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Chapter 4
Probability
Definitions
Experiment. A process that generates well defined outcomes. For example, the
experiment of flipping a coin has 2 defined outcomes, heads or tails.
S = {H, T}.
Random Variable. An experimental outcome that generates exactly one
numerical value.
Sample Space. All of the possible outcomes of an experiment. For example, the
outcomes possible when rolling a single die are S = {1, 2, 3, 4, 5, 6}. The
outcomes possible when flipping a coin are S = {H, T}. The sum of all possible
outcomes is always equal to 1.
Probability. The likelihood that an event will occur. All probabilities always fall
between 0 and 1, and are usually expressed as percentages.
Event. The outcome of an experiment. For example, The outcome of rolling an
even number on a die is A = {2, 4, 6}, and P(A) = 1/2. The outcome of rolling a
five is A = {5}, and P(A) = 1/6 (or P(5)=1/6).
Types of Probability
Classical. (Also referred to as Theoretical). The number of
outcomes in the sample space is known, and each outcome is equally
likely to occur.
Empirical. (Also referred to as Statistical or Relative Frequency).
The frequency of outcomes is measured by experimenting.
Subjective. You estimate the probability by making an “educated
guess”, or by using your intuition.
Venn Diagram
Event A
~A
Complement of A
P( A)  P(~ A)  1
Total Sample Space (S). P(S) = 1
Example: Suppose you roll a die and the outcome you want to observe is that of rolling a 4.
Therefore, A = {4}, and ~A = (1, 2, 3, 5, 6}.
The Complement Rule
The complement rule is used to determine the
probability of an event occurring by subtracting
the probability of the event not occurring from 1.
If P(A) is the probability of event A and P(~A)
is the complement of A,
P(A) + P(~A) = 1 or
P(A) = 1 - P(~A).
Intersection of Events A and B
(AandB)
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The intersection of events A and B is
the event containing the sample points
belonging to both A and B.
Example: Suppose you roll a die and the outcome you want to observe is that of rolling a
number greater than 3 and rolling an odd number. Therefore, A = {4,5,6} and B = {1, 3, 5}.
The intersection is rolling a 5.
Union of Events A and B
(the addition law)
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The union of events A and B is the event
containing the points belonging to A or B, or
both.
Example: Suppose you roll a die and the outcome you want to observe is that of rolling a
number greater than 3 or rolling an odd number. Therefore, A = {4,5,6} and B = {1, 3, 5}.
The union is rolling a 1, 3, 4, 5, or 6. Therefore, P(A+B) = 1/6+1/6+1/6+1/6+1/6 = 5/6.
P( AORB)  P( A)  P( B)  P( AANDB)
Subtract the doublecounted outcome.
Mutually Exclusive Events
Experiment
Two or more events are mutually exclusive
if the events have no points (or outcomes)
in common.
In other words, it’s impossible for both
events to occur at the same time. Each
outcome is unique and has nothing in
common with the other.
P( AORB)  P( A)  P( B)  P( AANDB)
Example: Suppose you roll a die and the outcome you want to observe is that of rolling either
an even number or an odd number. Therefore, A = {2,4,6} and B = (1,3,5}, and the two events
have absolutely nothing in common.
P( AORB)  P( A)  P( B)  P( AANDB)  (.5)  (.5)  0  1
Multiplication Law
(joint probability)
The multiplication law is derived from the definition of conditional probability.
P( AANDB)  P( A) P( B | A)
•
Example: Suppose you choose two cards from a deck of 52 cards. What is the
probability of selecting a king [P(K)] from the deck, not replacing it, and then
immediately selecting a queen P[(Q|K)].
•
Answer: Selecting a king changes the probability of selecting the very next card. P(K)
= 4/52, after which the P(Q) then becomes 4/51.
•
Because the first card is not replaced, the events are dependent.
•
P(KandQ) = P(K)P(Q|K) = (4/52)(4/51) = .006
Conditional Probability
A conditional probability is the probability of an event occurring given that another event has
already occurred. The notation reads: P(A|B) = Probability of A given that B has already
occurred.
P( AANDB)
P( A | B) 
P( B)
Example: Suppose that event A is rolling a die = 5, or A = {5}. Suppose event B is rolling an
odd number, or B = {1,3,5}. So, P(A|B) is 1/3. The probability of rolling and odd number is
P(B) = 3/6. The intersection of A and B is 5, and P(5) = 1/6.
P( AANDB) 1 / 6
P( A | B) 

 .333
P( B)
3/ 6
Independent Events
Two events are independent if the occurrence of one event does not affect the occurrence
of another event.
Therefore, the probability of event A occurring given that event B has already occurred
equals the probability of event A.
P ( A | B )  P ( A)
The independent occurrence of event B does not change the occurrence of event A.
P( H | T )  P( H )
Dependent Events
Two events are dependent if the probability of one event changes given that
another event has occurred.
P ( A | B )  P ( A)
P(Q | K )  P(Q)
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Chapter 4
Contingency Tables
Contingency Tables
Often we tally the results of a survey into a two-way table, and then
use these results to determine various probabilities. We refer to this
two-way table as a Contingency Table.
Contingency Table. A table that classifies observations according to
2 or more characteristics.
Length of Service
Less Than 1
Year
B1
1 to 5 Years
B2
6 to 10 Years
B3
More Than 10
Year
B4
Total
Would Remain, A 1
10
30
5
75
120
Would Not Remain, ~A 1
25
15
10
30
80
35
45
15
105
200
Loyalty
Total
EXAMPLE
A company wishes to determine how loyal their employees are. The question asked
was “If you were given a slightly better offer by another company, would you
accept the offer?”
The responses of 200 employees are tallied below:
Length of Service
Less Than 1
Year
B1
1 to 5 Years
B2
6 to 10 Years
B3
More Than 10
Year
B4
Total
Would Remain, A 1
10
30
5
75
120
Would Not Remain, ~A 1
25
15
10
30
80
35
45
15
105
200
Loyalty
Total
What is the probability of randomly selecting an employee who is loyal and has
more than 10 years of service?
 120  75 
P( A1 ANDB4 )  P( A1 ) P( B4 | A1 )  

  .375
 200  120 
What is the probability of randomly selecting an employee who would remain (is
loyal) or has less than one year of service?
 120   35   10 
P( A1orB1 )  P( A1 )  P( B1 )  P( A1 ANDB1 )  

 

 200   200   200 
EXAMPLE
The Dean of the School of Business at Owens University
collected the following information about undergraduate
students in her college:
MAJOR
Male
Female
Total
Accounting
170
110
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
EXAMPLE continued
MAJOR
Male
Female
Total
Accounting
170
110
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
If a student is selected at random, what is the probability that the student
is a female (F) accounting major (A)
P(A and F) = 110/1000.
Given that the student is a female, what is the probability that she is an
accounting major?
P(A|F) = P(A and F)/P(F)
= [110/1000]/[400/1000] = .275