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Student’s Solutions Manual and Study Guide: Chapter 4
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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, specifically 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 joint
probability table, 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
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
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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
Counting the Possible Sequences
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
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 3
KEY TERMS
A Priori
Bayes' Rule
Classical Method
Collectively Exhaustive Events
Combinations
Complement of a Union
Complement of an Event
Conditional Probability
Elementary Events
Event
Experiment
Independent Events
Intersection
Joint Probability
Marginal Probability
mn Counting Rule
Mutually Exclusive Events
Permutations
Probability Matrix
Relative Frequency of Occurrence Method
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"
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 4
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}
A  B = _______________ and A  B = _______________.
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.
21.
If the P(AB) = P(A), then the events A, B are _______________ events.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 5
22.
23.
If the P(X) = .53, the P(Y) = .12, and the P(X  Y) = .07, then P(X  Y) = ______________.
If the P(X) = .26, the P(Y) = .31, and X, Y are mutually exclusive, then P(X  Y) =
_______________.
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
25.
Men
Women
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 __________.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
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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
P(EA) = .24 and P(EB) = .56. If E is known to have occurred, then the revised
probability of A occurring is _______________ and of B occurring is _______________.
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 7
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
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 8
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 elementary events in 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,
D1 D2 A2,
D1 A1 A2,
D1 A2 A3,
D2 A1 A2,
D2 A2 A3,
A1 A2 A3,
D1 D2 A3,
D1 A1 A3,
D1 A2 A4,
D2 A1 A3,
D2 A2 A4,
A1 A2 A4,
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 9
A1 A3 A4, A2 A3 A4
Combinations are used to counting the sample space because sampling is done
without replacement and the order of elements is not important.
6C3
=
6!
= 20 (sample size)
3!3!
Probability that one of three is defective is:
12/20 = 3/5 = .60
There are 20 elementary events in the sample space and 12 of them have exactly
1 defective part.
4.7
20!
= 38,760
6!14!
It is assumed here that 6 different (without replacement) employees are to be
selected.
b) 6 P6 = 6! = 6 *5* 4*3*2 *1= 720
a) 20C6 =
4.9
D
E
F
A
5
8
12
25
B
10
6
4
20
C
8
2
5
15
23
16
21
60
a) P(A 
b) P(E 
c) P(D 
d) P(C 
4.11
D) = P(A) + P(D) - P(A  D) = 25/60 + 23/60 - 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
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 10
P (ridden either a train or an airplane) =
P(A  T) = P(A) + P(T) - P(A  T) = .47 + .28 - P(A  T).
We 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.
4.13
Let C = have cable TV
Let T = have 2 or more TV sets
P(C) = .67, P(T) = .74, P(C  T) = .55
a) P(C  T) = P(C) + P(T) - P(C  T) = .67 + .74 - .55 = .86
b) P(C  T but not both) = P(C  T) - P(C  T) = .86 - .55 = .31
c) P(NC  NT) = 1 - P(C  T) = 1 - .86 = .14
d) The special law of addition does not apply because P(C  T) is not .0000.
Possession of cable TV and 2 or more TV sets are not mutually exclusive.
4.15
C
D
E
F
A
5
11
16
8
40
B
2
3
5
7
17
7
14
21
15
57
a)
b)
c)
d)
4.17
P(A 
P(D 
P(D 
P(A 
E) =
B) =
E) =
B) =
16/57 = .2807
3/57 = .0526
.0000
.0000
Let D = Defective part
a) (without replacement)
P(D1  D2) = P(D1)  P(D2 D1) =
b) (with replacement)
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
6 5
30


= .0122
50 49 2450
Student’s Solutions Manual and Study Guide: Chapter 4
Page 11
P(D1  D2) = P(D1)  P(D2) =
4.19
Let
6 6
36


= .0144
50 50 2500
T = employed in tourist industry,
PT  .10
U = under 35 years of age
PU T  .53 PU NT  .44
a) PNT  1  .10  .90
b) PT  U  PTPU T  .10.53  .053
c) PT  NU  PTPNU T  .101  .53  .10.47  .047
d) PNT  U  PNTPU NT  1  .10.44  .90.44  .396
e) PNT  NU  PNTPNU NT  1  .101  .44  .90.56  .504
f) PNT  not over 35  PNT  U   PNTPU NT  1  .10.44
 .90.44  .396 (same as d))
g) PNT  NU  PNTPNU NT  1  .101  .44  .90.56  .504 (same
as e) )
The matrix:
U
T
Yes
No
Yes
.053
.396
.449
No
.047
.504
.551
.10
.90
1.00
4.21
Let S = safety
Let A = age
P(S) = .30
P(A) = .39
P(AS) = .87
a) P(S  NA) = P(S) P(NAS)
but P(NAS) = 1 - P(AS) = 1 - .87 = .13
P(S  NA) = (.30)(.13) = .039
b) P(NS  NA) = 1 - P(S  A) = 1 - [P(S) + P(A) - P(S  A)]
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 12
but P(S  A) = P(S)  P(AS) = (.30)(.87) = .261
P(NS  NA) = 1 - (.30 + .39 - .261) = .571
c) PNS  A  PA  PA  S  .39  .261  .129
The matrix:
A
No
.039
.571
.30
.70
.39
.61
1.00
Yes
No
Yes
.261
.129
S
4.23
E
F
G
A
15
12
8
35
B
11
17
19
47
C
21
32
27
80
D
18
13
12
43
65
74
66
205
a) P(GA) = 8/35 = .2286
b) P(BF) = 17/74 = .2297
c) P(CE) = 21/65 = .3231
d) P(EG) = .0000
4.25
Calculator
Computer
Yes
No
Yes
46
3
49
No
11
15
26
57
18
75
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 13
Select a category from each variable and test
P(V1V2) = P(V1).
For example, P(Yes ComputerYes Calculator) = P(Yes Computer)?
46 49

?
57 75
.8070  .6533
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
Let Q = Qualified
P(E) = .46
P(Q) = .37
P(E  Q) = .15
a) P(EQ) = P(E  Q)/P(Q) = .15/.37 = .4054
b) P(QE) = P(E  Q)/P(E) = .15/.46 = .3261
c) P(QNE) = P(Q  NE)/P(NE)
but P(Q  NE) = P(Q) - P(Q  E) = .37 - .15 = .22
P(NE) = 1 - P(E) = 1 - .46 = .54
P(QNE) = .22/.54 = .4074
d) P(NE  NQ) = 1 - P(E  Q) = 1 - [P(E) + P(Q) - P(E  Q)]
= 1 - [.46 + .37 - .15] = 1 - (.68) = .32
The matrix:
Q
E
Yes
No
Yes
.15
.22
No
.31
.32
.46
.54
.37
.63
1.00
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
4.29
Let
H = purchase hardware on-line
S = purchase software on-line
PS  .52 PS H  .93
PH  .34
PNS H  1  PS H  1  .93  .07
a)
b) PS NH 
PS  NH PS  PS  H 


PNH
1  PH 
PS  PH PS H 
1  PH 
c) PNH S 
d)

Page 14

.52  .34.93
 .3088
1  .34
PNH  S PS  PH  S .52  .34.93


 .3919
PS
PS
.52
P NH NS 
PNH  NS 1  PH  S 1  PH   PS   PH  S



PNS
1  PS
1  PS
1  .34  .52  .34 .93
 .9504
1  .52
4.31
Let A = product produced on Machine A
B = product produces on Machine B
C = product produced on Machine C
D = defective product
P(A) = .10
P(B) = .40
P(C) = .50
P(DA) = .05
P(DB) = .12
Event
Ei
A
B
C
Prior
P(Ei)
.10
.40
.50
Conditional
P(DEi)
.05
.12
.08
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Joint
P(D  Ei)
.005
.048
.040
P(DC) = .08
Revised
.005/.093=.0538
.048/.093=.5161
.040/.093=.4301
Student’s Solutions Manual and Study Guide: Chapter 4
Page 15
P(D)=.093
Revised Probability:
P(AD) = .005/.093 = .0538
P(BD) = .048/.093 = .5161
P(CD) = .040/.093 = .4301
4.33
Let
P(M) = .72
M = lawn treated by Maritime Lawn Service
G = lawn treated by Greenchem
V = very healthy lawn
P(G) = .28
P(VM) = .30 P(VG) = .20
Event
Prior
Conditional
Ei
M
G
P(Ei)
.72
.28
P(VEi)
.30
.20
Revised Probability:
Joint
Revised
P(V  Ei)
.216
.056
P(V)=.272
P(EiV)
.216/.272=.7941
.056/.272=.2059
P(MV) = .216/.272 = .7941
P(GV) = .056/.272 = .2059
4.35
Variable 1
Variable 2
D
E
A
10
20
30
B
15
5
20
C
30
15
45
55
40
95
a) P(E) = 40/95 = .42105
b) P(B  D) = P(B) + P(D) - P(B  D)
= 20/95 + 55/95 - 15/95 = 60/95 = .63158
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 16
c) P(A  E) = 20/95 = .21053
d) P(BE) = 5/40 = .1250
e) P(A  B) = P(A) + P(B) = 30/95 + 20/95 =
50/95 = .52632
f) P(B  C) = .0000 (mutually exclusive)
g) P(DC) = 30/45 = .66667
h) P(AB) =
4.37
P( A  B) .0000
= .0000 (A and B are mutually exclusive)

P( B)
20 / 95
D
E
F
G
A
3
9
7
12
31
B
8
4
6
4
22
C
10
5
3
7
25
21
18
16
23
78
a) P(F  A) = 7/78 = .08974
b) P(AB) =
P( A  B)
.0000

= .0000 (A and B are mutually exclusive)
P( B)
22 / 78
c) P(B) = 22/78 = .28205
d) P(E  F) = .0000 (Mutually Exclusive)
e) P(DB) = 8/22 = .36364
f) P(BD) = 8/21 = .38095
g) P(D  C) = P(D) + P(C) – P(D  C) = 21/78 + 25/78 – 10/78
= 36/78 = .46154
h) P(F) = 16/78 = .20513
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
4.39
Let T = thoroughness
P(T) = .78
P(K) = .40
Page 17
Let K = knowledge
P(T  K) = .27
a) P(T  K) = P(T) + P(K) - P(T  K) =
.78 + .40 - .27 = .91
b) P(NT  NK) = 1 - P(T  K) = 1 - .91 = .09
c) P(KT) = P(K  T)/P(T) = .27/.78 = .3462
d) PNT  K  PK  PT  K  .40  .27  .13
The matrix:
K
T
Yes
No
Yes
.27
.13
No
.51
.09
.78
.22
.40
.60
1.00
4.41
Let T = technology driven, B = lives in British Columbia, M = lives in
Manitoba
PT  .16 PB  .16 PM  .037 PT B  .20 PT M  .17
a) PB  T  PBPT B  .16.20  .032
b) PM  T  PMPT M  .037.17  .00629
c) PB T  
PB  T  .032

 .20
PT 
.16
d) PM NT 
PM  NT PM   PM  T  .037  .00629


 .0366
PNT
PNT
1  .16
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 18
e) P NB  NM  T  

4.43
PNB  NM  T  PT   P B  T   PM  T 


P T 
PT 
.16  .032  .00629
 .7607
.16
Let C = cell phone only
Given: P(C) = .25
Let I = high-speed Internet
P(I) = .65
P(IC) = .80
a) P(C  I) = P(C)  P(IC) = (.25)(.80) = .20
b) P(C  I) = P(C) + P(I) - P(C  I) = .25 + .65 - .20 = .70
c) P(C  NI) = P(C)  P(NIC)
Since P(IC) = .80, P(NIC)= 1 - P(I C) = 1 - .80 = .20
P(C  NI) = (.25)(.20) = .05
d) P(NC  NI) = 1 - P(C  I) = 1 - .70 = .30
e) P(NC  I) = P(I) - P(C  I) = .65 - .20 = .45
The joint probability table:
4.45
I
NI
C
.20
.05
.25
NC
.45
.30
.75
.65
.35
1.00
Let R = read
Let B = checked in the with boss
P(R) = .40
P(B) = .34
P(BR) = .78
a) P(B  R) = P(R)P(BR) = (.40)(.78) = .312
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Page 19
b) P(NR  NB) = 1 - P(R  B)
but P(R  B) = P(R) + P(B) - P(R  B) =
.40 + .34 - .312 = .428
P(NR  NB) = 1 - .428 = .572
c) P(RB) = P(R  B)/P(B) = (.312)/(.34) = .9176
d) P(NBR) = 1 - P(BR) = 1 - .78 = .22
e) P(NBNR) = P(NB  NR)/P(NR)
but P(NR) = 1 - P(R) = 1 - .40 = .60
P(NBNR) = .572/.60 = .9533
f) Probability matrix for problem 4.45:
B
NB
R
.312
.088
.40
NR
.028
.572
.60
.34
.66
1.00
4.47
Let
T = time to get through to a customer representative
R = impolite customer reps
C = customer service
P = payment disputes
B = incorrect billing
H = incompetent complaint handling
I = indifference to customers
PT  .17 PR   .14 PC  .14 PP  .11 PB  .10 PH  .08 PI  .07
a) Since P and B are mutually exclusive,
PP  B  PP  PB  .11  .10  .21
b) PI  H  .00 (mutually exclusive)
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
c) PR B 
Page 20
PR  B .00

 .00
PB
.10
d) Since H and P are mutually exclusive,
PNH  NP  1  PH  P  1  .08  .11  .81
4.49
Let M = mail/catalogues
P(M) = .38 P(M  S) = .0000
Let S = sales/sales rep
P(NM  NS) = .41
a) P(M  NS) = P(M) - P(M  S) = .38 - .00 = .38
b) Because P(M  S) = .0000, P(M  S) = P(M) + P(S)
Therefore, P(S) = P(M  S) - P(M)
but P(M  S) = 1 - P(NM  NS) = 1 - .41 = .59
Thus, P(S) = P(M  S) - P(M) = .59 - .38 = .21
c) P(SM) = P(S  M)/P(M) = .0000/.38 = .0000
d) P(NMNS) = P(NM  NS)/P(NS) = .41/.79 = .5190
where: P(NS) = 1 - P(S) = 1 - .21 = .79
Probability matrix for problem 4.49:
S
M
4.51
Y
N
Y
.00
.38
.38
N
.21
.41
.62
.21
.79
1.00
Let K = Khan Let C = Chan
Let J = Jackson
Let B = blood test
P(K) = .41
P(C) = .32
P(J) = .27
P(BK) = .05
P(BC) = .08
P(BJ) = .06
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Student’s Solutions Manual and Study Guide: Chapter 4
Event
Ei
K
C
J
Prior
Conditional
P(Ei)
.41
.32
.27
4.53
Page 21
P(BEi)
.05
.08
.06
Joint
P(B  Ei)
.0205
.0256
.0162
P(B) = .0623
Revised
P(EiB)
.3291
.4109
.2600
Let RB = Canadians who read a book or a magazine more than 10 hours per
week.
Event
Prior
Conditional
Ei
0-24
25-34
35-44
> 45
P(Ei)
.315
.137
.161
.386
P(RBEi)
.11
.24
.27
.39
Joint
P(RB  Ei)
.03465
.03288
.04347
.15054
P(RB) = .26154
Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition
Revised
P(EiRB)
.03465/.26154 = .13248
.03288/.26154 = .12572
.04347/.26154 = .16621
.15054/.26154 = .57559
Student’s Solutions Manual and Study Guide: Chapter 4
Page 22
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Black, Chakrapani, Castillo: Business Statistics, Second Canadian Edition