Unformatted text preview: Business Statistics (BUSA 3101)
Dr. Lari H. Arjomand
[email protected] Slide 1 Chapter 5 A Survey of Probability Concepts
s Experiments, Counting Rules, and Assigning Probabilities s Events and Their Probability Some Basic Relationships of Probability
s s Conditional Probability 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. .5
The occurrence
of the event is just as likely as
it is unlikely. 1
The event
is almost
certain
to occur. Slide 3 An Experiment and Its Sample Space An experiment is any process that generates
An experiment is any process that generates welldefined outcomes.
welldefined outcomes. The sample space for an experiment is the set of
The sample space for an experiment is the set of all experimental outcomes.
all experimental outcomes. An experimental outcome is also called a sample
An experimental outcome is also called a sample point.
point. Slide 4 Example 1: Bradley Investments
John has invested in two stocks, Markley Oil and Collins Mining. John has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 8 5
−2 0
−20 Slide 5 Some Principles of Counting
1. Multiplication
1.
2. Combination
3. Permutation Slide 6 A Counting Rule for MultipleStep 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). Slide 7 Example 1: Bradley Investments
John has invested in two stocks, Markley Oil and Collins Mining. John has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 8 5
−2 0
−20 Slide 8 A Counting Rule for MultipleStep Experiments John Investments can be viewed as a
twostep experiment. It involves two stocks, each
with a set of experimental outcomes.
Markley Oil:
Collins Mining:
Total Number of Experimental Outcomes: n1 = 4
n2 = 2
n1n2 = (4)(2) = 8 A helpful graphical representation of a multiplestep experiment is a tree diagram. Slide 9 Tree Diagram (Example 1 Continued)
Markley Oil
(Stage 1) Collins Mining
(Stage 2)
Gain 8 Gain 10 Gain 8
Gain 5 Lose 2 Even
Lose 20 Gain 8
Lose 2 Lose 2 Gain 8
Lose 2 Experimental
Outcomes
(10, 8) Gain $18,000
(10, 2) Gain $8,000
(5, 8) Gain $13,000 (5, 2) Gain $3,000
(0, 8) Gain $8,000
(0, 2) Lose $2,000
(20, 8) Lose $12,000
(20, 2) Lose $22,000 Slide 10 Thinking Challenge:
Example s How many unique identifiers can be provided by inventory labels consisting of two letters (ranging from AA to ZZ) followed by four numbers (digits 0 through 9) ? Student Slide 11 Thinking Challenge:
Solution
s s
s There are 26 ways (letter A through Z) to fill either the first or second positions and there are 10 ways (digits 0 through 9) to fill the third through sixth positions.
The number of unique inventory labels is: 26 X 26 X 10 X 10 X 10 X 10 = 6,760,000 Slide 12 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, where the order of
selection is not important (A, B, C = A, C, B = C, A, B = . .).
Number of Combinations of N Objects Taken n at a Time
N
Cn N!
N = = n n !( N − n ) ! where: N! = N(N − 1)(N − 2) . . . (2)(1) n! = n(n − 1)(n − 2) . . . (2)(1) 0! = 1 Slide 13 Thinking Challenge:
Example
s s Suppose that 5 customers (A, B, C, D, E) need service calls and the field technician can only service 3 of them today. Suppose that the dispatcher who assigns the sequence of customers does not care who gets serviced first, second, or third.
In other words, the dispatcher regards ABC, ACB, BAC, BCA, CAB, or CBA as being the same event because the same three customers (A, B, C) get serviced. Student
Solve for Combination Slide 14 Thinking Challenge:
Solution
s The number of combinations is 10:
N
Cn N!
N = = n n !(N − n ) ! 5!/3!(53)! = 10 Slide 15 Some Principles of Counting
Another Example The United Way has received funding applications from ten agencies. If only three will be funded, how many different groups of three are possible? Note in this case the order of selection is not important, hence you will have to use the combination formula. So, 10C3 = (10!)/[3!(10 3)!] = 120. Student Slide 16 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 (ABC # ACB # BCA, and so on).
Number of Permutations of N Objects Taken n at a Time
PnN N!
N = n ! = n (N − n ) ! where: N! = N(N − 1)(N − 2) . . . (2)(1) n! = n(n − 1)(n − 2) . . . (2)(1) 0! = 1 Slide 17 Thinking Challenge:
Example
s s Suppose that 5 customers (A, B, C, D, E) need service calls and the field technician can only service 3 of them today. The dispatcher must assign the sequence of customers. The order in which they are serviced is important to the customers, so each possible arrangement of 3 is different Student
Solve for Permutation Slide 18 Thinking Challenge:
Solution
s The possible permutations are:
PnN N!
N = n ! = n (N − n ) ! 5!/(53)! = 60 Slide 19 Assigning Probabilities There are three (3) approaches to solve for probabilities:
1 Classical Method Assigning probabilities based on the assumption of equally likely outcomes
2 Relative Frequency Method Assigning probabilities based on experimentation or historical data
3 Subjective Method Assigning probabilities based on judgment Slide 20 1 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 Slide 21 2 Relative Frequency Method Example: Lucas Tool Rental Lucas Tool Rental would like to assign
probabilities to the number of car polishers
it rents each day. Office records show the following
frequencies of daily rentals for the last 40 days.
Number of
Polishers Rented
0
1
2
3
4 Number
of Days 4 6
18
10 2 Slide 22 Example: Relative Frequency Method
Each probability assignment is given by
dividing the frequency (number of days) by
the total frequency (total number of days).
Number of
Polishers Rented
0
1
2
3
4 Number
of Days 4 6
18
10 2
40 Probability .10 .15
4/40 .45 .25 .05
1.00 Slide 23 3 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.
s Example: There is 75% chance that England will adopt the Euro currency by 2010. Slide 24 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 Complement of an Event
Union of Two Events
Union of Two Events Intersection of Two Events Intersection of Two Events Mutually Exclusive Events Mutually Exclusive Events Slide 25 Complement of an Event The complement of event A is defined to be the event
The complement of event A is defined to be the event consisting of all sample points that are not in A.
consisting of all sample points that are not in A. The complement of A is denoted by Acc.
The complement of A is denoted by A . Event A Venn
Diagram Ac Sample
Space S 1 = A +Ac
1 – A =Ac Slide 26 Union of Two Events The union of events A and B is the event containing
The union of events A and B is the event containing all sample points that are in A or B or both.
all sample points that are in A or B or both. The union of events A and B is denoted by A ∪ B..
The union of events A and B is denoted by A ∪ B Event A Event B Sample
Space S Slide 27 Intersection of Two Events The intersection of events A and B is the set of all
The intersection of events A and B is the set of all sample points that are in both A and B.
sample points that are in both A and B. The intersection of events A and B is denoted by A ∩ Β..
The intersection of events A and B is denoted by A ∩ Β Event A Event B Sample
Space S Intersection of A and B Slide 28 Addition Law The addition law provides a way to compute the
The addition law provides a way to compute the probability of event A,, or B, or both A and B occurring.
probability of event A or B, or both A and B occurring. The law is written as:
The law is written as:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) Slide 29 Mutually Exclusive Events Two events are said to be mutually exclusive if the
Two events are said to be mutually exclusive if the events have no sample points in common.
events have no sample points in common. Two events are mutually exclusive if, when one event
Two events are mutually exclusive if, when one event occurs, the other cannot occur.
occurs, the other cannot occur. Event A Event B Sample
Space S Slide 30 Mutually Exclusive Events If events A and B are mutually exclusive, P((A ∩ B) = 0.
If events A and B are mutually exclusive, P A ∩ B) = 0. The addition law for mutually exclusive events is:
The addition law for mutually exclusive events is:
P(A ∪ B) = P(A) + P(B) there’s no need to
include “− P(A ∩ B) ” Slide 31 Conditional Probability The probability of an event given that another event
The probability of an event given that another event has occurred is called a conditional probability..
has occurred is called a conditional probability The conditional probability of A given B is denoted
The conditional probability of A given B is denoted by P(AB).
by P(AB). A conditional probability is computed as follows :
A conditional probability is computed as follows : P( A ∩ B)
P ( A  B) =
P ( B) Slide 32 Multiplication Law The multiplication law provides a way to compute the
The multiplication law provides a way to compute the probability of the intersection of two events.
probability of the intersection of two events. The law is written as:
The law is written as:
P(A ∩ B) = P(B)P(AB) Slide 33 Independent Events If the probability of event A is not changed by the
If the probability of event A is not changed by the existence of event B, we would say that events A
existence of event B, we would say that events A and B are independent.
and B are independent. Two events A and B are independent if::
Two events A and B are independent if
P(AB) = P(A) or P(BA) = P(B) Slide 34 Multiplication Law
for Independent Events The multiplication law also can be used as a test to see
The multiplication law also can be used as a test to see if two events are independent.
if two events are independent. The law is written as:
The law is written as:
P(A ∩ B) = P(A)P(B) Slide 35 Assigning Probabilities There are three (3) approaches to solve for probabilities:
1 Classical Method Assigning probabilities based on the assumption of equally likely outcomes
2 Relative Frequency Method Assigning probabilities based on experimentation or historical data
3 Subjective Method Assigning probabilities based on judgment Slide 36 Summary of Probability Equations
1. Joint Probability:
A. When two events are dependent
P(A and B) = P(A) * P(B given A)
B. When two events are independent
P(A and B) = P(A) * P(B)
2. Conditional Probability:
A. When two events are dependent
P(B given A) = P(A and B)/P(A)
B. When two events are independent
P(A given B) = P(B) Slide 37 Summary of Probability Equations (Continued)
3. Additive Probability:
A. When two events are not mutually exclusive
P(A or B) = P(A) + P(B) P(A and B)
B. When two events are mutually exclusive
P(A or B) = P(A) + P(B) Slide 38 Thinking Challenge Examples 1
What’s the probability?
P(A) =
P(D) =
P(C and B) =
P(A or D) =
P(B and D) = Event
A Event
C
D
4
2 Total
6 B Student 1 3 4 Total 5 5 10 Slide 39 Solutions
The probabilities are:
P(A) = 6/10
P(D) = 5/10
P(C and B) = 1/10
P(A or D) =(6/10)+(5/10)
(2/10)= 9/10
P(B and D) = 3/10 Event
A Event
C
D
4
2 Total
6 B 1 3 4 Total 5 5 10 Slide 40 Thinking Challenge Example 1 (Continued)
Using the Addition Rule, what’s the probability? Assuming events A, B, C, and D are not mutually exclusive.
P(A or D) =
P(B or C) = Event
A Student Event
C
D
4
2 Total
6 B 1 3 4 Total 5 5 10 Slide 41 Solutions
Using the Addition Rule, the probabilities are: Event
A Event
C
D
4
2 Total
6 B 1 3 4 Total 5 5 10 P(A or D) = P(A) + P(D)  P(A and D)
6
5
2
9
=
+
−
=
10 10 10 10
P(B or C) = P(B) + P(C)  P(B and C)
4
5
1
8
=
+
−
=
10 10 10 10 Slide 42 Thinking Challenge Examples 2
New England Commuter Airways recently supplied the
following information to the U.S. Government on their
commuter flights from Boston to New York. Arrival Frequency Early
On Time
Late
Cancelled
To t a l 100
800
75
25
1000 Slide 43 Questions & Solutions Let A be the event that a flight arrives early. Then, P(A) = 100/1000 = 0.1. Let B be the event that a flight arrives late. Then, P(B) = 75/1000 = 0.075. Note: Events A and B are mutually exclusive. Why? What is the probability that a flight is either early or late? P(A or B) = P(A) + P(B) = 0.1 + 0.075 = 0.175. Student Slide 44 Questions & Solutions Let C be the event that a flight arrives on time. Then, P(C) = 800/1000 = 0.8. Let D be the event that a flight is cancelled. Then, P(D) = 25/1000 = 0.025. Note: Events C and D are mutually exclusive. Why? Use the complement rule to show that the probability of an early (A) or a late (B) flight is 0.175. Student Slide 45 Thinking Challenge Examples 1
(Continued)
Using the table then the formula, what’s the probability?
P(AD) =
P(CB) = Are C & B independent ? Event
A Event
C
D
4
2 Total
6 B Student 1 3 4 Total 5 5 10 Slide 46 Solutions Event
A Using the formula, the probabilities are: Event
C
D
4
2 Total
6 B 1 3 4 Total 5 5 10 P(A and D) 2 / 10 2
P(A  D) =
=
=
P(D)
5 / 10 5 P(C and B) 1 / 10 1
P(C  B) =
=
=
P(B)
4 / 10 4
P(C) = 5
1
≠
10 4 C & B are dependent, because: P(C) # P(C  B) Slide 47 Thinking Challenge: Example 2 A recent study found that 60% of mothers with children under the age of ten are employed fulltime. Three mothers are selected at random. We will assume that mothers are employed independent of each other. What is the probability they are all employed fulltime? P(all 3 are employed fulltime) = (0.6)(0.6)(0.6) = 0.216. Student Slide 48 Thinking Challenge: Example 2 (Continued) What is the probability that at least one of the mothers is employed fulltime? P(at least one) = 1 P(none are employed fulltime) = 1 – [(0.4)(0.4)
(0.4)] = 0.936. Student Slide 49 Bayes’ Theorem Optional
Readings Slide 50 End of Chapter 5 Slide 51 ...
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 Fall '09
 Statistics, Counting, Probability, Kompakt, Markley Oil

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