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1. 1
CHAPTER ONE
INTRODUCTION TO PROBABILITY
It is difficult to go very far in solving business problems
without a basic understanding of probability. The
‘laws of chance’ is the basis for many decisions that are
made in real life.
2. 2
For example, when we carry out a survey, we need to
know how the sample data relates to the target
population.
Our insurance premiums are determined by the chance
of some mishap occurring and
quality control systems are built around probability
laws.
There are also many games of chance including card
and dice games, horse racing and the national lottery.
•
3. 3
Basic Ideas
The value of a probability can be given either as a fraction, as a
decimal or as a percentage.
An event with probability zero is termed as impossible, while an event
with a probability of 1 or 100% is termed certain.
4. 4
Definition of Some Common Terms
• Experiment: An experiment is an operation whose outcome cannot be
predicted with certainty. Below are examples of an experiment.
•
• Outcome: The result of an experiment called outcome
5. 5
Examples of experiments and their associated outcomes are as follows:
Experiment Experimental outcomes
1. Toss a coin Head, Tail
2. Select a part for inspection Defective, Non-defective
3. Conduct a sales call Purchase, no purchase
4. Roll a die 1, 2, 3, 4, 5, 6,
5. Play a football game Win, lose, tie
6. 6
Random Experiment
• If we record the number of babies who are born in Koforidua every
week,
• If we record the number of customers who enter a bank in given time
interval when the bank is open.
Such experiments, because their outcomes are uncertain, are called
random experiments.
7. 7
• Trial: A trial of an experiment is a single performance
of an experiment.
•
• Equally Likely Outcomes: In an experiment, the
outcomes which have equal chance of occurring are
called equally likely outcomes. For example, if we
toss a coin, outcomes head or tail are equally likely.
Also, if we throw a die, the outcomes 1, 2, 3, 4, 5, 6
are equally likely.
•
• Random Sampling: This is choosing a sample from a
population without being bias.
8. 8
• Sample Space: The sample space of an experiment is
the set of all possible outcomes of a trial of the
experiment. Sample space is denoted by S.
• Each element of the sample space is called a sample
point and an event is a subset of the sample space.
• The probability of an event is the sum of the weights
of the sample points in the event’s subspace (subset of
S).
9. 9
• The sample space in which each sample point is
equally likely is called an equiprobable sample
space.
• Suppose the number of elements in a sample space is
n(S) = n, with each element representing an equally
likely outcome of a probability experiments. Then the
probability assigned to each element is
10. 10
• Event: An event is a collection of sample points with
common property. It is a subset of the sample space.
• For example, when a die is cast, the event E of
throwing an even number is E = {2, 4, 6}.
• Also when two coins are tossed, the event of getting
exactly one head is E = {HT, TH}.
11. 11
The empty set is a subset of S and S also a subset of S. and S are therefore events. We
call the impossible event and S the certain event. A subset of S containing one element of S
is called a simple event.
12. 12
• Example: Consider the experiment of rolling two dice
and observing the number which appears on the
uppermost face of each die. The sample space of the
experiments consists of the following array of 36
outcomes.
• The first coordinate of each point is the number which
appears on the 1st
die while the 2nd
coordinate is the
number which appears on the second die.
14. 14
• For any given experiment, we may be interested in the
outcome of certain events rather than in the outcome
of a specific element in the sample space.
• It is important to understand that an event occurs if
and only if any of its elements is the outcome of an
experiment.
15. 15
For instance,
P = {(1, 2), (2, 3), (6, 4)} ,
Q = {(1, 4), (4, 4), (3, 6)} and
R = {(6, 4), (2, 5), (1, 3)} are events.
If, when the dice is rolled, the number 6 appears on the
first die and the number 4 appears on the second die,
then the events P and R have occurred, since each of
these events has (6, 4) as an element.
Event Q did not occur since (6, 4) Q.
16. 16
Operations on Events
• Since an event is a subset of a sample space, we can combine two or
more events to form new events, using the various set operations.
•
• The sample space is considered as the universal set. If A and B are two
events defined on the same sample space, then:
17. 17
A B
Denotes the event “A or B or both”.
Thus event A B
occurs if either A occurs or B occurs or
both A and B occur.
A B
Denotes the event “both A and B”.
Thus event A B
occurs if both A and B occur.
( )
A orA
Denotes the event which occurs if and only if
A does not occur.
18. 18
Example1.1
Let P be the event that an employee selected at random
from oil drilling company smokes cigarettes. Let Q be
the event that an employee selected drinks alcoholic
beverages. Then
19. 19
COUNTING RULES
Being able to identify and count the experimental outcomes is a
necessary step in assigning probabilities. We now discuss three useful
counting rules.
20. 20
Multiple – Step Experiments
If an experiment can be described as a sequence of k steps with 1
n possible outcomes on the
first step, 2
n possible outcomes on the second step, and so on, then the total number
experimental outcomes is given by
1 2
( )( )...................( )
k
n n n
21. 21
Viewing the experiment of tossing two coins as a sequence of first tossing one coin 1
( 2)
n
and then tossing the other coin 2
( 2)
n , we can see from the counting rule that there are
(2)(2) 4
distinct experimental outcomes.
22. 22
A tree diagram is a graphical representation that helps in visualising a multiple-step experiment.
23. 23
Activity 1.1
How many lunches consisting of a soup, sandwich, dessert and a drink
are possible if we can select 4 soups, 3 kinds of sandwiches, 5 desserts
and 4 drinks?
24. 24
Solution
Here 1 2 3
4, 3, 5
n n n
and 4 4
n hence there are
1 2 3 4 4 3 5 4 24
n n n n
different ways to choose a lunch.
25. 25
Solution
Since the number must be even, we have 1
( 2)
n choices for the units’ position. For each of
these, we have 2 4
n choices for the tens position and then 3 3
n choices for the hundreds
position.
Therefore, we can form a total of 1 2 3 2 4 3 24
n n n
even three-digit numbers.
Activity 1.2
How many even three-digit numbers can be formed from the digits 3, 2, 5, 6 and 9 if
each digit can be used only once?
26. 26
Permutation
• A second counting rule that is sometimes useful is the
counting rule for permutations.
• It allows one to compute the number of experimental
outcomes when r objects are to be selected from a set
of n objects where the order of selection is
important.
• The same r objects selected in different order is
considered a different experimental outcome. The
different arrangements are called permutations.
27. 27
Activity 1.3
In how many ways can three different books A, B, and C, be arranged
on a shelf?
28. 28
Solution
The three books can be arranged on a shelf in 6 different ways, as follows:
ABC, ACB, BCA, BAC, CAB, CBA
It can be seen that the order of the books is important. Each of the 6 arrangements is therefore
a different arrangement of the books.
It can be seen that, the first book can be chosen in 3 different ways, the second book chosen in
2 different ways and the third book in one way. The three books can therefore be arranged on
a shelf in 3 x 2 x 1 = 6 ways.
In general, n distinct objects can be arranged in ( 1)( 2)......(3)(2)(1)
n n n
ways. We represent
this product by the symbol !
n , which is read “n factorial”. Three objects can be arranged in
3! 3 2 1 6
ways.
By definition 1! 1
and0! 1
.
29. 29
Permutation of n Different Things Taken r at a Time
The number of permutations of n different objects, taken r at a time, is denoted by n
r
P and is
given by
!
( )
( )!
n
r
n
P r n
n r
As an example, consider the quality control process in which an inspector selects two of five
parts to inspect for defects. How many permutations may be selected?
The counting rule above shows that with 5
n and 2
r , we have
5
2
5! 5 4 3 2 1 120
20
(5 2)! 3 2 1 6
P
Thus, 20 outcomes are possible for the experiment of randomly selecting two parts from a
group of five when the order of selection must be taken into account.
30. 30
Activity 1.4
a.Three different Mathematics books, four different French books and
two different Physics books are to be arranged on a shelf. How many
different arrangements are possible if
i.the books in each particular subject must all stand together,
ii.only mathematics books must stand together?
b. In how many ways can 9 people be seated on a bench if only 3
seats are available?
31. 31
Solution
i. The maths books can be arranged among themselves in 3! 6
ways
the French books in 4! 24
ways
the Physics book in 2! 2
ways
and the three groups in 3! 6
ways
Hence, the required number of arrangements is 6 24 2 6 1728
32. 32
ii. Consider the Maths books as one big book. Then we have 7 books which can be
arranged in 7!ways. In all of these arrangements, the maths books are together.
But the maths books can be arranged among themselves in 3! ways. Hence the
required number arrangements is
7!3! 30240
a. The first seat can be selected in 9 ways and when this has been done, there are 8 ways
of selecting the second seat and 7 ways of selecting the third seat. Hence the required
number of arrangements is 9 8 7 504
33. 33
Activity 1.5
In how many ways can 6 boys and 2 girls be arranged in a row if
the two girls are together
the two girls are not together
34. 34
Solution:
Since the girls are together, they can be considered as one unit. We then have 7 objects (6 boys
and one unit of 2 girls) to be arranged. This can be done in 7! ways.
In all these arrangements, the two girls are together. But the two girls can be arranged among
themselves in 2! ways. Hence the required number of arrangements is 2! 7! 10080
The number of arrangements without restriction is 8!
The number of arrangements with the girls together 2! 7!
Therefore, the number of arrangements with the girls not together is 8! 2! 7! 30240
36. 36
Solution
We must arrange the 5 marbles in 5 positions thus: --- --- --- --- ---
The first position can be occupied by any one of five marbles i.e., there are 5 ways of filling
the first position. When this has been done there are 4 ways of filling the second position. Then
there are 3 ways of filling the third position, 2 ways of filling the fourth position and finally,
only 1way of filly the last position.
Therefore;
Number of arrangements of 5 marbles in a row 5 4 3 2 1 5! 120!
This is also called the number of permutations of n different objects taken n at a time and is
denoted by .
n
n
P
37. 37
Activity 1.7
In how many ways can 10 people be seated on a bench if only 4 seats
are available?
38. 38
Activity 1.7
Solution
Number of arrangements of 10 people taken 4 at a time 10 9 8 7 5040
This is also called the number of permutations of n different objects taken r at a time and is
denoted by .
n
r
P
39. 39
Self-Test 1.1
1. How many 4-digit numbers can be formed with the 10 digits 0,1,2,3,......,9 if
a) repetitions are allowed
b) repetitions are not allowed
c) the last digit must be zero and repetitions are not allowed
2. Four different mathematics books, six different physics books, and two different
chemistry books are to be arranged on a shelf. How many different arrangements are
possible if
a) the books in each particular subject must all stand together,
b) only maths books must stand together.
40. 40
Circular Permutation
Permutations that occur by arranging objects in a circle are called
circular permutation. It can be shown that the number of permutations
of n distinct objects arranged in a circle is
41. 41
Activity 1.8
In how many ways can 7 people be seated at a round table if
a) they can sit anywhere,
b) 2 particular people must not sit-next to each other?
42. 42
Activity 1.8
Solution
a. Let 1 of them be seated anywhere. Then the remaining 6 people can be seated in
6! 720
ways, which is the total number of ways of arranging the 7 people in a circle.
b. Consider the two particular people as one person. Then there are 6 people altogether
and they can be arranged in 5! ways. But the two people considered as 1 can be arranged
among themselves in 2! ways.
Thus the number of ways of arranging 7 people at a round table with 2 particular people sitting
together 5!2! 240
Then, the total number of ways in which 7 people can be seated at a round table so that the 2
particular people do not sit together 720 240 480
ways.
43. 43
Permutation with Repetitions
So far we have considered permutations of distinct objects. That is, all the objects were
completely different. The following gives permutations with repetitions.
Given a set of n objects having 1
n , elements alike of one kind, and 2
n elements alike of another
kind, and 3
n elements alike of a third kind, and so on for k kinds of objects, then the number
of different arrangements of the n objects, taken all together is
1 2
!
! !.... !
k
n
n n n
where
1 2 ...... k
n n n n
44. 44
Activity 1.9
How many different ways can 3 red, 4 yellow and 2 blue bulbs be
arranged in a string of Christmas tree lights with 9 sockets?
46. 46
COMBINATION-order not important
A third useful counting rule allows one to count the number of
experimental outcomes when the experiment involves selecting r
objects from a (usually larger) set of n objects.
It is called the counting rule for combinations.
47. 47
• The number of combinations of n objects taken r at a time is
48. 48
Activity 1.11
Out of 5 mathematicians and 7 physicists, a committee consisting of 2
mathematicians and 3 physicists is to be formed. In how many ways can
this be done if;
a)any mathematician and any physicists can be included,
b)one particular physicist must be on the committee
c)two particular mathematicians cannot be on the committee?
49. 49
Activity 1.11
Solution
a. 2 mathematicians out of 5 can be selected in 5
2
C ways
3 physicists out of 7 can be selected in 7
3
C ways
Total number of possible selections 5 7
2 3 10 35 350
C C
50. 50
b.
a. 2 maths out of 5 can be done in 5
2
C ways
2 physicists out of 6 can be selected in 6
2
C ways
Total number of possible selections = 5 6
2 2 10 15 150
C C
51. 51
c.
a. 2 maths out of 3 can be done in 3
2
C ways
3 physicists out of 7 can be done in 7
3
C ways
Total number of possible selections = 3 7
2 3 3 35 105
C C
.
52. 52
ASSIGNING PROBABILITIES
Now let us see how probabilities can be assigned to experimental
outcomes. The three approaches most frequently used are the classical,
relative frequency and subjective methods. Regardless of the method
used, two basic requirements for assigning probabilities must be met.
53. 53
Basic Requirement for Assigning Probabilities
The probability assigned to each experimental outcome must be between 0 and 1 inclusively.
If we let i
E denotes the ith experimental outcome and ( )
i
p E its probability, then it
requirement can be written as
0 ( ) 1
i
p E
for all i.
The sum of probabilities for all the experimental outcomes must be equal to 1. For n
experimental outcomes, this requirement can be written as
2
( ) ( ) ..... ( ) 1
i n
p E p E p E
54. 54
[A] The classical method
The classical method of assigning probabilities is appropriate when all
the experimental outcomes are equally likely. If n experimental
outcomes are possible, a probability of
is assigned to each experimental outcome.
When using this approach, the two basic requirements for assigning
probabilities are automatically satisfied.
•
55. 55
That is, if a trail of an experiment can results in m
mutually exclusive and equally likely outcomes, and if
exactly h of these outcomes corresponds to an event A,
then the probability of event A is given by
56. 56
Thus, if all the simple events in S are equally likely, then
( )
( )
( )
n A
P A
n S
for all AS
Where ( )
n A denotes the number of sample points or elements in A, and ( )
n S denotes the
number of sample points or elements in S.
57. 57
Activity 1.12
The following table shows 100 patients classified according to blood group and sex.
Blood group
A B O
Male 30 20 17
Female 15 10 8
If a patient is selected at random from this group, find the probability that the patient selected.
a. is a male and has blood group B
b. is a female and has blood group A
58. 58
Solution
There are 100 ways in which we can select a patient from the 100 patients. Since the patient is
selected at random, all the 100 ways of selecting a patient are equally likely.
a. There are 20 male with blood group B. Therefore the probability that the patient
selected is a male and has blood group B is
20
0.2
100
b. There are 15 females with blood group A. Therefore the probability that the patient
selected is a female and has blood group A is
15
0.15
100
59. 59
[B] The relative frequency method
The relative frequency method of assigning probabilities is appropriate
when data are available to estimate the proportion of the times the
experimental outcome will occur if the experiment is repeated a large
number of times.
60. 60
Example
Consider a study of waiting times in the X-ray
department for a local hospital. A clerk recorded the
number of patients waiting for service at 9:00 am on 20
successive days and obtained the following results.
Number waiting Number of days outcome occurred
0 2
1 5
2 6
3 4
4 3
Total 20
61. 61
This data show that on 2 of the 20 days zero patients
were waiting for service; on 5 of the days, one patient
was waiting for service; and so on. Using the relative
frequency method, we would assign a probability of
2/20 = 0.10 to the experimental outcome of zero
patients waiting for service, 5/20 = 0.25 to the
experimental outcome of one patient waiting; and so on.
As with the classical method, using the relative
frequency method automatically satisfies the two basic
requirements.
62. 62
That is,
If some process is repeated a large number of times n, and if some resulting event with the
characteristic A occurs m times, the relative frequency of occurrence of A,
m
n
will be
approximately equal to the probability of A. Thus,
lim ( )
( )
n A
P A
n n
The disadvantage of this process is that the experiment must be repeatable. Any probability
obtained this way is an approximation.
63. 63
Activity 1.13
The following table gives the frequency distribution of the heights of
150 students. If a student is selected at random from this group, find the
probability that the student selected is taller than the modal height of the
students.
Height (cm) 130 140 150 160 170 180 190
Frequency 8 16 28 44 33 17 4
64. 64
Solution
The modal height of the students is 160 cm. This is the height with the highest frequency. The
number of students who are taller than 160 cm (33 + 17 + 4) = 54
An estimate of the required probability is the relative frequency
54
0.36
150
65. 65
[C] The subjective method
The subjective method of assigning probabilities is most
appropriate when one cannot realistically assume that the
experimental outcomes are equally likely and when little
relevant data are available.
When this method is used to assign probabilities to the
experimental outcomes, we may use any information
available, such as our experience or intuition. Because
subjective probability expresses a person’s degree of belief,
it is personal.
Using this method, different people can be expected to assign
different probabilities to the same experimental outcome.
66. 66
Some Basic Relationship of Probabilities
1. Complement Of an Event
Given an event A, the complement of A is defined to be the event consisting of all sample
points that are not in A. The complement of A is denoted by ,
A or c
A or A.
In any probability application, either event A or its complement ,
A must occur. Therefore, we
have
( ) ( ) 1
p A p A
67. 67
2. Addition Law
The additional law is helpful when we are interested in knowing the probability that at least
one of the two events occurs. That is, with events A and B we are interested in knowing the
probability that event A or event B or both occur. The addition law is written as follows:
( ) ( ) ( ) ( )
p A B p A p B p A B
The Venn diagram above depicts the union of events A and B.
68. 68
A special case arises for mutually exclusive events. Two events are said
to be mutually exclusive if the events have no sample points in common.
69. 69
Events A and B are mutually exclusive if, when one event occurs, the other cannot occur. Thus,
a requirement for A and B to be mutually exclusive is that their intersection must contain no
sample points. The figure above shows two mutually exclusive events A and B. In this case
( ) 0
p A B
and the addition law can be written as follows:
( ) ( ) ( )
p A B p A p B
If A, B, and C are three events defined on the same sample space, then:
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
p A B C p A p B p C p A B p A C p B C p A B C
But if events A, B and C are mutually exclusive, then
( ) ( ) ( ) ( )
p A B C p A p B p C
70. 70
3. Conditional Probability
Often, the probability of an event is influenced by whether a related event already occurred.
Suppose we have an event A with probability ( )
p A .
If we obtain new information and learn that a related event, denoted by B, already occurred.
We will want to take advantage of this information by calculating a new probability for event
A. This new probability of event A is called conditional probability and is written ( | )
p A B .
We use the notation | to indicate that we are considering the probability of event A given the
condition that event B has occurred. Hence the notation ( | )
p A B reads “the probability of A
given B.
71. 71
As an illustration of the application of conditional
probability, consider the situation of the promotion
status of male and female officers of Koforidua police
force in the eastern region of Ghana. The police force
consists of 1200 officers, 960 men and 240 women.
Over the past two years, 324 officers on the police force
received promotions. The specific breaks down of the
promotions are shown in the table below.
72. 72
Promotion status of police officers over the past two years
Men Women Total
Promoted 288 36 324
Not promoted 672 204 876
Total 960 240 1200
Joint probability table for promotions
Men Women Total
Promoted 0.24 0.03 0.27
Not promoted 0.56 0.17 0.73
Total 0.80 0.20 1.00
Let M = event an officer is a man
W = event an officer is a women
A = event an officer is promoted
A= event an officer is not promoted
( ) 288 /1200 0.24
p M A
Is the probability that a randomly selected officer is a man and is promoted
73. 73
Is the probability that a randomly selected officer is a man and is not promoted.
( ) 36 /1200 0.03
p W A
( ) 204 /1200 0.17
p W A
Because each of these values gives the probability of the intersection of two events, the
probabilities are called joint probabilities.
74. 74
The values in the margins of the joint probability table provide
the probabilities of each event separately.
That is ( ) 0.80, ( ) 0.20, ( ) 0.27
p M p W p A
and ( ) 0.73
p A
These probabilities are referred to as marginal probabilities
because of their location in the margins of the joint probability table.
75. 75
The probability that an officer is promoted given that that officer is a man is given by
( )
( | ) , ( ) 0
( )
p A M
p A M p M
p M
The fact that the conditional probabilities can be computed as the ratio of a joint probability to
a marginal probability provides the following general formula for conditional probability
calculations for two events A and B.
( )
( | )
( )
p A B
p A B
p B
, ( ) 0
p B
( )
( | )
( )
p A B
p B A
p A
, ( ) 0
p A
76. 76
Independent Events
From the table on joint probabilities, ( ) 0.27,
p A ( | ) 0.30
p A M and ( | ) 0.15
p A W
we see that the probability of a promotion (event A) is affected or influenced by whether the
officer is a man or a woman.
77. 77
Particularly, because ( | ) ( )
p A M p A
, we could say that events A and M are dependent
events. That is, the probability of event A (promotion) is altered or affected by knowing that
event M (the officer is a man) exists
Similarly, with ( | ) ( )
p A W p A
, we would say the events A and W are dependent events.
However, if the probability of event A is not changed by the existence of event M that is
( | ) ( )
p A M p A
we would say that the events A and M are independent events.
That is, two events A and B are independent if
( | ) ( )
p A B p A
or ( | ) ( )
p B A p B
otherwise, the events are dependent.
78. 78
Multiplication Law
The multiplication law is based on the definition of conditional
probability.
•
( ) ( ) ( | )
p A B p A p B A
or
( ) ( ) ( | )
p A B p B p A B
79. 79
for independent events A and B, ( | ) ( )
p B A p B
or ( | ) ( )
p B A p B
Note that the multiplication law for independent events provides another way to determine
whether A and B are independent. That is if ( ) ( ) ( )
p A B p A p B
then A and B are
independent;
If ( ) ( ) ( )
p A B p A p B
then A and B are dependent.
80. 80
Activity 1.14
Refer to example on blood group. If a patient is selected at random from the 100 patients, find
the probability that the patient selected:
a. is a male or has blood group A,
b. does not have blood group A
c. is a female or does not have blood group B
The following table shows 100 patients classified according to blood group and sex.
Blood group
A B O
Male 30 20 17
Female 15 10 8
81. 81
Solution
a. Let M denotes the event “a patient is a male” and A the event “a patient has blood
group A”. We wish to find ( )
p M A
.
By the addition law of probability,
( ) ( ) ( ) ( )
p M A p M p A p M A
( ) ( ) ( )
100 100 100
n M n A n A M
67 45 30 82
0.82
100 100 100 100
b. We wish to find ( )
p A
( ) 1 ( )
p A p A
( ) 45
1 1 0.55
100 100
n A
82. 82
c.a. Let F denote the event “a patient selected is a female” and B the event “a patient
selected has blood group B”. We wish to find ( )
p F B
( ) ( ) ( ) ( )
p F B p F p B p F B
( ) {1 ( )} ( )
33 30 23
1
100 100 100
20
1 0.8
100
p F p B p F B
83. 83
Activity 1.16
The events A, B and C have probabilities
1 1
( ) , ( )
2 3
p A p B
and
1
( )
4
p C Furthermore,
,
A C B C
and
1
( )
6
p A B
, find:
a) [( ) ]
p A B
b) ( )
p A B
c) [( ) ]
p A B
d) ( )
p A B
e) ( )
p A B C
84. 84
Solution
a. [( ) ] 1 ( )
p A B p A B
1 5
1
6 6
b. ( ) ( ) ( )
p A B p A p A B
1 1 1
2 6 3
c. [( ) ] 1 ( )
p A B p A B
1 { ( ) ( ) ( )}
p A p B p A B
1 1 1 2 1
1 1
2 3 6 3 3
d. ( ) [( ) ]
p A B p A B
1
3
e. ( ) ( ) ( ) ( )
p A B C p A p B p C
( ) ( ) ( ) ( )
p A B p A C p B C p A B C
1 1 1 1 11
2 3 4 6 12
85. 85
Self-Test 1.3
1. Samples of a cast aluminium part are classified on the basis of surface finish (in micro
inches) and lengths measurements. The results of 100 parts are summarised below.
LENGTH
Excellent Good
Surface Excellent 75 7
Finish Good 10 8
Let A denote the event that a sample has excellent surface finish, and let B denote the event
that a sample has excellent length. Find a) ( )
p A , b) ( )
p B , c) ( )
p A , d) ( )
p A B
e) ( )
p A B
.
2. In a certain population of women, 4% have breast cancer, 20% are smokers and 3% are
both smokers and have breast cancer. If a woman is selected at random from this
population, find the probability that the person selected is:
a. a smoker or has breast cancer
b. a smoker and does not have breast cancer
c. not a smoker and does not have breast cancer
86. 86
Activity 1.17 (conditional probability)
The probabilitythat a regularlyscheduled flight departs on time is ( ) 0.83
p D ; the probability
that it arrives on time ( ) 0.82
p A ; and the probability that it departs and arrives on time
( ) 0.78
p D A
. Find the probability that a plane
a. arrives on time given that it departed on time
b. departed on time given that it has arrived on time;
c. arrives on time, given that it did not depart on time.
87. 87
Solution
a. The probability that a plane arrives on time given that it departed on time is
( ) 0.78
( | ) 0.94
( ) 0.83
p A D
p A D
p D
b. We wish to find
( ) 0.78
( | ) 0.95
( ) 0.82
p D A
p D A
p A
c. We wish to find
( ') 0.82 0.78
( | ') 0.24
( ') 1 0.83
p A D
p A D
p D
NB: The notion of conditional probability provides the capability of re-evaluating the idea of
probability of an event in the light of additional information, that is, when it is known that
another event has occurred. The probability ( | )
p A B is an updating of ( )
p A based on the
knowledge that event B has occurred.
88. 88
Activity 1.18
Bag 1 contains 4 white balls and 3 green balls, and Bag 2 contains 3
white balls and 5 green balls. A ball is drawn from bag 1 and placed
unseen in bag 2. Find the probability that a ball now drawn from bag 2
is (a) green, (b) white.
89. 89
Solution
Let G1, G2, W1, and W2 represent, respectively, the events of drawing a green ball from bag 1,
a green ball from bag 2, a white ball from bag 1 and a white ball from bag 2.
a. We wish to find 2
( )
p G
And so,
2
2 1 1 2
( ) ( ) ( )
p G p G G p W G
1 2 1 1 2 1
( ) ( | ) ( ) ( | )
3 6 4 5 38
7 9 7 9 63
p G p G G p W p G W
90. 90
b.
a. We wish to find 2
( )
p W
2
2 1 1 2
( ) ( )
W G W W W
These two events are also mutually exclusive and so.
2
2 1 1 2
( ) ( ) ( )
p W p G W p W W
1 2 1 1 2 1
'
( ) ( | ) ( ) ( | )
3 3 4 4 25
7 9 7 9 63
( ) ( ) 1
p G p W G p W p G W
p A B C p A B C
92. 92
Activity 1.19
(Independents events)
The probability that Kofi hits a target is ¼ and the corresponding probabilities for Kojo and
Kwame are 1/3 and 2/5, respectively. If they all fire together, find the probability that
a. they all miss
b. exactly one shot hits the target
c. at least one shot hits the target
d. Kofi hits the target given that exactly one hit is registered.
93. 93
Solution
Let A, B, and C denotes the events that the target is hit by Kofi, Kojo and Kwame, respectively.
Then 1 1 2
( ) , ( ) , ( )
4 3 5
p A p B p C
Since they all fire together, the events A, B, and C are independent.
a. If E is the event “they all miss”, then E A B C
and so
( ) ( ) ( ) ( ) ( )
p E A B C p A p B p C
1 1 2 3
1 1 1
4 3 5 10
94. 94
b.
a. Let F denote the event “exactly one shot hits the target” then,
( ) ( ) ( )
F A B C A B C A B C
The events are mutually exclusive, so
( ) ( ) ( ) ( )
p F p A B C p A B C p A B C
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )
1 1 2 1 1 2 1 1 2
1 1 1 1 1 1
4 3 5 4 3 5 4 3 5
1 3 1 9
10 20 5 20
p A p B p C p A p B p C p A p B p C
95. 95
c.
a. Let G denote the event “at least one shot hits the target. The ( )
G A B C
By de Morgan’s Law( )
A B C A B C
'
( ) ( ) 1
p A B C p A B C
( ) 1 ( )
p G p A B C
3 7
1
10 10
96. 96
d.
a. We wish to find ( | )
p A F , the probability that Kofi hits the target given that exactly one
hit is registered.
Now,
( )
( | )
( )
p A F
p A F
p F
and ( )
p A F A B C
Therefore,
( )
( | )
( )
p A B C
p A F
p F
1
2
10
9 9
20
97. 97
TREE DIAGRAMS
A very useful diagram to use when solving compound events,
particularly when conditional probability is involved, is the tree
diagram. This diagram represents different outcomes of an experiment
by means of branches.
98. 98
Activity 1.20
The demand for gas is dependent on the weather and much
research has been undertaken to forecast the demand
accurately. This is important since it is quite difficult (and
expensive) to increase the supply at short notice. If on any
particular day, the air temperature is below normal, the
probability that the demand will be high is 0.6. However,
at normal temperatures the probability of high demand
occurring is only 0.2, and if the temperature is above
normal the probability of high demand drops to 0.05. What
is the probability of a high demand occurring if, over a
period of time, the temperature is below normal on 20% of
occasions and above normal on 30% of occasions?
100. 100
Activity 1.21
A company purchases electronic components in batches of
100 and the supplier guarantees that there will be no more
than 5 defective components in each batch. Before the
acceptance of a particular batch the company has a policy of
selecting without replacement two components for testing. If
both components are satisfactory the batch is accepted and if
both are defective the batch is rejected. However, if only one
is defective another component is selected and if this is
satisfactory the batch is accepted, while if defective, the
batch is rejected. If the probability that a component is
defective is 5%, what is the probability that the batch will be
accepted?
102. 102
There are three routes where the decision is to accept, and the addition law can be used to give
the probability that the batch will be accepted. That is
0.90202 0.04602 0.0462 0.99424
The probability that the batch will be rejected is 1 - 0.99424 = 0.00576. It is unlikely that a
batch would be rejected.
103. 103
EXPECTED VALUE
In general it is a long-run average, which means it is the value you
would get if you repeat an experiment long enough. It is calculated by
multiplying the value of a particular variable by the probability of its
occurrence and repeating this for all possible values. In symbols this can
be represented as:
Expected value = px
104. 104
Activity 1.22
Over a long period of time a sales person recorded the number of sales
she achieved per day. From an analysis of her records it was found that
she made no sales 20% of the time, one sale 50% of the time and 2 sales
30% of the time. What is her expected number of sales?
105. 105
Solution
The x in this case takes on values of 0, 1 and 2
Expected value is 0.2 0 0.5 1 0.3 2 1.1
sales
Expected values are frequently used to calculate expected monetary values, or EMV
106. 106
Activity 1.22
An investor buys £1000 of shares with the object of making capital gain
after 1 year. She believes that there is a 5% chance that the shares will
be double in value, a 25% chance that they will be worth £1500, a 30%
chance that they will only be worth £500 and a 40% chance that they
will not change in value. What is the expected monetary value of this
investment, ignoring dealing costs?
107. 107
Solution
The EMV is found in a similar manner to the expected number of sales in the previous activity.
That is
0.05 2000 0.25 1500 0.3 500 0.4 1000
EMV
100 375 150 400
£1025
So the expected monetary value is £1025, and expected profit of £25.
108. 108
Self-Test 1.4
1. A light bulb is to be selected at random from a box of 100 bulbs, details of which are
given below
Type of bulb Defective Satisfactory
60 watts 20 40
100 watts 10 30
Find:
i. P(bulb is defective)
ii. P(bulb is defective | selected bulb is 60 watts)
Suppose that the company receiving the bulbs has a policy of selecting, without
replacement, two bulbs from each box when it is delivered. It then applies the following
decision rules:
If both bulbs are satisfactory then the box will be accepted.
If both bulbs are defective, the box is returned to the suppliers
If one bulb is defective, a third bulb is selected and the box is only accepted if this bulb
is satisfactory.
What is the probability that a box, referred to above, will be accepted?
2. The probability that a married man watches a certain television show is 0.4 and the
probability that a married woman watches the show is 0.5. The probability that a man
watches the show, given that his wife does is 0.7. Find the probability that
a) a married couple watched the show;
b) a wife watches the show given that her husband does
c) at least 1 person of a married couple will watch the show
3. A town has 2 fire engines operating independently. The probability that a specific
engine is available when needed is 0.96.
a) What is the probability that neither is available when needed?
b) What is the probability that exactly one fire engine is available when needed?
109. 109
BAYES THEOREM
The total probability rule
In the figure below, the events 1 2
,
A A , 3
A , 4
A , 5
A and 6
A are mutually exclusive and
1 2 3 4 5 6
S A A A A A A
These events are said to form a partition of the sample space S. By a partition of S, we mean a
collection of mutually exclusive events whose union is S. In general, the events
1 2 3
, , ,....., n
A A A A form a partition of the sample space S if
i
A
( 1,2,...., )
i n
i j
A A
( , , 1,2,...., )
i j i j n
1
n
i
i
S U A
110. 110
In the figure below it can be seen that if B is an event defined on the sample space S such that
( ) 0
p B , then
1 2 3 4 5
( ) ( ) ( ) ( ) ( )
B A B A B A B A B A B
Since the events are mutually exclusive,
1 2 3 4 5
( ) ( ) ( ) ( ) ( ) ( )
p B p A B p A B p A B p A B p A B
1 1 2 2 3 3 4 4 5 5
( ) ( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | ) ( ) ( | )
p B p A p B A p A p B A p A p B A p A p B A p A p B A
In general, if 1 2 3
, , ,....., n
A A A A form a partition of a sample space S, then for any event B
defined on S such that ( ) 0
p B ,
This result is called the total probability rule.
1
( ) ( ) ( )
n
i i
i
p B p A p B A
111. 111
Activity 1.23
In a certain assembly plant, three machines A, B, and C make 30%, 45%
and 25%, respectively of the products. It is known from past experience
that 2% of the products made by machine A, 3% of the products made
by machine B and 2% of the products made by machine C are defective.
If a finished product is selected at random, what is the probability that it
is defective?
•
112. 112
Solution
Let A1 denote the event “the finished product was made by machine A”
A2 denote the event “the finished product was made by machine B”
A3 denote the event “the finished product was made by machine C”
and let D denote the event “the finished product is defective. We wish to find ( )
p D .We are
given that 1
( ) 0.3,
p A 2
( ) 0.45,
p A 3
( ) 0.25,
p A 1
( ) 0.02,
p D A 2
( ) 0.03,
p D A and
3
( ) 0.02
p D A
1 2
,
A A and 3
A form a partition of the sample space. Hence,
1 1 2 2 3 3
( ) ( ) ( | ) ( ) ( | ) ( ) ( | )
p D p A p D A p A p D A p A p D A
0.3 0.02 0.45 0.03 0.25 0.02 0.0245
The probability that a finished product selected at random is defective is 0.0245.
113. 113
Given this new information, we update the prior
probability values by calculating revised probabilities,
referred to as posterior probabilities.
•
Bayes’ theorem provides a means for making these
probability calculations.
114. 114
Application
Consider a manufacturing firm that receives shipments of parts from two different suppliers.
Let A1 denote the event that a part is from supplier 1 and A2 denote the event that a part is from
supplier 2. Currently 65% of the parts purchased by the company are from supplier 1 and the
remaining 35% are form supplier 2.
Hence if a part is selected at random, we would assign the prior probabilities 1
( ) 0.65
p A and
2
( ) 0.35
p A
The quality of the purchased parts varies with the source of supply. The quality ratings of the
two suppliers are as shown below.
Historical quality levels of two suppliers
Percentage good parts Percentage bad parts
Supplier 1 98 2
Supplier 2 95 5
If we let G denote the event that a part is good and B denote the event that a part is bad, then
we have the following conditional probabilities.
1 2
1 2
( | ) 0.02 ( | ) 0.95
( | ) 0.98 ( | ) 0.05
p B A and p G A
p G A p B A
116. 116
To find the probabilities of each experimental outcome,
we simply multiply the probabilities on the branches
leading to the outcome.
117. 117
• Suppose now that the parts from the two suppliers are used in the
firms manufacturing process and that a machine breaks down because
it attempts to process a bad part.
• Given the information that the part is bad, what is the probability that
it came from supplier 1 and what is the probability that it came from
supplier 2?
118. 118
Let B denote the event that the part is bad, we are looking for the posterior probability
1 2
( | ) ( | )
p A B and p A B
From the law of conditional probability, 1
1
( )
( | )
( )
p A B
p A B
p B
Referring to the probability tree, we see that 1 1 1
( ) ( ) ( | )
p A B p A p B A
To find ( )
p B , we note that event B can occur in only two ways: 1
( )
A B
and 2
( )
A B
Therefore, we have
1 2
( ) ( ) ( )
p B p A B p A B
1 1 2 2
( ) ( | ) ( ) ( | )
p A p B A p A p B A
Bayes’ Theorem (two-event case)
1 1
1
1 1 2 2
( ) ( | )
( | )
( ) ( | ) ( ) ( | )
p A p B A
p A B
p A p B A p A p B A
Similarly,
2 2
2
1 1 2 2
( ) ( | )
( | )
( ) ( | ) ( ) ( | )
p A p B A
p A B
p A p B A p A p B A
i.e.
1
(0.65)(0.02) 0.0130
( | ) 0.4262
(0.65)(0.02) (0.35)(0.05) 0.0305
p A B
2
(0.35)(0.05) 0.0175
( | ) 0.5738
(0.65)(0.02) (0.35)(0.05) 0.0305
p A B
119. 119
Tabular Approach
A tabular approach is helpful in conducting the Bayes theorem calculations. Such an approach
is shown below for the parts supplier problem. The computations are done in the following
steps.
Step 1: Prepare the following three columns:
Column 1 The mutually exclusive events i
A for which posterior probabilities are desired.
Column 2 The prior probabilities ( )
i
p A for the events
Column 3 The conditional probabilities ( | )
i
p B A of the new information B given each
event.
Step 2: In column 4, compute the joint probabilities ( )
i
p A B
for each event and the new
information B by using the multiplication law. These joint probabilities are found by
multiplying the prior probabilities in column 2 by the corresponding conditional probabilities
in column 3.
Step 3: Sum the joint probabilities in column 4. The sum is the probability of the new
information, B.
Step 4: In column 5, compute the posterior probabilities using the basic relationship of
conditional probability.
( )
( )
( )
i
i
p A B
p A B
p B
121. 121
Self-Test1.5
1. The prior probabilities for events 1
A and 2
A are 1
( ) 0.40
p A and 2
( ) 0.60
p A
It is also known that 1 2
( ) 0
p A A
Suppose 1 2
( | ) 0.20 ( | ) 0.05
p B A and p B A
a) Are 1
A and 2
A mutually exclusive? Explain.
b) Compute 1
( )
p A B
and 2
( )
p A B
c) Compute ( )
p B
d) Apply Bayes’ theorem to compute 1 2
( | ) ( | ).
p A B and p A B
2. The prior probabilities for events 1 2 3,....., n
A A A A are 1
( ) 0.30
p A , 2
( ) 0.50
p A and
3
( ) 0.20
p A . The conditional probabilities of event B given 1 2 3,....., n
A A A A are
1 2 3
( | ) 0.50, ( | ) 0.40 ( | ) 0.30
p B A p B A and p B A
a. Compute 1
( )
p B A
, 2
( )
p B A
and 3
( )
p B A
b. Apply Bayes’ theorem to compute the posterior probability 2
( | ).
p A B
c. Use the tabular approach to applying Bayes’ theorem to compute
1 2 3
( | ) , ( | ) ( | ).
p A B p A B and p A B
122. 122
Activity 1.24
A consulting firm rents car from three agencies: 30% from agency A,
20% from agency B and 50% from agency C. 15% of the cars from A,
10% of the cars from B and 6% of the cars from C have bad tyres. If a
car rented by the firm has bad tyres, find the probability that it came
from agency C.
123. 123
Solution
LetA1 denote the event “the car came from agency A”.
A2denote the event “the car came from agency B”.
A3 denote the event “the car came from C”, and
T denote the event “a car rented by the firm has bad tyres.
We wish to find 3
( | ).
p A T
We are given 1
( ) 0.3
p A , 2
( ) 0.2
p A , 3
( ) 0.5
p A ,
1 2 3
( | ) 0.15 , ( | ) 0.1 ( | ) 0.06.
p T A p T A and p T A
1 2
,
A A and 3
A are mutually exclusive and 1 2 3
( ) ( ) ( ) 1
p A p A p A
and so 1 2
,
A A and 3
A form a partition of the sample space.
Hence by Bayes’ theorem;
3 3
3
1 1 2 2 3 3
( ) ( | )
( | )
( ) ( | ) ( ) ( | ) ( ) ( | )
p A p T A
p A T
p A p T A p A p T A p A p T A
0.5 0.06
0.3158
0.3 0.15 0.2 0.1 0.5 0.06
124. 124
1. In a certain assembly plant, three machines B1, B2, and B3 make 30%, 45% and 25%,
respectively of the products. It is known from past experience that 2%, 3% and 2% of
the products made by B1, B2 and B3, respectively, are defective.
a) If a finished product is selected at random, what is the probability that it is defective?
b) If a finished product is found to be defective, what is the probability that it was produced
by B3?
2. A certain construction company buys 20%, 30% and 50% of their nails from hardware
suppliers A, B and C respectively. Suppose it is known that 0.05%, 0.02% and 0.01%
of the nails from A, B and C respectively, are defective.
a) What percentage of the nails purchased by the construction company is defective?
If a nail purchased by the construction company is defective, what is the probability that it
came from
125. 125
PROBABILITY USING
COMBINATORIALANALYSIS
Activity 1.25
1. A box contains 8 red, 3 white and 9 blue balls. If 3 balls are drawn at random without
replacement, determine the probability that
a) all 3 are red
b) all 3 are white
c) 2 are red and 1 is blue
d) at least 1 is white
e) 1 of each colour is drawn
f) the balls are drawn in the order red, white, blue
126. 126
Solution
a. Let R1 = ‘red ball on 1st
draw”
R2 = “red ball on 2nd
draw”
R3 = “red ball on 3rd
draw ’’
Then 1 2 3
R R R
denotes the event that “all 3 balls drawn are red”.
We wish to find
1 2 3 1 2 1 3 1 2
( ) ( ) ( ) ( )
p R R R p R p R R p R R R
8 7 6
20 19 18
Alternatively,
1 2 3
3 8
( )
3 20
nmberof selectionsof out of the red balls
p R R R
numberof selectionsof out of the balls
8
3
20
3
14
285
C
C
127. 127
a. P(all 3 are write)
3
3
20
3
1
1140
C
C
b.
( 8 )( 1 3 )
(2 1 )
3 20
selectionof twoout of red balls seletionof out of whiteballs
p arered and white
numberof selections of out of balls
8 3
2 1
20
3
( ) 7
95
C C
C
c. p (none is white)
17
3
20
3
34
57
C
C
Then p (at least 1 is white) = 1 – p (none is white)
34 23
1
57 57
Note that, the simplest method of working the at least one …...’ type of problem is
p (at least one …..) = 1 – p (None of them …..)
E.g. P (at least one of them solves the problem) = 1 – P (none of them solves the problem)
128. 128
a. p (1 of each colour is drawn)
8 3 9
1 1 1
20
3
18
95
C C C
C
b. (balls drawn in order red, white, blue)
1
3!
P(1 of each colour is drawn)
1 18 3
6 95 95
Another method,
1 2 3 1 2 1 3 1 2
( ) ( ) ( ) ( )
p R W B p R p W R p B R W
8 3 9 3
20 19 18 95
129. 129
Activity 1.26
In the game of poker, five cards are drawn from a pack of 52 well-
shuffled cards. Find the probability that
a)4 are aces
b)4 are aces and 1 is king
c)3 are tens and 2 are jacks
d)a nine, ten, jack, queen, king are obtained in any order
e)3 are of any one suit and 2 are of another
f)at least 1 ace is obtained
130. 130
Solution
a. P (4 aces)
4 48
4 1
52
5
1
54145
C C
C
b. P (4 aces and 1 king)
4 4
4 1
52
5
1
649,740
C C
C
c. P (3 aces and 2 are jacks)
4 4
3 2
52
5
1
108,290
C C
C
d. P(nine, ten, jack, queen, king in any order)
4 4 4 4 4
1 1 1 1 1
52
5
64
162435
C C C C C
C
e. P(3 or anyone suit, 2 of another)
13 13
3 2
52
5
(4 ) (3 ) 429
4165
C C
C
Since there are 4 ways of choosing the first suit and 3 ways of choosing the second suit.
f. P (no ace)
48
5
52
5
35673
54145
C
C
Then P(at least one ace) = 1 – P (no ace)
35673 18472
1
54145 54145
![54
[A] The classical method
The classical method of assigning probabilities is appropriate when all
the experimental outcomes are equally likely. If n experimental
outcomes are possible, a probability of
is assigned to each experimental outcome.
When using this approach, the two basic requirements for assigning
probabilities are automatically satisfied.
•](https://image.slidesharecdn.com/presentation1-250623170734-914673b4/85/Presentation1-pptx-Learn-and-understand-everything-54-320.jpg)
![59
[B] The relative frequency method
The relative frequency method of assigning probabilities is appropriate
when data are available to estimate the proportion of the times the
experimental outcome will occur if the experiment is repeated a large
number of times.](https://image.slidesharecdn.com/presentation1-250623170734-914673b4/85/Presentation1-pptx-Learn-and-understand-everything-59-320.jpg)
![65
[C] The subjective method
The subjective method of assigning probabilities is most
appropriate when one cannot realistically assume that the
experimental outcomes are equally likely and when little
relevant data are available.
When this method is used to assign probabilities to the
experimental outcomes, we may use any information
available, such as our experience or intuition. Because
subjective probability expresses a person’s degree of belief,
it is personal.
Using this method, different people can be expected to assign
different probabilities to the same experimental outcome.](https://image.slidesharecdn.com/presentation1-250623170734-914673b4/85/Presentation1-pptx-Learn-and-understand-everything-65-320.jpg)
![83
Activity 1.16
The events A, B and C have probabilities
1 1
( ) , ( )
2 3
p A p B
and
1
( )
4
p C Furthermore,
,
A C B C
and
1
( )
6
p A B
, find:
a) [( ) ]
p A B
b) ( )
p A B
c) [( ) ]
p A B
d) ( )
p A B
e) ( )
p A B C
](https://image.slidesharecdn.com/presentation1-250623170734-914673b4/85/Presentation1-pptx-Learn-and-understand-everything-83-320.jpg)
![84
Solution
a. [( ) ] 1 ( )
p A B p A B
1 5
1
6 6
b. ( ) ( ) ( )
p A B p A p A B
1 1 1
2 6 3
c. [( ) ] 1 ( )
p A B p A B
1 { ( ) ( ) ( )}
p A p B p A B
1 1 1 2 1
1 1
2 3 6 3 3
d. ( ) [( ) ]
p A B p A B
1
3
e. ( ) ( ) ( ) ( )
p A B C p A p B p C
( ) ( ) ( ) ( )
p A B p A C p B C p A B C
1 1 1 1 11
2 3 4 6 12
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