Transcript No Slide Title

Dr. Ka-fu Wong
ECON1003
Analysis of Economic Data
Ka-fu Wong © 2003
Chap 5- 1
Chapter Five
A Survey of Probability Concepts
GOALS
1.
2.
3.
4.
5.
6.
7.
l
Define probability.
Describe the classical, empirical, and subjective
approaches to probability.
Understand the terms: experiment, event, outcome,
permutations, and combinations.
Define the terms: conditional probability and joint
probability.
Calculate probabilities applying the rules of addition
and the rules of multiplication.
Use a tree diagram to organize and compute
probabilities.
Calculate a probability using Bayes’ theorem.
Ka-fu Wong © 2003
Chap 5- 2
Definitions
A probability is a measure of the likelihood that
an event in the future will happen.

It can only assume a value between 0 and 1.

A value near zero means the event is not likely to
happen. A value near one means it is likely.

There are three definitions of probability: classical,
empirical, and subjective.
Ka-fu Wong © 2003
Chap 5- 3
Definitions continued
 The classical definition applies when there are n
equally likely outcomes.
 The empirical definition applies when the
number of times the event happens is divided by
the number of observations, based on data.
 Subjective probability is based on whatever
information is available, based on subjective
feelings.
Ka-fu Wong © 2003
Chap 5- 4
Example 1
(to be used to illustrate the definitions)
 A fair die is rolled once.
 Peter is concerned with whether the resulted number
is even, i.e., 2, 4, 6.
 Paul is concerned with whether the resulted number
is less than or equal to 3, i.e., 1, 2, 3.
 Mary is concerned with whether the resulted number
is 6.
 Sonia is concerned with whether the resulted number
is odd, i.e., 1, 3, 5.
 A fair die is rolled twice.
 John is concerned with whether the resulted number
of first roll is even, i.e., 2, 4, 6.
 Sarah is concerned with whether the resulted number
of second roll is even, i.e., 2, 4, 6.
Ka-fu Wong © 2003
Chap 5- 5
Definitions continued
 An experiment is the observation of some activity or the
act of taking some measurement.
 The experiment is rolling the die.
 An outcome is the particular result of an experiment.
 The possible outcomes are the numbers 1, 2, 3, 4, 5, and
6.
 An event is the collection of one or more outcomes of an
experiment.
 For Peter: the occurrence of an even number, i.e., 2, 4, 6.
 For Paul: the occurrence of a number less than or equal
to 3, i.e., 1, 2, 3.
 For Mary: the occurrence of a number 6.
 For Sonia: the occurrence of an odd number, i.e., 1, 3, 5.
Ka-fu Wong © 2003
Chap 5- 6
Mutually Exclusive, Independent and
Exhaustive events
 Events are mutually exclusive if the occurrence of any one
event means that none of the others can occur at the
same time.
 Peter’s event and Paul’s event are not mutually
exclusive – both contains 2.
 Peter’s event and Mary’s event are not mutually
exclusive – both contains 6.
 Paul’s event and Mary’s event are mutually exclusive –
no common numbers.
 Peter’s event and Sonia’s event are mutually
exclusive – no common numbers.
Ka-fu Wong © 2003
Chap 5- 7
Mutually Exclusive, Independent and
Exhaustive events
 Events are independent if the occurrence of one event
does not affect the occurrence of another.
 P(A&B) = P(A)*P(B)
 Not independent:
Peter’s event and Paul’s event.
Peter’s event and Mary’s event.
Paul’s event and Mary’s event.
Peter’s event and Sonia’s event.
 Independent:
John’s event and Sarah’s event are independent.
P(John & Sarah) = P(John)*P(Sarah)
Ka-fu Wong © 2003
Chap 5- 8
Mutually Exclusive, Independent and
Exhaustive events
 Events are collectively exhaustive if at least one of the
events must occur when an experiment is conducted.
 Peter’s event (even numbers) and Sonia’s event (odd
numbers) are collectively exhaustive.
 Peter’s event (even numbers) and Mary’s event
(number 6) are not collectively exhaustive.
Ka-fu Wong © 2003
Chap 5- 9
Example 2
 Throughout her teaching career Professor Jones has
awarded 186 A’s out of 1,200 students. What is the
probability that a student in her section this semester will
receive an A?
 This is an example of the empirical definition of
probability.
 To find the probability a selected student earned an A:
186
P(A) 
 0.155
1200
This number may be interpreted as “unconditional probability”.
In most cases, we are interested in the probability of earning
an A for a selected student who study 10 hours or more per
week. We call this “conditional probability”.
P(A | study 10 or more hours per week)
Ka-fu Wong © 2003
Chap 5- 10
Subjective Probability
 Examples of subjective probability are:
 Estimating the probability mortgage rates for
home loans will top 8 percent this year.
 Estimating the probability that HK’s economic
growth will be 3% this year.
 Estimating the probability that HK will solve its
deficit problem in 2006.
Ka-fu Wong © 2003
Chap 5- 11
Learning exercise 1:
University Demographics
 Current enrollments by college and by sex appear in the
following table.
College
Ag-For
Arts-Sci
Bus-Econ
Educ
Engr
Law
Undecl
Totals
Female
500
1500
400
1000
200
100
800
4500
Male
900
1200
500
500
1300 200
900
5500
Totals
1400
2700
900
1500
1500 300
1700
10000
 If I select a student at random, answer the following:
 Find P(Female or Male)
 Find P(not-Ag-For)
 Find P(Female |BusEcon)
 Find P(Male and Arts-Sci)
 Are “Female” and “Educ” Statistical independent? Why
or Why not?
Ka-fu Wong © 2003
Chap 5- 12
Learning exercise 1:
University Demographics
College
Ag-For
Arts-Sci
Bus-Econ
Educ
Engr
Law
Undecl
Totals
Female
500
1500
400
1000
200
100
800
4500
Male
900
1200
500
500
1300 200
900
5500
Totals
1400
2700
900
1500
1500 300
1700
10000
P(Female or Male) =(4500 + 5500)/10000 = 1
P(not-Ag-For) =(10000 – 1400) /10000 = 0.86
P(Female | Bus Econ) = 400 /900 = 0.44
P(Male and Arts-Sci) =1200 /10000 = 0.12
Are Female and Educ Statistical independent?
NO!
P(female and Educ)=1000 /10000 = 0.1
P(Educ) =1500 /10000 = 0.15
P(female)=4500 /10000 = 0.45
P(female and Educ)> P(female)*P(Educ) = 0.0675
Ka-fu Wong © 2003
Chap 5- 13
Learning exercise 2:
Predicting Sex of Babies
 Many couples take advantage of ultrasound exams to
determine the sex of their baby before it is born. Some
couples prefer not to know beforehand. In any case,
ultrasound examination is not always accurate. About 1 in 5
predictions are wrong.
 In one medical group, the proportion of girls correctly
identified is 9 out of 10 and
 the number of boys correctly identified is 3 out of 4.
 The proportion of girls born is 48 out of 100.
 What is the probability that a baby predicted to be a girl
actually turns out to be a girl?
Formally, find P(girl | test says girl).
Ka-fu Wong © 2003
Chap 5- 14
Learning exercise 2:
Predicting Sex of Babies
 P(girl | test says girl)
 In one medical group, the proportion of girls correctly
identified is 9 out of 10 and
 the number of boys correctly identified is 3 out of 4.
 The proportion of girls born is 48 out of 100.
 Think about the next 1000 births handled by this medical group.
 480 = 1000*0.48 are girls
 520 = 1000*0.52 are boys
 Of the girls, 432 (=480*0.9) tests indicate that they are girls.
 Of the boys, 130 (=520*0.25) tests indicate that they are
girls.
 In total, 562 (=432+130) tests indicate girls. Out of these
462 babies, 432 are girls.
 Thus P(girl | test says girl ) = 432/562 = 0.769
Ka-fu Wong © 2003
Chap 5- 15
Learning exercise 2:
Predicting Sex of Babies
 480 = 1000*0.48 are girls 1000*P(girls)
 520 = 1000*0.52 are boys 1000*P(boys)
 Of the girls, 432 (=480*0.9) tests indicate that they are
1000*P(girls)*P(test says girls|girls)
girls.
 Of the boys, 130 (=520*0.25) tests indicate that they are
1000*P(boys)*P(test says girls | boys)
girls.
 In total, 562 tests indicate girls.
1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]
 Out of these 562 babies, 432 are girls.
 Thus P(girls | test syas girls ) = 432/562 = 0.769
1000*P(girls)*P(test says girls|girls)
1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]
Ka-fu Wong © 2003
Chap 5- 16
Learning exercise 3:
Putting in Extra Trunk Lines Between [insert
local names for Town A and Town B]
 Given recent flooding (or other condition more appropriate
to your area) between Town A and Town B, the local
telephone company is assessing the value of adding an
independent trunk line between the two towns. The second
line will fail independently of the first because it will
depend on different equipment and routing (we assume a
regional disaster is highly unlikely).
 Under current conditions, the present line works 98 out of
100 times someone wishes to make a call. If the second
line performs as well, what is the chance that a caller will
be able to get through? Formally,
 P( Line 1 works ) = 98/100
 P( Line 2 works ) = 98/100
 Find P( Line 1 or Line 2 works ).
Ka-fu Wong © 2003
Chap 5- 17
Learning exercise 3:
Putting in Extra Trunk Lines Between [insert
local names for Town A and Town B]
 P( Line 1 works ) = 98/100
 P( Line 2 works ) = 98/100
 Find P( Line 1 or Line 2 works ).
P( Line 1 or Line 2 works )
= 1 – P(Both line1 and Line 2 fail)
= 1 – P(Line 1 fails)*P(line 2 fails)
= 1 – 0.02*0.02
= 0.9996.
Line 2 works
Ka-fu Wong © 2003
Line 2 fails
Line 1 works
0.98*0.98
0.98*0.02
Line 1 fails
0.02*0.98
0.02*0.02
Chap 5- 18
Learning exercise 4:
Part-time Work on Campus
 A student has been offered part-time work in a laboratory. The
professor says that the work will vary from week to week. The
number of hours will be between 10 and 20 with a uniform
probability density function, represented as follows:
8
10
Ka-fu Wong © 2003
12
14
16
18
20
22
 How tall is the rectangle?
 What is the probability of
getting less than 15 hours
in a week?
 Given that the student gets
at least 15 hours in a week,
what is the probability that
more than 17.5 hours will
be available?
Chap 5- 19
Learning exercise 4:
Part-time Work on Campus
8
10
12
14
16
18
20
P(hour>17.5)/P(hour>15)
Ka-fu Wong © 2003
22
 How tall is the rectangle?
 (20-10)*h = 1
 h=0.1
 What is the probability of
getting less than 15 hours in
a week?
 0.1*(15-10) = 0.5
 Given that the student gets
at least 15 hours in a week,
what is the probability that
more than 17.5 hours will
be available?
 0.1*(20-17.5) = 0.25
 0.25/0.5 = 0.5
Chap 5- 20
Learning exercise 5:
Customer Complaints
 You are the manager of the complaint department for a large
mail order company. Your data and experience indicate that the
time it takes to handle a single call has the following probability
density function,
h=2/15
0
5
Ka-fu Wong © 2003
10
15
 Show that the area under the
triangle is 1.
 Find the probability that a call will
take longer than 10 minutes. That
is, find P( Time > 10 ).
 Given that the call takes at least 5
minutes, what is the probability
that it will take longer than 10
20 minutes? That is, find P( Time > 10
| Time > 5 ).
 Find P( Time < 10 ).
Chap 5- 21
Learning exercise 5:
Customer Complaints
h=2/15
0
5
Ka-fu Wong © 2003
10
15
 Show that the area under the
triangle is 1.
 (2/15)*(15-0)/2 = 1
20
Chap 5- 22
Learning exercise 5:
Customer Complaints
h=2/15
x
0
5
Ka-fu Wong © 2003
10
15
 Find the probability that a call will
take longer than 10 minutes. That
is, find P( Time > 10 ).
 (2/15)/x = 10/5
 X=2/30 = 0.067
 P(time>10)=(2/30)*5/2=1/6
20
Chap 5- 23
Learning exercise 5:
Customer Complaints
h=2/15
 Find P( Time > 10 | Time > 5 ).
 P(time>10)=(2/30)*5/2=1/6
 P(time>5) = 2/15*10/2=2/3
 P( Time > 10 | Time > 5 )
= (1/6)/(2/3) = 1/4.
x
0
5
Ka-fu Wong © 2003
10
15
20
 Find P( Time < 10 ).
 P( Time < 10 )
= 1 – P(time >10)
= 1-1/6
= 5/6.
Chap 5- 24
Learning exercise 6:
Clutch Sizes in Boreal Owl Nests
 The number of eggs in Boreal owl nests has a probability
mass function with
 P(0) = 0.2 ,
 P(1) = 0.1 ,
 P(2) = 0.1 ,
 P(3) = 0.3 , and
 P(4) = 0.3 .
 Draw the probability mass function (pmf) correctly
labeling the axes.
 What is the probability that a randomly chosen nest will
have no eggs?
 If you examine two nests and they are independent, what
is the probability that neither nest will have eggs?
Ka-fu Wong © 2003
Chap 5- 25
Learning exercise 6:
Clutch Sizes in Boreal Owl Nests
Draw the probability mass
function (pmf) correctly
labeling the axes.
0.35
0.3
probability
P(0) = 0.2 , P(1) = 0.1 ,
P(2) = 0.1 , P(3) = 0.3 ,
and P(4) = 0.3 .
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
Number of eggs
What is the probability that a randomly chosen nest will
have no eggs?
P(0) = 0.2
If you examine two nests and they are independent, what is
the probability that neither nest will have eggs?
P(0 & 0) = P(0)*P(0) = 0.04.
Ka-fu Wong © 2003
Chap 5- 26
Basic Rules of Probability
If two events A and B are mutually
exclusive, the special rule of addition states
that the probability of A or B occurring
equals the sum of their respective
probabilities:
P(A or B) = P(A) + P(B)
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Chap 5- 27
EXAMPLE 3
 New England Commuter Airways recently
supplied the following information on their
commuter flights from Boston to New York:
Ka-fu Wong © 2003
Arrival
Frequency
Early
100
On Time
800
Late
75
Canceled
25
Total
1000
Chap 5- 28
EXAMPLE 3 continued
Arrival
Early
Frequency
100
On Time
800
Late
75
Canceled
25
Total
1000
 If A is the event that a
flight arrives early, then
P(A) = 100/1000 = .10.
 If B is the event that a
flight arrives late, then P(B)
= 75/1000 = .075.
 The probability that a
flight is either early or late
is:
P(A or B) = P(A) + P(B)
= .10 + .075 =.175.
Ka-fu Wong © 2003
Chap 5- 29
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).
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Chap 5- 30
The Complement Rule continued
 A Venn diagram illustrating the complement rule
would appear as:
A
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~A
Chap 5- 31
EXAMPLE 4
 Recall EXAMPLE 3. Use the complement rule to find the
probability of an early (A) or a late (B) flight
Arrival
Early
Frequency
100
On Time
800
Late
75
Canceled
25
Total
1000
 P(A or B) = 1 - P(C or D)
 If C is the event that a flight
arrives on time, then P(C) =
800/1000 = .8.
 If D is the event that a flight
is canceled, then P(D) =
25/1000 = .025.
P(A or B) = 1 - P(C or D)
= 1 - [.8 +.025] =.175
Ka-fu Wong © 2003
Chap 5- 32
EXAMPLE 4 continued
 P(A or B) = 1 - P(C or D)
= 1 - [.8 +.025] =.175
C
.8
D
.025
~(C or D) = (A or B)
.175
Ka-fu Wong © 2003
Chap 5- 33
The General Rule of Addition
 If A and B are two events that are not mutually
exclusive, then P(A or B) is given by the
following formula:
P(A or B) = P(A) + P(B) - P(A and B)
Ka-fu Wong © 2003
Chap 5- 34
The General Rule of Addition
 The Venn Diagram illustrates this rule:
B
A and B
A
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Chap 5- 35
EXAMPLE 5
 In a sample of 500 students, 320 said they
had a stereo, 175 said they had a TV, and 100
said they had both:
Stereo
320
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Both
100
TV
175
Chap 5- 36
EXAMPLE 5 continued
 In a sample of 500 students, 320 said they had a stereo,
175 said they had a TV, and 100 said they had both.
 If a student is selected at random, what is the
probability that the student has only a stereo, only a
TV, and both a stereo and TV?
P(S) = 320/500 = .64.
P(T) = 175/500 = .35.
P(S and T) = 100/500 = .20.
Ka-fu Wong © 2003
Chap 5- 37
EXAMPLE 5 continued
 In a sample of 500 students, 320 said they had a stereo,
175 said they had a TV, and 100 said they had both.
P(S) = 320/500 = .64.
P(T) = 175/500 = .35.
P(S and T) = 100/500 = .20.
 If a student is selected at random, what is the
probability that the student has either a stereo or a TV
in his or her room?
P(S or T)
= P(S) + P(T) - P(S and T)
= .64 +.35 - .20 = .79.
Ka-fu Wong © 2003
Chap 5- 38
Joint Probability
 A joint probability measures the likelihood that
two or more events will happen concurrently.
 An example would be the event that a student
has both a stereo and TV in his or her dorm room.
 P(dice=5, and dice=6) =0 because the two
outcomes are mutually exclusive.
Ka-fu Wong © 2003
Chap 5- 39
Special Rule of Multiplication
 The special rule of multiplication requires that
two events A and B are independent.
 Two events A and B are independent if the
occurrence of one has no effect on the
probability of the occurrence of the other.
 This rule is written:
Ka-fu Wong © 2003
P(A and B) = P(A)P(B)
Chap 5- 40
EXAMPLE 6
 Chris owns two stocks, IBM and General
Electric (GE). The probability that IBM
stock will increase in value next year is .5
and the probability that GE stock will
increase in value next year is .7. Assume
the two stocks are independent. What is
the probability that both stocks will
increase in value next year?
P(IBM and GE) = (.5)(.7) = .35.
Ka-fu Wong © 2003
Chap 5- 41
EXAMPLE 6 continued
 The probability that IBM stock will increase in value
next year is .5 and the probability that GE stock will
increase in value next year is .7.
 What is the probability that at least one of these
stocks increase in value during the next year? (This
means that either one can increase or both.)
Approach 1:
P(at least one)
= (.5)(.3) + (.5)(.7) +(.7)(.5)
= .85.
Ka-fu Wong © 2003
Approach 2:
P(at least one)
= 0.5 + 0.7 – 0.35
= .85.
Chap 5- 42
Conditional Probability
 A conditional probability is the probability of a
particular event occurring, given that another
event has occurred.
 The probability of the event A given that the
event B has occurred is written P(A|B).
 P(female | BusEcon)
 P(girl | test says girl)
Ka-fu Wong © 2003
Chap 5- 43
General Multiplication Rule
 The general rule of multiplication is used to
find the joint probability that two events
will occur.
 It states that for two events A and B, the
joint probability that both events will
happen is found by multiplying the
probability that event A will happen by the
conditional probability of B given that A has
occurred.
Ka-fu Wong © 2003
Chap 5- 44
General Multiplication Rule
 The joint probability, P(A and B) is given by the
following formula:
P(A and B) = P(A)P(B/A)
or
P(A and B) = P(B)P(A/B)
 P(test says girl and girl)
= P(girls) * P(test says girls | girls)
 P(test says boy and boy)
= P(boys) * P(test says boys | boys)
Ka-fu Wong © 2003
Chap 5- 45
EXAMPLE 7
The Dean of the School of Business at Owens
University collected the following information
about undergraduate students in her college:
Ka-fu Wong © 2003
MAJOR
Male
Female
Total
Accounting
170
110
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
Chap 5- 46
EXAMPLE 7 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?
Approach 1:
P(A|F) = P(A and F)/P(F)
= [110/1000]/[400/1000]
= .275
Ka-fu Wong © 2003
Approach 2:
P(A|F)
= 110/400
= .275
Chap 5- 47
Tree Diagrams
 A tree diagram is useful for portraying
conditional and joint probabilities. It is
particularly useful for analyzing business
decisions involving several stages.
 EXAMPLE 8: In a bag containing 7 red chips and
5 blue chips you select 2 chips one after the
other without replacement. Construct a tree
diagram showing this information.
Ka-fu Wong © 2003
Chap 5- 48
EXAMPLE 8 continued
6/11
7/12
5/12
R1
5/11
B2
7/11
R2
B1
4/11
Ka-fu Wong © 2003
R2
B2
Chap 5- 49
EXAMPLE 8 continued
The tree diagram is very illustrative about the relation
between joint probability and conditional probability
Let A (B) be the event of a red chip in the first (second) draw.
6/11
P(B|A) = 6/11
R2
P(A) = 7/12
7/12
5/12
R1
5/11
B2
7/11
R2
B1
4/11
Ka-fu Wong © 2003
P(A and B)
= P(A)*P(B|A)
= 6/11 * 7/12
B2
Chap 5- 50
Bayes’ Theorem
 Bayes’ Theorem is a method for revising a
probability given additional information.
 It is computed using the following formula:
P(A1 )P(B/A 1 )
P(A1 | B) 
P(A1 )P(B/A 1 )  P(A 2 )P(B/A 2 )
Bayes Theorem is an essential tool to understand
options and real options.
Ka-fu Wong © 2003
Chap 5- 51
Bayes’ Theorem
P(A1 )P(B/A 1 )
P(A1 | B) 
P(A1 )P(B/A 1 )  P(A 2 )P(B/A 2 )
 Bayes’ Theorem can be derived based on simple
manipulation of the general multiplication rule.
P(A1|B)
= P(A1 & B) /P(B)
= [P(A1) P(B|A1)] / P(B)
= [P(A1) P(B|A1)] / [P(A1 & B) + P(A2 & B)]
= [P(A1) P(B|A1) ]/ [P(A1) P(B|A1) + P(A2) P(B|A2)
Ka-fu Wong © 2003
Chap 5- 52
EXAMPLE 9
 Duff Cola Company recently received several
complaints that their bottles are under-filled. A
complaint was received today but the production
manager is unable to identify which of the two
Springfield plants (A or B) filled this bottle. What is
the probability that the under-filled bottle came from
plant A?
 The following table summarizes the Duff production
experience.
Ka-fu Wong © 2003
% of Total Production
% of under-filled bottles
A
55
3
B
45
4
Chap 5- 53
Example 9 continued
% of Total Production
% of under-filled bottles
A
55
3
B
45
4
 What is the probability that the under-filled bottle came
from plant A?
P(A)P(U/A)
P(A)P(U/A)  P(B)P(U/B)
.55(.03)

 .4783
.55(.03) .45(.04)
P(A/U) 
The likelihood the bottle was filled in Plant A is reduced from
.55 to .4783.
Without the information about U, the manager will say the
under-filled bottle is likely from plant A. With the additional
information about U, the manager will say the under-filled
bottle is likely from plant B.
Ka-fu Wong © 2003
Chap 5- 54
Some Principles of Counting
 The multiplication formula indicates that if
there are m ways of doing one thing and n
ways of doing another thing, there are m x
n ways of doing both.
 Example 10: Dr. Delong has 10 shirts and 8
ties. How many shirt and tie outfits does he
have?
(10)(8) = 80
Ka-fu Wong © 2003
Chap 5- 55
Some Principles of Counting
 A permutation is any arrangement of r
objects selected from n possible objects.
 Note: The order of arrangement is
important in permutations.
n!
n Pr 
(n  r)!
Ka-fu Wong © 2003
Chap 5- 56
Some Principles of Counting
A combination is the number of ways
to choose r objects from a group of n
objects without regard to order.
n!
n Cr 
r!(n  r)!
Ka-fu Wong © 2003
Chap 5- 57
EXAMPLE 11
 There are 12 players on the Carolina Forest
High School basketball team. Coach
Thompson must pick five players among the
twelve on the team to comprise the starting
lineup. How many different groups are
possible?
12!
 792
12 C 5 
5!(12  5)!
Ka-fu Wong © 2003
Chap 5- 58
Example 11 continued
 Suppose that in addition to selecting the group,
he must also rank each of the players in that
starting lineup according to their ability.
12!
 95,040
12 P 5 
(12  5)!
Ka-fu Wong © 2003
Chap 5- 59
Chapter Five
A Survey of Probability Concepts
- END -
Ka-fu Wong © 2003
Chap 5- 60
Probabilities for n Dice
 Suppose we roll n regular balanced six-sided
dice. If the event X is the sum of the n values
which appear, what are the probabilities
associated for each value of X, for the possible
values X = n, ... , 6n?
 In the case n =1, these probabilities are all 1/6.
 For two dice , it is easiset to
consider a table of possible outcomes:
Ka-fu Wong © 2003
Chap 5- 61
Probabilities for n Dice
 For two dice , it is easiest to consider a table of
possible outcomes:
Ka-fu Wong © 2003
(1,1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
(3,1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
(4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
(5,1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
(6,6)
Chap 5- 62
Probabilities for n Dice
 Next, we can consider the sums associated with
these outcomes::
Ka-fu Wong © 2003
2
3
4
5
6
7
3
4
5
6
7
8
4
5
6
7
8
9
5
6
7
8
9
10
6
7
8
9
10
11
7
8
9
10
11
12
Chap 5- 63
Probabilities for n Dice
 Since there are 36 outcomes, each equally likely,
we see that the probabilities are:
Ka-fu Wong © 2003
X
Count
P(X)
X
Count
P(X)
2
1
1/36
8
5
5/36
3
2
2/36
9
4
4/36
4
3
3/36
10
3
3/36
5
4
4/36
11
2
2/36
6
5
5/36
12
1
1/36
7
6
6/36
Chap 5- 64
Probabilities for n Dice
 If there are more than two dice, it is difficult to make
tables such as these. The number
of outcomes will be 6n, so to calculate the probabilities it
is sufficient to count the
number of times each sum occurs among the 6n
possible outcomes. This is easily done
by considering the generating function for the number of
times each sum appears:
f(x) = (x + x2 + x3 + x4 + x 5+x6)n
 In our example, n=2
(x + x2 + x3 + x4 + x 5+x6)2
=x2 + 2x3 +3x4 + 4x5+5x6+6x7+5x8+4x9+3x10+2x11+x12
 The coefficient of x3 is number of times the sum of 3
occurs.
Ka-fu Wong © 2003
Chap 5- 65