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.
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.
l
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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.
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Chap 5- 3
Definitions
continued
The classical definition applies when there are equally likely outcomes.
n
The empirical definition applies when the number of times the event happens is divided by the number of observations, based on data .
Subjective information is available, based on subjective feelings.
probability is based on whatever
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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.
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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.
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Chap 5- 6
Mutually Exclusive, Independent and Exhaustive events
Events are event means that none of the others can occur at the same time.
mutually exclusive if the occurrence of any one
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.
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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)
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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.
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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:
P(A) 186 1200 0.155
This number may be interpreted as “unconditional probability”.
In most cases, we are interested in week the probability of earning an A for a selected student who study 10 hours or more per . We call this “conditional probability”. P(A | study 10 or more hours per week)
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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.
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Chap 5- 11
Learning exercise 1: University Demographics
Current enrollments by college and by sex appear in the following table. College Female Male Totals Ag-For Arts-Sci Bus-Econ Educ Engr Law 500 900 1400 1500 1200 2700 400 500 900 1000 200 500 100 1300 200 1500 1500 300 Undecl Totals 800 900 1700 4500 5500 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)
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Find P(Male and Arts-Sci) Are “Female” and “Educ” Statistical independent? Why or Why not?
Chap 5- 12
Learning exercise 1: University Demographics
College Female Male Totals Ag-For Arts-Sci Bus-Econ Educ Engr Law 500 900 1400 1500 1200 2700 400 500 900 1000 200 500 100 1300 200 1500 1500 300 Undecl Totals 800 900 1700 4500 5500 10000 P(Female or Male) P(not-Ag-For) P(Female | Bus Econ) P(Male and Arts-Sci) =(4500 + 5500)/10000 = 1 =(10000 – 1400) /10000 = 0.86
= 400 /900 = 0.44
=1200 /10000 = 0.12
Are Female and Educ Statistical independent? P(female and Educ) =1000 /10000 = 0.1
P(female) =4500 /10000 = 0.45
P(Educ) NO!
=1500 /10000 = 0.15
P(female and Educ) > P(female)*P(Educ) = 0.0675
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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).
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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
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Chap 5- 15
Learning exercise 2: Predicting Sex of Babies
480 = 1000*0.48 are girls 520 = 1000*0.52 are boys 1000*P(girls) 1000*P(boys) Of the girls, 432 (=480*0.9) tests indicate that they are girls.
1000*P(girls)*P(test says girls|girls) Of the boys, 130 (=520*0.25) tests indicate that they are girls.
1000*P(boys)*P(test says girls | boys)
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)]
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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 ).
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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 Line 1 works Line 1 fails 0.98*0.98
0.02*0.98
Line 2 fails 0.98*0.02
0.02*0.02
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Chap 5- 18
Learning exercise 4: Part-time Work on Campus
8
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: 10 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?
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Chap 5- 19
8 10
Learning exercise 4: Part-time Work on Campus
12 14 16 18 20 P(hour>17.5)/P(hour>15)
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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
0
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
Show that the area under the triangle is 1. 5 10 15
20
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 minutes? That is, find P( Time > 10 | Time > 5 ).
Find P( Time < 10 ).
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Chap 5- 21
Learning exercise 5: Customer Complaints
h=2/15
Show that the area under the triangle is 1.
(2/15)*(15-0)/2 = 1 0 5 10 15 20
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Chap 5- 22
0 5
Learning exercise 5: Customer Complaints
x h=2/15
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 10 15 20
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Chap 5- 23
0 5
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 10 15 20
Find P( Time < 10 ).
P( Time < 10 ) = 1 – P(time >10) = 1-1/6 = 5/6.
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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?
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Chap 5- 25
Learning exercise 6: Clutch Sizes in Boreal Owl Nests
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.
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0 0 1 2 Number of eggs 3 What is the probability that a randomly chosen nest will have no eggs?
P(0) = 0.2
4 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.
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Chap 5- 26
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Basic Rules of Probability
If two events
A
and
B
are mutually exclusive, the special rule of addition that the probability of equals the sum of their respective probabilities:
A
or
B
states occurring P(
A
or
B
) = P(
A
) + P
(B)
Chap 5- 27
EXAMPLE 3
New England Commuter Airways recently supplied the following information on their commuter flights from Boston to New York:
Arrival Early On Time Late Canceled Total Frequency 100 800 75 25 1000 Ka-fu Wong © 2003
Chap 5- 28
EXAMPLE 3
continued
Arrival Early On Time Late Canceled Total
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Frequency 100 800 75 25 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 = 75/1000 = .075.
P ( B )
The probability that a flight is either early or late is: P ( A or B ) = P( A ) + P( = .10 + .075 =.175.
B ) Chap 5- 29
The Complement Rule
The complement rule probability of an event occurring by subtracting the probability of the event occurring from 1. is used to determine the
not
If
P
(A) is the probability of event is the complement of
A
,
A
P(
A
) +
P
(~
A
) = 1 or
P
(
A
) = 1 and
P
(~
A
)
P
(~
A
).
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Chap 5- 30
The Complement Rule
continued
A Venn diagram illustrating the complement rule would appear as:
A
~A
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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
P ( A or B ) = 1 P ( C or D ) Arrival Early On Time Late Canceled Total Frequency 100 800 75 25 1000
If C is the event that a flight arrives on time, then P( 800/1000 = .8.
C ) = If D is the event that a flight is canceled, then P( 25/1000 = .025.
D ) = P ( A or B ) = 1 P ( C or = 1 - [.8 +.025] =.175
D )
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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
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Chap 5- 33
The General Rule of Addition
If
A
and
B
are two events that are not mutually exclusive, then
P
( following formula:
A
or
B
) is given by the
P
(
A
or
B
) =
P
(
A
) +
P
(
B
) -
P
(
A
and
B
)
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Chap 5- 34
The General Rule of Addition
The Venn Diagram illustrates this rule: A
A and B
B
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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
Both 100
TV 175
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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.
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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 P ( or S T ) ) + P ( T ) P ( S and = .64 +.35 - .20 = .79.
T )
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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.
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Chap 5- 39
Special Rule of Multiplication
The special rule of multiplication two events
A
and
B
are requires that
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:
P
(
A
and
B
) =
P
(
A
)
P
(
B
)
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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.
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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.
Approach 2: P(at least one) = 0.5 + 0.7 – 0.35 = .85.
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Chap 5- 42
Conditional Probability
A conditional probability particular event occurring, given that another event has occurred.
is the probability of a The probability of the event event
B A
given that the has occurred is written
P
(
A
|
B
).
P(female | BusEcon)
P(girl | test says girl)
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Chap 5- 43
General Multiplication Rule
The general rule of multiplication find the joint probability that two events will occur. is used to
It states that for two events joint probability that both events will happen is found by multiplying the probability that event conditional probability of occurred.
A
will happen by the
B A
and
B
, the given that
A
has
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Chap 5- 44
General Multiplication Rule
The joint probability, following formula: P ( A and B ) is given by the P ( A P ( A and B ) = P ( A ) P ( B/A ) or 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)
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Chap 5- 45
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EXAMPLE 7
The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college:
MAJOR Accounting Finance Marketing Management Total Male 170 120 160 150 600 Female 110 100 70 120 400 Total 280 220 230 270 1000
Chap 5- 46
EXAMPLE 7
continued
MAJOR Accounting Finance Marketing Management Total Male Female 170 120 110 100 160 150 600 70 120 400 Total 280 220 230 270 1000 If a student is selected at random, what is the probability that the student is a female ( major ( A ) F) accounting 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
Approach 2: P(A|F) = 110/400 = .275
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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.
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Chap 5- 48
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EXAMPLE 8
continued
7/12 5/12 6/11 R1 B1 5/11 7/11 4/11 R2 B2 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.
R2 P(B|A) = 6/11 P(A) = 7/12 R1 5/11 7/11 B2 R2 P(A and B) = P(A)*P(B|A) = 6/11 * 7/12 5/12 B1 4/11 B2
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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(A
1
| B)
P(A
1
P(A
1
)P(B/A
1
) )P(B/A
1
P(A
2
) )P(B/A
2
)
Bayes Theorem is an essential tool to understand options and real options.
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Chap 5- 51
Bayes
’
Theorem
P(A
1
| B)
P(A
1
P(A
1
)P(B/A
1
) )P(B/A
P(A
1 2
) )P(B/A
2
)
Bayes ’ Theorem can be derived based on simple manipulation of the general multiplication rule.
P(A 1 |B) = P(A 1 & B) /P(B) = [P(A 1 ) P(B|A 1 )] / P(B) = [P(A 1 ) P(B|A 1 )] / [P(A 1 & B) + P(A 2 & B)] = [P(A1) P(B|A 1 ) ]/ [P(A 1 ) P(B|A 1 ) + P(A 2 ) P(B|A 2 )
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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 ( the probability that the under-filled bottle came from plant A ?
A or B ) filled this bottle. What is The following table summarizes the Duff production experience. A B % of Total Production % of under-filled bottles 55 45 3 4
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Chap 5- 53
Example 9
continued
% of Total Production % of under-filled bottles A B 55 45 3 4
What is the probability that the under-filled bottle came from plant A ?
P(A/U) P(A)P(U/A) P(A)P(U/A) P(B)P(U/B) .55(.03) .55(.03) .45(.04) .4783
The likelihood the bottle was filled in Plant
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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.
Chap 5- 54
Some Principles of Counting
The there are ways of doing another thing, there are
n
multiplication formula
m
ways of doing both.
indicates that if ways of doing one thing and
m n
x
Example 10: ties. How many shirt and tie outfits does he have? Dr. Delong has 10 shirts and 8 (10)(8) = 80
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Chap 5- 55
Some Principles of Counting
A permutation is any arrangement of objects selected from
n r
possible objects.
Note :
The order of arrangement is important in permutations.
n P r (n n!
r)!
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Chap 5- 56
Some Principles of Counting
A
combination
to choose
r
is the number of ways objects from a group of objects without regard to order.
n
n
C
r
r!
(n n!
r)!
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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 C 5 12!
5!
(12 5)!
792 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
P
5
12!
(12
5)!
95,040
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Chap 5- 59
Ka-fu Wong © 2003
Chapter Five
A Survey of Probability Concepts
- END -
Chap 5- 60
Probabilities for
n
Dice
Suppose we roll
n
dice. If the event
X
regular balanced six-sided is the sum of the associated for each value of
X, n
values which appear, what are the probabilities for the possible values
X = n
, ... , 6
n
? In the case
n
=1, these probabilities are all 1/6.
For two dice , it is easiset to consider a table of possible outcomes:
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Chap 5- 61
Probabilities for
n
Dice
For two dice , it is easiest to consider a table of possible outcomes: (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,4) (2,4) (3,4) (4,4) (5,4) (6,4) (1,5) (2,5) (3,5) (4,5) (5,5) (6,5) (1,6) (2,6) (3,6) (4,6) (5,6) (6,6)
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Chap 5- 62
Probabilities for
n
Dice
Next, we can consider the sums associated with these outcomes:: 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
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Chap 5- 63
Probabilities for
n
Dice
Since there are 36 outcomes, each equally likely, we see that the probabilities are: X 2 3 4 5 6 7 Count 1 2 3 4 5 6 P(X) 1/36 2/36 3/36 4/36 5/36 6/36 X 8 9 10 11 12 Count 5 4 3 2 1 P(X) 5/36 4/36 3/36 2/36 1/36
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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 + x 2 + x 3 + x 4 + x 5 +x 6 ) n
In our example, n=2 (x + x 2 + x 3 + x 4 + x 5 +x 6 ) 2 =x
2 + 2x 3 The coefficient of x occurs. +3x 4 + 4x 5 +5x 6 +6x 7 +5x 8 +4x 9 +3x 10 +2x 11 +x 12 3 is number of times the sum of 3
Ka-fu Wong © 2003
Chap 5- 65