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Chapter 4
Probability
LEARNING OBJECTIVES
The main objective of Chapter 4 is to help you understand the basic
principles of probability, thereby enabling you to:
1.
2.
3.
4.
5.
6.
7.
8.
Describe what probability is and when one would use it
Differentiate among three methods of assigning probabilities: the classical
method, relative frequency of occurrence, and subjective probability
Deconstruct the elements of probability by defining experiments, sample
spaces, and events, classifying events as mutually exclusive, collectively
exhaustive, complementary, or independent, and counting possibilities
Compare marginal, union, joint, and conditional probabilities by defining
each one.
Calculate probabilities using the general law of addition, along with a
probability matrix, the complement of a union, or the special law of
addition if necessary
Calculate joint probabilities of both independent and dependent events
using the general and special laws of multiplication
Calculate conditional probabilities with various forms of the law of
conditional probability, and use them to determine if two events are
independent.
Calculate conditional probabilities using Bayes’ rule
CHAPTER OUTLINE
4.1
Introduction to Probability
4.2
Methods of Assigning Probabilities
Classical Method of Assigning Probabilities
Relative Frequency of Occurrence
Subjective Probability
4.3
Structure of Probability
Experiment
Event
Elementary Events
Sample Space
Unions and Intersections
Mutually Exclusive Events
Independent Events
Collectively Exhaustive Events
Complimentary Events
Counting the Possibilities
The mn Counting Rule
Sampling from a Population with Replacement
Combinations: Sampling from a Population Without Replacement
4.4
Marginal, Union, Joint, and Conditional Probabilities
4.5
Addition Laws
Probability Matrices
Complement of a Union
Special Law of Addition
4.6
Multiplication Laws
General Law of Multiplication
Special Law of Multiplication
4.7
Conditional Probability
Independent Events
4.8
Revision of Probabilities: Bayes' Rule
KEY TERMS
A Priori
Bayes' Rule
Classical Method of Assigning Probabilities
Collectively Exhaustive Events
Combinations
Complement of a Union
Complement
Conditional Probability
Elementary Events
Event
Experiment
Independent Events
Intersection
Joint Probability
Marginal Probability
mn Counting Rule
Mutually Exclusive Events
Probability Matrix
Relative Frequency of Occurrence
Sample Space
Set Notation
Subjective Probability
Union
Union Probability
STUDY QUESTIONS
1. The _______________ method of assigning probabilities relies on the insight or feelings of
the person determining the probabilities.
2.
If probabilities are determined "a priori" to an experiment using rules and laws, then the
_______________ method of assigning probabilities is being used.
3. The range of possibilities for probability values is from _______________ to
_______________.
4.
Suppose a technician keeps track of all defects in raw materials for a single day and uses this
information to determine the probability of finding a defect in raw materials the next day.
She is using the _______________ method of assigning probabilities.
5. The outcome of an experiment is called a(n) _______________. If these outcomes cannot
be decomposed further, then they are referred to as _______________ _______________.
6.
A computer hardware retailer allows you to order your own computer monitor. The
store carries five different brands of monitors. Each brand comes in 14", 15" or 17"
models. In addition, you can purchase either the deluxe model or the regular model in
each brand and in each size. How many different types of monitors are available
considering all the factors? _____________ You probably used the _______ rule
to solve this.
7.
Suppose you are playing the Lotto game and you are trying to “pick” three numbers.
For each of the three numbers, any of the digits 0 through 9 are possible (with
replacement). How many different sets of numbers are available? _____________
8.
A population consists of the odd numbers between 1 and 9 inclusive. If a researcher
randomly samples numbers from the population three at a time, the sample space is
_______________. Using combinations, how could we have determined ahead of time how
many elementary events would be in the sample space? ________
9.
Let A = {2,3,5,6,7,9} and B = {1,3,4,6,7,9}
_____________.
10. If the occurrence of one event does not affect the occurrence of the other event, then the
events are said to be _______________.
11. The outcome of the roll of one die is said to be _______________ of the outcome of the roll
of another die.
12. The event of rolling a three on a die and the event of rolling an even number on the same
roll with the same die are _______________.
13.
If the probability of the intersection of two events is zero, then the events are said to be
_______________.
14.
If three objects are selected from a bin, one at a time with replacement, the outcomes of
each selection are _______________.
15.
Suppose a population consists of a manufacturing facility's 1600 workers. Suppose an
experiment is conducted in which a worker is randomly selected. If an event is the selection
of a worker over 40 years old, then the event of selecting a worker 40 years or younger is
called the _______________ of the first event.
16. The probability of selecting X given that Y has occurred is called a _______________
probability.
17. The probability of X is called a _______________ probability.
18. The probability of X or Y occurring is called a _______________ probability.
19. The probability of X and Y occurring is called a _______________ probability.
20.
Only one of the four types of probability does not use the total possible outcomes in the
denominator when calculating the probability. This type of probability is called
_______________ probability.
______________.
_______________.
24.
In a company, 47% of the employees wear glasses, 60% of the employees are women, and
28% of the employees are women and wear glasses. Complete the probability matrix below
for this problem.
Wear Glasses?
Yes
No
Gender
Men
Women
25.
Suppose that in another company, 40% of the workers are part time and 80% of the part
time workers are men. The probability of randomly selecting a company worker who is
both part time and a man is _______________.
26. The probability of tossing three coins in a row and getting all tails is _______________.
This is an application of the _______________ law of multiplication because each toss is
_______________.
27.
Suppose 70% of all cars purchased in America are U.S.A. made and that 18% of all cars
purchased in America are both U.S.A. made and are red. The probability that a randomly
selected car purchased in America is red given that it is U.S.A. made is _______________.
Use the matrix below to answer questions 28-37:
A
B
C
.35
.14
.49
D
.31
.20
.51
.66
.34
1.00
28. The probability of A and C occurring is __________.
29. The probability of A or D occurring is __________.
30. The probability of D occurring is __________.
31. The probability of B occurring given C is __________.
32. The probability of B and D occurring is __________.
33. The probability of C and D occurring is __________.
34. The probability of C or D occurring is __________.
35. The probability of C occurring given D is __________.
36. The probability of C occurring given A is __________.
37. The probability of C or B occurring is __________.
38.
Suppose 42% of all people in a county have characteristic X. Suppose 17% of all people in
this county have characteristic X and characteristic Y. If a person is randomly selected
from the county who is known to have characteristic X, then the probability that they have
characteristic Y is _________________.
39. Suppose 22% of all parts produced at a plant have flaw X and 37% have flaw Y. In
addition, suppose 53% of the parts with flaw X have flaw Y. If a part is randomly selected,
the probability that it has flaw X or flaw Y is ________________.
40.
Another name for revision of probabilities is _______________.
41.
Suppose the prior probabilities of A and B are .57 and .43 respectively. Suppose that
probability of A occurring is _______________ and of B occurring is _______________.
ANSWERS TO STUDY QUESTIONS
1. Subjective
23. .57
2. Classical
24.
3. 0, 1
4. Relative Frequency
Wear Glasses
Yes
No
Men .19
.21 .40
Women .28
.32 .60
.47
.53 1.00
5. Event, Elementary Events
25. .32
6. 30, mn counting rule
26. 1/8 = .125, Special, Independent
7. 103 = 1000 numbers
27. .2571
8. {(1,3,5), (1,3,7), (1,3,9), (1,5,7),
(1,5,9), (1,7,9), (3,5,7), (3,5,9),
(3,7,9), (5,7,9)}, 5C3 = 10
28. .35
9. {1,2,3,4,5,6,7,9}, {3,6,7,9}
30. .51
29. .86
10. Independent
31. .2857
11. Independent
32. .20
12. Mutually Exclusive
33. .0000
13. Mutually Exclusive
34. 1.00
14. Independent
35. .0000
15. Complement
36. .5303
16. Conditional
37. .69
17. Marginal
38. .4048
18. Union
39. .4734
19. Joint or Intersection
40. Bayes’ Rule
20. Conditional
41. .3623, .6377
21. Independent
22. .58
SOLUTIONS TO PROBLEMS IN CHAPTER 4
4.1
Enumeration of the six parts: D1, D2, D3, A4, A5, A6
D = Defective part
A = Acceptable part
Sample Space:
D1 D2,
D1 D3,
D1 A4,
D1 A5,
D1 A6,
D2 D3,
D2 A4,
D2 A5,
D2 A6,
D3 A4,
D3 A5
D3 A6
A4 A5
A4 A6
A5 A6
There are 15 members of the sample space
The probability of selecting exactly one defect out of
two is:
9/15 = .60
4.3
If A = {2, 6, 12, 24} and the population is the positive even numbers through 30,
A’ = {4, 8, 10, 14, 16, 18, 20, 22, 26, 28, 30}
4.5
Enumeration of the six parts: D1, D2, A1, A2, A3, A4
D = Defective part
A = Acceptable part
Sample Space:
D1 D2 A1,
D1 D2 A4,
D1 A1 A4,
D1 A3 A4,
D2 A1 A4,
D2 A3 A4,
A1 A3 A4,
D1 D2 A2,
D1 A1 A2,
D1 A2 A3,
D2 A1 A2,
D2 A2 A3,
A1 A2 A3,
A2 A3 A4
D1 D2 A3,
D1 A1 A3,
D1 A2 A4,
D2 A1 A3,
D2 A2 A4,
A1 A2 A4,
Combinations are used to counting the sample space because sampling is done without replacement.
6C3 = = 20
Probability that one of three is defective is:
12/20 = 3/5
.60
There are 20 members of the sample space and 12 of them have exactly
1 defective part.
4.7
20C6 = = 38,760
It is assumed here that 6 different (without replacement) employees are to be selected.
A
B
C
4.9
D
5
10
8
23
E
8
6
2
16
b) P(E
c) P(D
d) P(C
4.11
F
12
4
5
21
25
20
15
60
- 5/60 = 43/60 = .7167
B) = P(E) + P(B) - P(E B) = 16/60 + 20/60 - 6/60 = 30/60 = .5000
E) = P(D) + P(E) = 23/60 + 16/60 = 39/60 = .6500
F) = P(C) + P(F) - P(C
F) = 15/60 + 21/60 - 5/60 = 31/60 = .5167
A = event of having flown in an airplane at least once
T = event of having ridden in a train at least once
P(A) = .47
P(T) = .28
P (ridden either a train or an airplane) =
= P(A) + P(T) -
-
Cannot solve this problem without knowing the probability of the intersection.
We need to know the probability of the intersection of A and T, the proportion
who have ridden both or determine if these two events are mutually exclusive.
4.13 Let C = have cable TV
Let T = have 2 or more TV sets
P(C) = .67, P(T) = .74, P(C
a) P(C
T) = .55
T) = P(C) + P(T) - P(C
-
T) = .67 + .74 - .55 = .86
- .55 = .31
- .86 = .14
not .0000.
Possession of cable TV and 2 or more TV sets are not mutually exclusive.
4.15
A
B
C
5
2
7
D
11
3
14
E
16
5
21
F
8
7
15
40
17
57
.0000
4.17
Let D = Defective part
a) (without replacement)
b) (with replacement)
4.19 Let A = investor is 35 years old or less
P(A) = .37
P(E)=.47
a) P(NA) = 1-.37 = .63
Let E = postgraduate education
P(E|A)=.10
b) P(A ∩ E) = P(A)∙P(E|A) = (.37)(.10) = 0.037
c) P(A U E) = P(A)+P(B)-P(A∩E)= .37+.47-.037 = .803
d) P(NA ∩ NE) = 1-P(A U E) = 1-.803 = .197
e) P(NA U NE) = P(NA)+P(NE)-P(NS ∩ NE) = .63+.53-.197=.963
f) P(A ∩ NE) = P(A)-P(A ∩ E) = .37-.037=.333
E
Yes
No
.037
.433
.47
Let S = stockholder
.333
.197
A
Yes
No
.37
.63
.53 1.00
Let C = college
a) P(NS) = 1 - .43 = .57
-
- .3225 = .4775
- .4775 = .5225
- .5225 = .6775
- .3225 = .0475
The matrix:
C
S
Yes
No
4.21
Yes
No
.3225
.0475
.37
.1075
.5225
.43
.57
.63
1.00
Let N = regular exercise in the park or out in nature
P(N) = .51
P(F)=.15
P(F|N)=.35
a) P(N∩NF) = P(N)∙P(NF|N)
but P(NF|N)=1-P(F|N)=1-.22=.78
P(N∩NF)=.51*.78=.3978
b) P(NN∩NF)=1-P(NUF)=1-[P(N) + P(F) - P(N ∩ F)]
but P(N ∩ F)]=P(N)∙P(F|N)=(.51)(.22)=.1122
P(NN∩NF)= 1-(.51+.15-.1122)=.4522
c) P(NN∩F)=P(NN)-P(NN∩NF)
but P(NN)=1-P(N)=1-.51=.49
P(NN∩F)=.49-.4522=.0378
The matrix:
Let F = fitness centre
F
N
Yes
No
Yes
No
.1122
.0378
.15
.3978
.4522
Let S = safety
P(S) = .30
.51
.49
.85
1.00
Let A = age
P(A) = .39
-
- .87 = .13
-
- [P(S) + P(A) (.87) = .261
- (.30 + .39 - .261) = .571
-
but P(NS) = 1 - P(S) = 1 - .30 = .70
- 571 = .129
The matrix:
A
Yes
S
A
B
C
D
Yes
No
4.23
E
15
11
21
18
65
No
.261
.129
.39
.039
.571
F
12
17
32
13
74
G
8
19
27
12
66
b)
.30
.70
.61
35
47
80
43
205
1.00
4.25
Yes
Calculator
No
Computer Yes
No
11
57
46
15
18
3
26
75
49
Select a category from each variable and test
For example,
?
Since this is one example that the conditional does not equal the marginal in
is matrix, the variable, computer, is not independent of the variable,
calculator.
4.27
Let E = Economy
P(E) = .46
P(Q) = .37
Let Q = Qualified
6 = .3261
-
- .15 = .22
P(NE) = 1 - P(E) = 1 - .46 = .54
-
- [P(E) + P(Q) -
= 1 - [.46 + .37 - .15] = 1 - (.68) = .32
The matrix:
Q
E
Yes
No
Yes
No
.15
.22
.37
.31
.32
.46
.54
.63
1.00
4.29
Let N = regular exercise in the park or out in nature
P(N) = .51
P(F)=.15
Let F = fitness centre
P(F|N)=.35
a) P(N∩NF) = P(N)∙P(NF|N)
but P(NF|N)=1-P(F|N)=1-.22=.78
P(N∩NF)=.51*.78=.3978
b) P(NN∩NF)=1-P(NUF)=1-[P(N) + P(F) - P(N ∩ F)]
but P(N ∩ F)]=P(N)∙P(F|N)=(.51)(.22)=.1122
P(NN∩NF)= 1-(.51+.15-.1122)=.4522
c) P(NN∩F)=P(NN)-P(NN∩NF)
but P(NN)=1-P(N)=1-.51=.49
P(NN∩F)=.49-.4522=.0378
The matrix:
F
N
Yes
No
4.22
Yes
No
.1122
.0378
.15
.3978
.4522
.51
.49
.85
1.00
Let M = microwave
Let D = dishwasher
P(M) = .84
P(D) = .47
P(M ∩ D) = .45
a) P(M U D)= P(M) + P(D) - P(M ∩ D) = .84 + .47 - .45 = .86
b) P(NM ∩ ND) = 1 - P(M U D)= 1 - .86 = .14
c) P(NM ∩ D) = P(D) - P(M ∩ D) = .47 - .45 = .02
d) P(M ∩ ND) = P(M) - P(M ∩ D) = .84 - .45 = .39
The matrix:
D
M Yes
No
A
B
4.23
E
15
11
Yes
No
.45
.39
.84
.02
.14
F
12
17
G
8
19
.47
.53
.16
35
47
1.00
C
D
21
18
65
32
13
74
27
12
66
D
.44
.09
.53
.80
.20
1.00
Yes
Calculator
No
80
43
205
4.24
A
B
C
.36
.11
.47
4.25
Computer Yes
No
11
57
46
15
18
3
26
75
49
Select a category from each variable and test
For example,
?
Since this is one example that the conditional does not equal the marginal in
is matrix, the variable, computer, is not independent of the variable,
calculator.
4.26
Let C = construction
Let N = North West
19,946 total failures
3,267 failures in construction
2,382 failures in North West
458 failures in construction and North West
a) P(N) = 2,382/19,946 = .1194
b) P(C U N) = P(C) + P(N) - P(C ∩ N) =
3,267/19,946+ 2,382/19,946 - 458/19,946= 5,191/19,946= .2603
but NC = 19,946- 3,267 = 16,679
and P(NC) = 16,679/19,946= .8362
Th
- .2603)/(.8362) = .8846
but P(C) = 3,267/19,946= .1638
P(C ∩ N) = 458/19,946= .023
- .023)/.1638 = .8596
4.27
Let E = Economy
P(E) = .46
P(Q) = .37
Let Q = Qualified
-
- .15 = .22
P(NE) = 1 - P(E) = 1 - .46 = .54
-
- [P(E) + P(Q) -
= 1 - [.46 + .37 - .15] = 1 - (.68) = .32
The matrix:
Q
Yes
No
E
4.28
Yes
No
.15
.22
.37
.31
.32
.46
.54
.63
1.00
Let TV = internet on handset while watching TV
P(TV) = .86
Let SN = social networking site
- .20 = .80
c) P(not SN ∩ TV) = P(TV) - P(TV ∩ SN) = .86 - .172 = .688
(.86)(.80) = .688
The matrix:
SN
Yes
No
TV
Yes
.172
No
4.29
.688
.14
1.00
.86
Let LI = life insurance
Let PS = pension scheme
P(LI) = .49
P(PS) = .16
P(PS LI) = .27
- .27 = .73
= P(PS ∩ NLI)/P(NLI)
so P(NLI ∩ PS) = P(PS) - P(LI ∩ PS) = .16 - .1323 = .0277
P(NLI) = 1 - P(LI) = 1 - .49 = .51
P(PS NLI) = (.0277)/(.51) = .054
c) P(NLI PS) = P(NLI ∩ PS)/P(PS) = .0277/.16 = .1731
d) P(NLI NPS) = P(NLI ∩ NPS)/P(NPS)
but P(NLI ∩ NPS) = P(NLI) - P(NLI ∩ PS) = .51 - .0277 = .4823
and P(NPS) = 1 - P(PS) = 1 - .16 = .84
The matrix:
Yes
PS
No
LI Yes
No
.29
.1323
.0277
.16
.3577
.4823
Let H = hardware
P(H) = .37
.49
.51
.84
1.00
Let S = software
P(S) = .54
)=1-
- .97 = .03
-
- .3589 = .1811
P(NH) = 1 - P(H) = 1 - .37 = .63
.63) = .2875
and P(NS) = 1 - P(S) = 1 - .54 = .46
The matrix:
22
S
H
Yes
No
4.31
Yes
No
.3589
.1811
.54
.0111
.4489
Let
.37
.63
.46
1.00
A = product produced on Machine A
B = product produces on Machine B
C = product produced on Machine C
D = defective product
- .1811 = .4489
P(A) = .10
Event
Prior Conditional
P(B) = .40
Joint
P(C) = .50
Revised
P(Ei)
A
B
C
.10
.40
.50
.05
.12
.08
.005 .005/.093=.0538
.048 .048/.093=.5161
.040 .040/.093=.4301
P(D)=.093
Revise:
D) = .040/.093 = .4301
4.33
Let
G = lawn treated by GreenGhumb
L = lawn treated by LawnMasGer
V = very healthy lawn
N = not very healthy lawn
P(G) = .72
Event
Prior
Conditional
P(L) = .28
Joint
Revised
P(Ei)
A
B
.72
.28
.30
.20
.216 .216/.272=.7941
.056 .056/.272=.2059
P(V)=.272
Revised:
Let
T = lawn treated by Tri-state
G = lawn treated by Green Chem
V = very healthy lawn
N = not very healthy lawn
P(T) = .72
Event
P(Ei)
Prior
Conditional
P(G) = .28
Joint
Revised
A
B
.72
.28
.30
.20
.216 .216/.272=.7941
.056 .056/.272=.2059
P(V)=.272
Revised:
P
Variable 1
D
E
10
20
B
15
30
15
55
40
30
5
4.35
A
Variable 2
C
20
45
95
a) P(E) = 40/95 = .42105
= P(B) + P(D) = 20/95 + 55/95 - 15/95 = 60/95 = .63158
= 30/95 + 20/95 =
50/95 = .52632
Does 30/95 = 10/95 ??
A
B
C
D
3
8
10
21
E
9
4
5
18
F
7
6
3
16
G
12
4
7
23
31
22
25
78
4.37
<35
Gender Male .11
Female .07
.18
4.37
Age(years)
35-44 45-54 55-64
>65
.20
.08
.28
.16
.01
.17
.19
.04
.23
.12
.02
.14
.78
.22
1.00
a) P(35-44) = .28
-54) = .04
-44) = P(Man) + P(35-44) -44) = .78 + .28 - .20 = .86
-64) = P(<35) + P(55-64) = .18 + .14 = .32
-54)/P(45-54) = .04/.23= .1739
-64) = .11 + .20 + .19 + .16 = .66
a) P(36-40) = .135
b) P(Woman ∩ 41-50) = .105
c) P(Man U 36-40) = P(Man) + P(36-40) - P(Man ∩ 36-40) = .462 + .135 - .064 = .533
d) P(<20 U 51-60) = P(<20) + P(51-60) = .229 + .162 = .391
-50) = P(Woman ∩ 41-50)/P(41-50) = .105/.208= .505
f) P(not W ∩ not 51-60) = .076 + .073 + .064 + .103 + .039 + .01= .365
4.39
Let R = retirement
P(R) = .42
Let L = life insurance
P(L) = .61
-
The matrix:
L
R
Yes
No
Yes
No
.33
.28
.61
.09
.30
.42
.58
.39
- .33 = .70
- .33 = .09
1.00
Let E = sent/received email
Let L = listened to web radio or watched web television
P(E) = .9
P(E ∩ L) = .50
P(L) = .52
a) P(E L) = P(E ∩ L)/P(L) = .50/.52 = .96
b) P(L E) = P(E ∩ L)/P(E) = .50/.90 = .56
c) P(L U E) = P(L) + P(E) - P(L ∩ E) = .9 + .52 - .53 = .89
d) P(E ∩ NL) = P(E) - P(E ∩ L) = .9 - .50 = .40
e) P(NL E) = P(NL∩E)/P(E) = .4/.9 = .444
The matrix:
L
E
Yes
No
4.41
Yes
No
.50
.02
.52
.40
.08
.9
.1
.48
1.00
Let M = MasterCard
P(M) = .30
A = American Express
P(A) = .20
V = Visa
P(V) = .25
-
- .06 = .39
Possession of Visa is not independent of possession of MasterCard
e) American Express is not mutually exclusive of Visa
4.43
Let M = expect to save more
R = expect to reduce debt
NM = don't expect to save more
NR = don't expect to reduce debt
P(M) = .43
- .81 = .19
P(NM) = 1 - P(M) = 1 - .43 = .57
P(NR) = 1 - P(R) = 1 - .45 = .55
= .43 + .45 - .3483 = .5317
c) P(neither save nor reduce debt) =
1- .5317 = .4683
Probability matrix for problem 4.43:
Yes
Save
Yes
No
Reduce
No
.3483 .0817 .43
.1017 .4683 .57
.45
.55 1.00
Let I = iPhone
B = Blackberry
P(I) = .3
a) P(I
b) P(I
B) = P(I) P(B I) = (.3)(.098) = .0294
B) = P(I) + P(B) - P(I B)
= . 3 + .4 - .0294 = .6706
c) P(neither I nor B) =
1 - P(I B) = 1 - .6706 = .3294
d) P(I NB) = P(I) P(NB I) =
but P(NB|I)=1-P(B|I)=1-.098=.902
.2706
Probability matrix for problem 4.43:
Yes
BlackBerry
No
Yes
.2706
.3
iPhone
No
.0294 .3706
.3294 .6
.7
1.00
.4
4.45
Let Q = keep quiet when they see co-worker misconduct
Let C = call in sick when they are well
P(Q) = .35
P(NQ) = 1 -
but P(C) must be solved for:
.2625 = P(C) ( .40)
Therefore, P(C) = .2625/.40 = .65625
- .2625 = .74375
-
- .2625 = ..39375
-
- .74375 = .25625
- .2625 = .0875
Probability matrix for problem 4.45:
C
Y
Q
Y
N
4.47
N
.2625 .0875 .35
.39375 .25625 .65
.65625 .34375 1.00
Let R = retention
P(R) = .56
Let P = process improvement
-
.56 - .36 = .20
c) P(P) = ??
.56 + .40 - .36 = .60
e) P(N
-
- .60 = .40
but P(NP) = 1 - P(P) = 1 - .40 = .60
P
R
Y
N
4.49
Y
N
.36
.04
.40
.20
.40
.60
.56
.44
1.00
From P(R), we can get P(NR) by taking 1 - .56 = .44.
However, only these two marginal values can be computed directly.
To solve for P(P), using what is given, since we know that 90% of P lies
in the intersection and that the intersection is .36, we can set up an
equation to solve for P:
.90P = .36
Solving for P = .40.
Let F = Flexible Work Let V = Gives time off for Volunteerism
P(F) = .41
from this, P(NF) = 1 - .41 = .59
-
so, P
-
- .246 = .469
-
- .246 = .164
P(NV) = 1 – P(V) = 1 - .305 = .695.
P(NF) = .59 P(NV) = .695
Solve for P(NF NV) = P(NV) – P(F NV) = .695 - .164 = .531
P(NF NV) = P(NF) + P(NV) - P(NF NV) = .59 + .695 - .531 = .754
Probability matrix for problem 4.49:
V
Y
F
Y
N
4.51
.246
.059
.305
N
.164 .41
.531 .59
.695 1.00
Let R = regulations
T = tax burden
P(R) = .30
b) P(R
P(T) = .35
T) = P(R) + P(T) - P(R
T) =
.30 + .35 - .2130 = .4370
c) P(R T) - P(R T) = .4370 - .2130 = .2240
d) P(R T) = P(R T)/P(T) = .2130/.35 = .6086
e) P(NR T) = 1 - P(R T) = 1 - .6086 = .3914
f) P(NR NT) = P(NR NT)/P(NT) = [1 - P(R
T)]/P(NT) =
(1 - .4370)/.65 = .8662
Probability matrix for problem 4.51:
T
Y
R
Y
N
4.53
.213
.137
.35
N
.087 .30
.563 .70
.65 1.00
Let B = Believe plastic shopping bags should be Banned
Let R = Recycle aluminum cans
P(B) = .54
but P(NB) = 1 – P(B) = 1 -
b) P(R
-
- .41 = .8160
-
But P(NB) =.46, P(NR) = 1 - P(R) = 1 - .6860 = .3140, and
- .8160 = .1840, therefore
- .1840 = .5900
0 and P(R) = .6860, therefore
Initial probability matrix for problem 4.53:
B
Y
R
Y
N
N
.41
.54
.46 1.00
Multiply .60 by the .46 to get the intersection of R and not B = .2760.
Fill in the rest of the cells.
Final probability matrix for problem 4.53:
B
R
Y
N
Y
N
.41
.13
.54
.276 .686
.184 .314
.46 1.000