IS 310
Business
Statistics
CSU
Long Beach
IS 310 – Business Statistics
Slide 1
Chapter 4 Introduction to Probability
Few events are certain to happen.
Probability offers a numeric measure of the
likelihood of an event occurring
IS 310 – Business Statistics
Slide 2
Probability as a Numerical Measure
of the Likelihood of Occurrence
Increasing Likelihood of Occurrence
Probability:
0
The event
is very
unlikely
to occur.
IS 310 – Business Statistics
.5
The occurrence
of the event is
just as likely as
it is unlikely.
1
The event
is almost
certain
to occur.
Slide 3
Probability Concepts

To understand probability, one needs to understand
the following concepts:

Experiment
Experimental outcome
Sample space
Sample point
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IS 310 – Business Statistics
Slide 4
An Experiment and Its Sample Space
An experiment is any process that generates
well-defined outcomes.
The sample space for an experiment is the set of
all experimental outcomes.
An experimental outcome is also called a sample
point.
IS 310 – Business Statistics
Slide 5
Experiment and Outcomes

Let’s look at the following experiments and their
outcomes.

Experiment

Tossing a coin
Selecting a part
for inspection
Playing ball game
Conducting a sales
call
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IS 310 – Business Statistics
Outcome
Head, tail
Defective, non-defective
Win, lose, tie
Sale, no sale
Slide 6
Counting Outcomes

In order to calculate probability, we must be able to
count the total number of outcomes in an experiment.
In simple experiments, it is easy to know the number
of outcomes. In complex experiments, we need some
rules for counting the number of outcomes.

There are three rules:
Rule for multiple-step experiments
Combination
Permutation
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IS 310 – Business Statistics
Slide 7
A Counting Rule for
Multiple-Step Experiments
 If an experiment consists of a sequence of k steps
in which there are n1 possible results for the first step,
n2 possible results for the second step, and so on,
then the total number of experimental outcomes is
given by (n1)(n2) . . . (nk).
 A helpful graphical representation of a multiple-step
experiment is a tree diagram.
IS 310 – Business Statistics
Slide 8
Counting Rule for Multi-step Experiments

Let’s take the experiment of tossing two coins. How
many total outcomes are possible in this experiment?


The first coin has two outcomes (H, T)
The second coin has also two outcomes (H, T)

Total outcomes is: (2) (2) = 4

Often, a tree diagram helps. Look at Figure 4.2 (10Page 145; 11-Page 152).
IS 310 – Business Statistics
Slide 9
Counting Rule for Multi-step Experiments

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Let’s take another example.
Kentucky Power & Light Company is starting an
expansion project. The project consists of two stages:
design and construction.
Design work can be completed in 2, 3 or 4 months
(three outcomes)
Construction work can be completed in 6, 7 or 8
months (three outcomes)
Total number of outcomes in this experiment is:
(3) (3) = 9
Look at Table 4.1 and Figure 4.3 (10-Page 146; 11Page 153).
IS 310 – Business Statistics
Slide 10
Counting Rule for Combinations
A second useful counting rule enables us to count the
number of experimental outcomes when n objects are to
be selected from a set of N objects.
Number of Combinations of N Objects Taken n at a Time
CnN
where:
N!
 N
  
 n  n !(N  n )!
N! = N(N  1)(N  2) . . . (2)(1)
n! = n(n  1)(n  2) . . . (2)(1)
0! = 1
IS 310 – Business Statistics
Slide 11
Counting Rule for Permutations
A third useful counting rule enables us to count the
number of experimental outcomes when n objects are to
be selected from a set of N objects, where the order of
selection is important.
Number of Permutations of N Objects Taken n at a Time
PnN
where:
N!
 N
 n !  
 n  (N  n )!
N! = N(N  1)(N  2) . . . (2)(1)
n! = n(n  1)(n  2) . . . (2)(1)
0! = 1
IS 310 – Business Statistics
Slide 12
Example Problems
A home builder offers four different floor plans, three
different exterior designs and two types of roof.
How many ways can a home buyer choose his/her
home?
4 x 3 x 2 = 24
IS 310 – Business Statistics
Slide 13
Examples on Combinations and Permutations
Combinations:
How many ways can three Republicans be chosen as members of a
committee from a group of seven Republicans?
7
7!
C = ----------- = 35
3 3! (7 – 3)!
Permutations:
Automobile license plates in a state consists of seven letters of the
alphabets; a letter may not be repeated. How many license plates are
possible?
26
26!
P
= --------- = 26 x 25 x 24 x 23 x 22 x 21 x 20
7
(26-7)!
IS 310 – Business Statistics
Slide 14
Examples on Combinations and Permutations
Automobile license plates in a state consist of five letters and two
digits. A letter or a digit may be repeated. How many license
plates are possible?
26
26!
The first position may be filled in P = -------------- = 26
1
(26 – 1)!
26 x 26 x 26 x 26 x 26 x 10 x 10
Automobile license plates in CA consist of any combination of
seven letters and/or digits. A letter or a digit may be repeated
seven times. How many license plates are possible?
36 x 36 x 36 x 36 x 36 x 36 x 36
IS 310 – Business Statistics
Slide 15
Examples on Combinations and Permutations
Ten individuals are interested in serving on a board of directors. There are
three positions available. How many ways can the selection be made?
10
10!
C
= -------------- = 120
3
3! (10 – 3)!
In a class of 40 students, there are 25 women and 15 men. A committee is
to be selected with 5 women and three men. How many ways can this
selection be made?
25
15
25!
15!
C
x C
= -------------- x ------------5
3
5! (25 – 5)!
3! (15 – 3)!
IS 310 – Business Statistics
Slide 16
Examples on Combinations and Permutations
Five different scholarships are to be awarded to five
students. How many ways can these scholarships be
awarded?
5
5!
P = -------- = 5! = 5 x 4 x 3 x 2 = 120
5
(5 – 5)!
IS 310 – Business Statistics
Slide 17
Assigning Probabilities
Classical Method
Assigning probabilities based on the assumption
of equally likely outcomes
Relative Frequency Method
Assigning probabilities based on experimentation
or historical data
Subjective Method
Assigning probabilities based on judgment
IS 310 – Business Statistics
Slide 18
Classical Method
If an experiment has n possible outcomes, this method
would assign a probability of 1/n to each outcome.
Example
Experiment: Rolling a die
Sample Space: S = {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a
1/6 chance of occurring
IS 310 – Business Statistics
Slide 19
Relative Frequency Method
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Assigns probabilities based on relative frequencies.
Let’s take the example (10-Page 149; 11-Page 156)
Number of patients waiting for service in the X-ray
department of a hospital:
Number Waiting
Number of Days
0
2
1
5
2
6
3
4
4
3
Total
20
IS 310 – Business Statistics
Slide 20
Relative Frequency Method

The probability of zero patient waiting for service is:
2/20 = 0.1
The probability of one patient waiting for service is:
5/20 = 0. 25
The probability of two patients waiting for service is:
6/20 = 0.30
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And so on.
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IS 310 – Business Statistics
Slide 21
Subjective Method
 When economic conditions and a company’s
circumstances change rapidly it might be
inappropriate to assign probabilities based solely on
historical data.
 We can use any data available as well as our
experience and intuition, but ultimately a probability
value should express our degree of belief that the
experimental outcome will occur.
 The best probability estimates often are obtained by
combining the estimates from the classical or relative
frequency approach with the subjective estimate.
IS 310 – Business Statistics
Slide 22
Sample Problems

a.
b.
Problem # 11 (10-Page 152; 11-Page 159)
858/(858+228) = 858/1,086 = 0.79
Yes.
Problem # 12 (10-Page 153; 11-Page 160)
a.
b.
c.
55!/(5!) x (55-5)!
1/[55!/(5!)x(55-50!]
1/[55!/(5!)x(55-5)!] x 1/[42!/(1!)x(42-1)!]
IS 310 – Business Statistics
Slide 23
Events and Their Probabilities
An event is a collection of sample points.
The probability of any event is equal to the sum of
the probabilities of the sample points in the event.
If we can identify all the sample points of an
experiment and assign a probability to each, we
can compute the probability of an event.
IS 310 – Business Statistics
Slide 24
Events and Their Probabilities

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Let’s go back to the example of Kentucky Power &
Light Company.
Suppose, we want to find out the probability of
completing the expansion project in 10 months or
less.
According to Table 4.3, the following sample points
(2,6), (2,7), (2,8), (3,6), (3,7) and (4,6) provide a
completion time of 10 months or less.
Thus, the event C that the project can be completed in
10 months or less is:
P(C) = P(2,6) + P(2,7) + P(2,8) + P(3,6) + P(3,7) + P(4,6)
IS 310 – Business Statistics
Slide 25
Sample Problem
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Problem #20 (10-Page 157; 11-Page 164)
Given:
State
Number of Companies
New York
54
California
52
Texas
48
Illinois
33
Ohio
30
Find: P(N) =
Find P(B) =
IS 310 – Business Statistics
Find: P(T) =
Slide 26
Some Basic Relationships of Probability
There are some basic probability relationships that
can be used to compute the probability of an event
without knowledge of all the sample point probabilities.
Complement of an Event
Union of Two Events
Intersection of Two Events
Mutually Exclusive Events
IS 310 – Business Statistics
Slide 27
Complement of an Event
The complement of event A is defined to be the event
consisting of all sample points that are not in A.
The complement of A is denoted by Ac.
Event A
Ac
Sample
Space S
Venn
Diagram
IS 310 – Business Statistics
Slide 28
Union of Two Events
The union of events A and B is the event containing
all sample points that are in A or B or both.
The union of events A and B is denoted by A B
Event A
IS 310 – Business Statistics
Event B
Sample
Space S
Slide 29
Intersection of Two Events
The intersection of events A and B is the set of all
sample points that are in both A and B.
The intersection of events A and B is denoted by A 
Event A
Event B
Sample
Space S
Intersection of A and B
IS 310 – Business Statistics
Slide 30
Addition Law
The addition law provides a way to compute the
probability of event A, or B, or both A and B occurring.
The law is written as:
P(A B) = P(A) + P(B)  P(A  B
IS 310 – Business Statistics
Slide 31
Mutually Exclusive Events
Two events are said to be mutually exclusive if the
events have no sample points in common.
Two events are mutually exclusive if, when one event
occurs, the other cannot occur.
Event A
IS 310 – Business Statistics
Event B
Sample
Space S
Slide 32
Mutually Exclusive Events
If events A and B are mutually exclusive, P(A  B = 0.
The addition law for mutually exclusive events is:
P(A B) = P(A) + P(B)
There is no need to
include “ P(A  B”
IS 310 – Business Statistics
Slide 33
Sample Problem

Problem #24 (10-Page 162; 11-Page 169)

a. Let S = Experience surpassed expectation
N = No response
M= Met expectation
F = Fell short of expectation
P(S) = 1 – P(N) – P(M) – P(F) = 1- 0.04 – 0.65 – 0.26 = 0.05

b. P(M or S) = P(M) + P(S) = 0.65 + 0.05 = 0.70
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IS 310 – Business Statistics
Slide 34
Sample Problem

Problem # 28 (10-Page 163; 11-Page 170)

a. P(B or P) = P(B) + P(P) – P(B and P)
= 0.458 + 0.54 – 0.3 = 0.698
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

b. P(neither B nor P) = 1 – P(B or P)
= 1 – 0.698
= 0.302
IS 310 – Business Statistics
Slide 35
Conditional Probability
The probability of an event given that another event
has occurred is called a conditional probability.
The conditional probability of A given B is denoted
by P(A|B).
A conditional probability is computed as follows :
P( A  B)
P( A|B) 
P( B)
IS 310 – Business Statistics
Slide 36
Sample Problem
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Problem #33 (10-Page 169-170; 11-Page 176-177)
Reason
Quality Cost/Conv Other Total
FT
0.22
0.20
0.04
0.46
PT
0.21
0.31
0.02
0.54
_________________________________________
0.43
0.51
0.06 1.00
b. School cost/convenience followed by quality
c. P(Q given FT) = 0.22/0.46 = 0.47
d. P(Q given PT) = 0.21/0.54 = 0.39
e. P(A) = 0.46 P(B) = 0.43 P(A and B) = 0.22
Since P(A and B) not equal to P(A) * P(B), the events are
NOT independent
IS 310 – Business Statistics
Slide 37
Multiplication Law
The multiplication law provides a way to compute the
probability of the intersection of two events.
The law is written as:
P(A B) = P(B)P(A|B)
IS 310 – Business Statistics
Slide 38
Independent Events
If the probability of event A is not changed by the
existence of event B, we would say that events A
and B are independent.
Two events A and B are independent if:
P(A|B) = P(A)
IS 310 – Business Statistics
or
P(B|A) = P(B)
Slide 39
Multiplication Law
for Independent Events
The multiplication law also can be used as a test to see
if two events are independent.
The law is written as:
P(A B) = P(A)P(B)
IS 310 – Business Statistics
Slide 40
Miscellaneous Problems
Problem # 50 (10-Page 180; 11-Page 187)
Rating
Frequency
Poor
4
Below Average
8
Average
11
Above Average
14
Excellent
13
a.
What is the probability that a randomly selected viewer will rate the
new show as average or better?
(11 + 14 + 13) / 50 = 0.76
b.
What is the probability that a viewer will rate the show as below
average or worse?
(8 + 4) / 50 = 0.24
IS 310 – Business Statistics
Slide 41
Miscellaneous Problems
Problem # 57 (10-Page 183; 11-Page 189)
Given: six percent employees suffered lost-time
accidents last year. Fifteen percent who had lost-time
accidents last year will experience the same this year.
a.
What percent employees will experience lost-time
accidents in both years?
(0.06) (0.15) = 0.009
b. (0.06) + (0.15) = 0.21
IS 310 – Business Statistics
Slide 42
Miscellaneous Problems
A box contains 20 units of a product of which four are defective and 16 are
good. Four units are selected at random. Calculate the following
probabilities:
a.
All four units are defective
P(4D) = P(D) . P(D). P(D). P(D) = (4/20) (3/19) (2/18) (1/17) = 1/4845
b.
Two units are defective and two units are good
Two defective units must come
from the four defective units:
4
C
= 6 ways
2
IS 310 – Business Statistics
Two good units must come from
the sixteen good units:
16
C
= 120 ways
2
Slide 43
Miscellaneous Problems Continued
Four units can be selected from the original 20 units in
20
C
= 4845 ways
4
(6) (120)
P(2D and 2G) = ----------- = 144/969
4845
IS 310 – Business Statistics
Slide 44
Miscellaneous Problems Continued
c.
At least three units are defective
P(at least 3D) means P(3D or 4D) = P(3D) + P(4D)
4
16
4
C .C
C
3
1
4
P(3D) = ----------------P(4D) = -----------------20
20
C
C
4
4
IS 310 – Business Statistics
Slide 45
Miscellaneous Problems
In how many ways can a television director schedule six different
commercials during the six time slots allocated to commercials during
the telecast of the first quarter of a football game?
6
P
= 6! = 720
6
Seven wooden dowels of varying lengths are arranged in a row. What is
the probability that the dowels will be arranged in order of size?
2
P(arranged in order of size) = ----------------7!
IS 310 – Business Statistics
Slide 46
End of Chapter 4
IS 310 – Business Statistics
Slide 47