.10
 Example 4.2.5: The probability of selecting a person who is a female and
under 30 is P(F and U) = 40/200 = 0.20.
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4.2 Basic Concepts (cont.)
 Either of Two Events
 The probability of either event A or event B occurring is written as P(A
or B).
 Example 4.2.6: In Datacomp Survey, the probability of selecting a
person who is male or under 30 is P(M or U) = (120 + 40) / 200 = 0.80.
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 Conditional Probability
 Whenever you are given information and are asked to find a probability
based on this information, the result is a conditional probability.
 This probability is written as P(A|B) and read as “probability of A given
B”.
 Example 4.2.7: In Datacomp Survey, what is the probability that a
randomly selected customer is male given that he is under 30?
P(M | U) = 60/100 = 0.60.
4.2 Basic Concepts (cont.)
 Independent Events
 Events A and B are independent if the probability of event A is
unaffected by the occurrence or nonoccurrence of event B.
 Events A and B are independent if and only if:
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Example 4.2.8: In Datacomp Survey, are events M and U independent?
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P(A | B) = P(A) (assuming P(B) ≠ 0), or
P(B | A) = P(B) (assuming P(A) ≠ 0), or
P(A and B) = P(A) • P(B)

P(M) = 120/200 = 0.6, P(M | U) = 60/100 = 0.6, so they are independent.
P(U) = 100/200 = 0.5, P(U | M) = 60/120 = 0.5, so they are independent.
P(M and U) = 60/200 = 0.3, P(M) • P(U) = 0.6 • 0.5 = 0.3, so they are
independent.
4.2 Basic Concepts (cont.)
 Dependent Events
 Events that are not independent are dependent events.
 P(A | B) = P(A and B) /P(B)
 P(B | A) = P(A and B) /P(A)
 P(A and B) = P(A | B) • P(B) = P(B | A) • P(A)
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4.2 Basic Concepts (cont.)
 Mutually Exclusive Events
 If an event can not occur when another event has occurred the two events
are said to be mutually exclusive.
 Events A and B are mutually exclusive if their joint probability is zero,
that is, P(A and B) = 0.
 P(A or B) = P(A) + P(B)
 Example 4.2.9: Consider an experiment of randomly selecting a card fro
a deck of 52 cards. Is the event of selecting a queen mutually exclusive
from the event of selecting a heart? No. Why?
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 Non-mutually Exclusive Events
 P(A and B) ≠ 0.
 P(A or B) = P(A) + P(B) - P(A and B).
4.3 Going Beyond the Contingency Table
 Venn Diagram
 In Venn diagram, a rectangle represents all possible outcomes of an
experiment.
 Event are shown in the rectangle as circles.
 The probability of an event occurring is its corresponding area in the
Venn diagram.
A
B
A
0.4
A
0.6
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Venn diagram for events A and B.
The rectangle represents all
possible outcomes of an experiment
Venn diagram for P(A) = 0.4.
4.3 Going Beyond the Contingency Table (cont.)
A
Venn diagram for P(A and B).
The points in the shaded area are in A and B
B
A
Venn diagram for P(A or B).
The points in the shaded area are in A or B
B
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A
B
Venn diagram of mutually exclusive events.
P(A and B) = 0 and P(A or B) = P(A) + P(B).
4.3 Going Beyond the Contingency Table (cont.)
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 Probability Rules
 General Additive Rule
P(A or B) = P(A) + P(B) - P(A and B)
 Special Additive Rule
If A and B are mutually exclusive then P(A or B) = P(A) + P(B)
 Example 4.3.1: In Datacomp Survey, what is the probability of selecting
a person who is male or under 30? That is, find P(M or U).
P(M or U) = P(M) + P(U) – P(M and U) = 120/200 + 100/200 – 60/200
= 0.80
 Example 4.3.2: The probability of event A is 0.5 and the probability of
event B is 0.2. If P(A and B) is 0.1, what is P(A or B)?
P(A or B) = P(A) + P(B) – P(A and B) = 0.5 + 0.2 – 0.1 = 0.6
 Example 4.3.3: In Datacomp Survey, find P(M or F).
P(M or F) = P(M) + P(F) – P(M and F) = 120/200 + 80/200 – 0/200
= 0.6 + 0.4 – 0 = 0.6 + 0.4 = P(M) + P(F) = 1.0
4.3 Going Beyond the Contingency Table (cont.)
 General Conditional Probability Rule
P( A and B)
P( B)
P( A and B)
P( B | A) 
P( A)
P( A | B) 
assuming
P(B)  0, and
assuming
P(A)  0
 Special Conditional Probability Rule
If events A and B are independent then:
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P ( A | B)  P ( A), and
P ( B | A)  P ( B)
4.3 Going Beyond the Contingency Table (cont.)
 Example 4.3.4: If P(A and B) = 0.4 and P(B) = 0.8, find P(A|B).
P( A and B) 0.4
P( A | B) 

 0.5
P( B)
0.8
 Example 4.3.5: If P(A) = 0.3 and P(B) = 0.4, and P(A and B) = 0.2, are events
A and B statistically independent? Use conditional probability rules.
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P( A and B) 0.2
P( A | B) 

 0.5  P( A)
P( B)
0.4
P( A and B) 0.2
P( B | A) 

 0.67  P( B)
P( A)
0.3
Events A and B are not independent.
4.3 Going Beyond the Contingency Table (cont.)
 Multiplicative Rule
P( A and B)  P( A | B)  P( B)  P( B | A)  P( A)
 Special Multiplicative Rule
If events A and B are independent then: P( A and B)  P( A)  P( B)
 Example 4.3.6: Let P(A) = 0.6, P(B) = 0.2, and P(A|B) = 0.1. Find P(A and B)
P( A and B)  P( A | B)  P( B )  0.1  0.2  0.02
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4.3 Going Beyond the Contingency Table (cont.)
 Sampling Without Replacement
 Assume that you select a card from a deck, examine it, and then discard
it. You then select another card. This procedure is called sampling
without replacement.
 Example 4.3.7: Let A = selecting a king on the first draw, and B =
selecting a king on the second draw. What is the probability of drawing
two kings [P(A and B)]?
If you select a king on the first draw, then, of the 51 cards remaining,
three are kings. So, P(A) = 4/52 and P(B|A) = 3/51. P(A and B) =
(4/52)(3/51) = 0.0045.
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4.3 Going Beyond the Contingency Table (cont.)
 Sampling With Replacement
 Assume that you select a card from a deck and replace it before selecting
the second card. This procedure is called sampling with replacement.
 Example 4.3.8: Let A = selecting a king on the first draw, and B =
selecting a king on the second draw. What is P(B|A)?
There are still 52 cards in the deck. So, P(B|A) = 4/52.
 Using Excel to Construct a Contingency Table
 KPK Data Analysis > Qualitative Data Charts > Contingency Table.
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4.4 Tree Diagrams
 A tree diagram shows all possible outcomes of an experiment and the
probabilities of each.
 A general form of a tree diagram is:
E1
B
E2
..
.
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B
En
B
 Example 4.4.1: Draw a tree diagram for the Datacomp Survey data.
4.4 Tree Diagrams (cont.)
 Rules for Tree Diagram
 Rule # 1: The probability of the event on the right side (say, event B) of
the tree is equal to the sum of the paths; that is, all probabilities along a
path leading to event B are multiplied, and then summed over all paths
leading to B.
 Rule # 2: The posterior probability for the ith path is:
i - th path
P( Ei | B) 
sum of paths
where, the " sum of paths" is found using Rule # 1.
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4.4 Tree Diagrams (cont.)
 Example 4.4.2: Zetadyne Corporation
50% of components produced
on shift 1
20% of components produced
on shift 2
30% of components produced
on shift 3
6% of the components
produced on shift 1 are
defective
8%
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of the components
produced on shift 2 are
defective
15% of the components
produced on shift 3 are
defective
Shift
1
(.06)
Defective
(.08)
Defective
(.15)
Defective
2
(.2)
3
4.4 Tree Diagrams (cont.)
Solution 1 – What percentage of the components are defective?
P(Defective)
= sum of paths
= (.5)(.06) + (.2)(.08) + (.3)(.15)
= .030 + .016 + .045
= .091
Solution 2 – Given that a defective component is found, what is the probability
that it was produced during shift 3?
P(shift 3 | defective)
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=
third path
sum of paths
= (.3)(.15)
.091
= (.045) = .495
.091