sample space, events, countable techniques

1. ‫الرحيم‬ ‫الرحمن‬ ‫هللا‬ ‫بسم‬
‫علما‬ ‫زدني‬ ‫ربي‬ ‫وقل‬
‫العظيم‬ ‫هللا‬ ‫صدق‬
STATISTICS
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2. STATISTICS
Dr. Ghada Mohammed Elgohary
Industrial Engineering Department
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3. Learning Objectives
After studying this chapter, you should be able to do the following:
1. Understanding and describe sample spaces and events for random experiments with graphs,
tables and tree diagram.
2. Interpret probabilities and use its outcomes to calculate probabilities of events in discrete
sample spaces.
3. Use permutations and combinations to count the number of outcomes in both an event and
the individual events.
4. Calculate the probabilities of joint events such as union and intersections from the
probabilities of individual events.
5. Interpret and calculate conditional probabilities of events.
6. Determine the independence of events and use independence to calculate probabilities.
7. Use Bayes’ theorem to calculate conditional probabilities
8. Understand random variables.
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4. Reference Book
Book: Applied Statistics and Probability for Engineers
“6th Edition”
Authors, Douglas C. Montgomery and George C. Runger.
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5. Course Contents
Chapter 1 The Role of Statistics in Engineering
Chapter 2 Probability
Chapter 3 Discrete Random Variables and Probability Distributions
Chapter 4 Continuous Random Variables and Probability Distributions
Chapter 5 Joint Probability Distributions
Chapter 8 Statistical Intervals for a Single Sample
Chapter 9 Tests of Hypotheses for a Single Sample
Revision
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6. Chapter Outline
❑Sample Spaces and Events
❑Random Experiments
❑Sample Spaces
❑Events
❑Counting Techniques
❑Addition Rules
❑Multiplication and Total Probability Rules
❑Independence
❑Bayes’ Theorem
❑Random Variables
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7. Sample Space and Events
Random Experiment
▪An experiment that can result in different outcomes, even though it is
repeated in the same manner every time, is called random experiment.
▪The set of all possible outcomes of a random experiment is called the
sample space of experiment and is denoted as 𝑺.
“A system designed without considering variation will be inadequate for practical use.”
“the sample space is defined according to the objective of analysis”
System
Input output
Controlled variables
System variables
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8. Presenting of Sample Space
1. 𝑺 = 𝑨, 𝑩, 𝑪, 𝑫, 𝑬, … . .
2. Tree diagram, such as the number of trails/experiment and the given outputs for each
trail/experiment.
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9. Continue
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10. Illustrative examples
Example 2-1 Camera Flash
An experiment that selects a cell phone camera and records the recycle time of a flash (the time taken to
ready the camera for another flash). The possible values for this time is positive it is convenient to define the
sample space as simply the positive real numbers
𝑺 = 𝑹+
= 𝒙 𝒙 > 𝟎}
If it is known that all recycle times are between 1.5 and 5 seconds, the sample space can be
𝑺 = 𝑹+ = 𝒙 𝟏. 𝟓 < 𝒙 < 𝟓}
If the objective of the analysis is to consider only whether the recycle time is low, medium, or high, the
sample space can be taken to be the set of three outcomes
𝑺 = {𝒍𝒐𝒘, 𝒎𝒆𝒅𝒊𝒖𝒎, 𝒉𝒊𝒈𝒉}
If the objective of the analysis is only to evaluate whether or not a particular camera conforms to a minimum
recycle time specification, the sample space can be simplified to a set of two outcomes
𝑺 ={yes, no}
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11. Discrete and continuous sample spaces
A sample space is discrete if it consists of a finite or countable
infinite set of outcomes.
A sample space is continuous if it contains an interval.
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12. Example 2-3 Message Delays
Each message in a digital communication system is classified as to whether it is
received within the time specified by the system design. If three messages are
classified, use a tree diagram to represent the sample space of possible
outcomes.
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13. Example 2-4 Automobile Options
An automobile manufacturer provides vehicles equipped with selected options.
Each vehicle is ordered as:
• With or without an automobile transmission
• With or without a sunroof
• With one of three choices of a stereo system
• With one of four choices exterior colors
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14. Events
Event
An event is a subset of the sample space of a random experiment.
Basic set operations are:
• The union denoted as 𝐸1 ∪ 𝐸2
•The intersection denoted as 𝐸1 ∩ 𝐸2
•The complement denoted as 𝐸𝑐 or 𝑬′
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15. Example 2-8 Hospital Emergency Visits
Table below summarizes visits to emergency departments at four hospitals. People may leave
without being seen by a physician, and those visits are denoted as LWBS. The remaining visits
are serviced at the emergency department, and the visitor may or may not be admitted for a
stay in the hospital.
Let A denote the event that a visit is going to Hospital 1, 𝑨 = 𝟓𝟐𝟗𝟐 𝒗𝒊𝒔𝒕𝒐𝒓𝒔
B denote the event that the result of the visit is LWBS, 𝑩 = 𝟗𝟓𝟑 𝒗𝒊𝒔𝒕𝒐𝒓𝒔
Calculate the number of outcomes in 𝑨 ∩ 𝑩, 𝑨′
, 𝒂𝒏𝒅 𝑨 ∪ 𝑩.
𝑨 ∩ 𝑩 = 𝟏𝟗𝟓𝒗𝒊𝒔𝒕𝒐𝒓𝒔,
𝑨′ = 𝟐𝟐𝟐𝟓𝟐 − 𝟓𝟐𝟗𝟐 = 𝟏𝟔, 𝟔𝟗𝟎 𝒗𝒊𝒔𝒕𝒐𝒓𝒔, 𝑨 ∪ 𝑩 = 𝟓𝟐𝟗𝟐 + 𝟗𝟓𝟑 − 𝟏𝟗𝟓 = 𝟔𝟎𝟓𝟎𝒗𝒊𝒔𝒕𝒐𝒓𝒔,
Hospital
People Visited Hospital 1 Hospital 2 Hospital 3 Hospital 4 Total
LWBS 195 270 246 242 953 B
Admitted 1277 1558 666 984 4485
Not admitted 3820 5163 4728 3103 16814
Total 5292 A 6991 5640 4329 22252
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195

16. Mutually Exclusive Events
Two events, denoted as 𝑬𝟏 𝒂𝒏𝒅 𝑬𝟐, such that
𝑬𝟏 ∩ 𝑬𝟐 = ∅ are said to be mutually exclusive events.
•The complement of an event: (𝑬′)′ = 𝑬
•The distributive law for set operations implies that:
𝑨 ∪ 𝑩 ∩ 𝑪 = (𝑨 ∩ 𝑪) ∪ (𝑩 ∩ 𝑪) and 𝑨 ∩ 𝑩 ∪ 𝑪 = 𝑨 ∪ 𝑪 ∩ 𝑩 ∪ 𝑪
•DeMorgan’s laws imply that: (𝑨 ∪ 𝑩)′= 𝑨′ ∩ 𝑩′ and (𝑨 ∩ 𝑩)′=𝑨′ ∪ 𝑩′
•And also, 𝑨 ∩ 𝑩 = 𝑩 ∩ 𝑨 and 𝑨 ∪ 𝑩 = 𝑩 ∪ 𝑨
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17. DR/ GHADA ELGOHARY
Venn Diagram
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18. Venn diagrams
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19. Counting Techniques
Multiplication Rule:-
Assume an operation can be described as a sequence of 𝑘 steps, and
◦ The number of ways of completing step 1 is 𝑛1, and
◦ The number of ways of completing step 2 is 𝑛2 for each way of completing
step 1, and
◦ The number of ways of completing step 3 is 𝑛3 for each way of completing
step 2, and so forth.
The total number of ways of completing the operation is
𝒏𝟏 × 𝒏𝟐 × ⋯ × 𝒏𝒌
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20. Permutation
Consider a set of elements, such as 𝑺 = 𝒂, 𝒃, 𝒄
So the orders of elements may be have the permutation as
𝒂𝒃𝒄, 𝒃𝒂𝒄, 𝒄𝒃𝒂, 𝒃𝒄𝒂, 𝒂𝒄𝒃, 𝒂𝒏𝒅 𝒄𝒂𝒃
The number of permutations = 𝟑 × 𝟐 × 𝟏 = 𝟑!
So
The number of permutations of 𝒏 different elements
= 𝒏 × 𝒏 − 𝟏 × 𝒏 − 𝟐 × ⋯ × 𝟐 × 𝟏 = 𝒏!
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21. Permutation of Subsets
The number of permutations of subsets of 𝒓 elements selected from a set of 𝒏 different
elements is written as 𝑷𝒓
𝒏
𝑷𝒓
𝒏 =
𝒏!
(𝒏−𝒓)!
Example 2-10 Printed Circuit Board, A printed circuit board has eight different
locations in which a component can be placed. If four different components are
to be placed on the board, how many different designs are possible?
𝑷𝒓
𝒏
= 𝟖 × 𝟕 × 𝟔 × 𝟓 =
𝟖!
𝟒!
= 1680 different designs are possible
1 2 3 4 5 6 7 8
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22. Permutation of Similar Objects
The number of permutations of 𝒏 = 𝒏𝟏 + 𝒏𝟐 + ⋯ + 𝒏𝒓 objects of which 𝒏𝟏are of
one type, 𝒏𝟐 are of a second type,…, and 𝒏𝒓 are of an 𝒓th type is
𝒏!
𝒏𝟏!𝒏𝟐!𝒏𝟑!…𝒏𝒓!
Example 2-11 Hospital Schedule, A hospital operating room needs to schedule
three knee surgeries and two hip surgeries in a day. We denote a knee and hip
surgery as k and h, respectively. The number of possible sequences of three knee
and two hip surgeries is
𝟓!
𝟑!𝟐!
= 𝟏𝟎
The 10 sequences are easily summarized:
{kkkhh ,kkhkh ,kkhhk, khkkh, khhkk, khhkk, hkkkh, hkkhk, hkhkk, hhkkk}
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23. Combinations “Without order consideration”
Is the number of subsets of r elements that can be selected from a
set of n elements is denoted as
𝑛
𝑟
or 𝐶𝑟
𝑛
“Without order consideration”
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𝐶𝑟
𝑛=
𝑛
𝑟
=
𝑛!
𝑟! (𝑛 − 𝑟)!

24. Examples
Example 2-13 Printed Circuit Board Layout
A printed circuit has 8 different locations in which a component can be placed.
If 5 identical components are to be placed on the board, how many different
designs are possible?
Solution
Each design is a subset of size five from the eight locations that are to contain
the components so using the following equation for combinations
𝐶𝑟
𝑛
=
𝑛
𝑟
=
𝑛!
𝑟!(𝑛−𝑟)!
𝑪𝒓
𝒏=
𝟖
𝟓
=
𝟖!
𝟓!(𝟖−𝟓)!
= 56 designs
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25. Example 2-14 Sampling without Replacement
A bin of 50 manufactured parts contains 3 defective parts and 47 non defective parts.
A sample of 6 parts is selected from the 50 parts without replacement. That is, each part
can be selected only once, and the sample is a subset of the 50 parts.
How many different samples are there of size 6 that contain exactly 2 defective parts? Solution
Multiplication Principle
If two operations are performed in order, with different outcomes for the both, so
The total number of combining these operations is doing by Multiplication Rule
𝑪𝑹
𝑵
=
𝒏
𝒓
∗
𝑵 − 𝒏
𝑹 − 𝒓
=
𝒏!
𝒓! 𝒏−𝒓 !
∗
(𝑵−𝒏)!
(𝑹−𝒓)! 𝑵−𝒓−𝑹+𝒓 !
𝐶6
50
=
3
2
∗
47
4
= 3 ∗ 178,365 = 535,095 different ways
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26. Addition Rules
Probability of a Union
𝑷 𝑨 ∪ 𝑩 = 𝑷 𝑨 + 𝑷 𝑩 − 𝑷(𝑨 ∩ 𝑩)
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Three or More Events
𝑷 𝑨 ∪ 𝑩 ∪ 𝑪 = 𝑷 𝑨 ∪ 𝑩 ∪ 𝑪 = 𝑷 𝑨 ∪ 𝑩 + 𝑷(𝑪) − 𝑷[ 𝑨 ∪ 𝑩 ∩ 𝑪]
𝑷 𝑨 ∪ 𝑩 ∪ 𝑪 = 𝑷 𝑨 + 𝑷 𝑩 − 𝑷 𝑨 ∩ 𝑩 + 𝑷 𝑪 − 𝑷[ 𝑨 ∩ 𝑪 ∪ 𝑩 ∩ 𝑪 − 𝑷]
𝑷 𝑨 ∪ 𝑩 ∪ 𝑪
= 𝑷 𝑨 + 𝑷 𝑩 + 𝑷 𝑪 − 𝑷 𝑨 ∩ 𝑩 − 𝑷 𝑨 ∩ 𝑪 − 𝑷 𝑩 ∩ 𝑪 + 𝑷(𝑨 ∩ 𝑩 ∩ 𝑪)

27. Mutually Exclusive Events
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In general, a collection of events, 𝑬𝟏, 𝑬𝟐, 𝑬𝟑,…., 𝑬𝒌, is said to be
mutually exclusive if there is no overlap (intersect) among of them.
Mutually Exclusive Events
A collection of events, 𝑬𝟏, 𝑬𝟐, 𝑬𝟑,…., 𝑬𝒌, is said to be mutually
exclusive if for all pairs,
𝑬𝟏 ∩ 𝑬𝒋 = ∅
So, the addition rule is
𝑷 𝑬𝟏 ∪ 𝑬𝟐∪ 𝑬𝟑 ∪ …. ∪ 𝑬𝒌 = 𝑷(𝑬𝟏)+P( 𝑬𝟐)+…. 𝑷(𝑬𝒌)

28. Conditional Probability
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𝑷 𝑺𝒖𝒏𝒏𝒚 𝑨𝑵𝑫 𝑪𝒍𝒐𝒖𝒅𝒚 = 𝑷 𝑺𝒖𝒏𝒏𝒚 ∗ 𝑷(𝑪𝒍𝒐𝒖𝒅𝒚/𝑺𝒖𝒏𝒏𝒚)
𝑷 𝑺𝒖𝒏𝒏𝒚 ∩ 𝑪𝒍𝒐𝒖𝒅𝒚 = 𝑷 𝑺𝒖𝒏𝒏𝒚 ∗ 𝑷(𝑪𝒍𝒐𝒖𝒅𝒚/𝑺𝒖𝒏𝒏𝒚)
Sunny
Not
Sunny
50%
50%
Cloudy
Dusty
30%
70%
Cloudy
Rainy
40%
60%
𝑃(𝑪𝒍𝒐𝒖𝒅𝒚/𝑺𝒖𝒏𝒏𝒚)
𝑃(𝑪𝒍𝒐𝒖𝒅𝒚/𝑵𝒐𝒕 𝑺𝒖𝒏
The Conditional Probability
The Conditional Probability of an event B given
an event A, denoted as P(B/A), is
Probability of B given A, is P(B/A)
P(B/A)=𝑷(𝑨 ∩ 𝑩)/𝑷(𝑨) for P(A)>0

29. Assignment 2
Represent the outcomes of these experiments:
1. The tossing of coin twice,
2. The passing of you in your exams in four
subjects
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