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STAT 111
Chapter Zero
A Review of Set Notation
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A Review of Set Notation
Definition:A set is a well-defined collection (possible empty) of objects, such as the set of letters in the alphabet, or the set of students in a classroom. We denote a set by a capital letter such as A,B,C and an element by a lower case letter such a,b,c.
Notes: If an element a belongs to a set C we write a ϵ C. If an element a does not belong to C we write a ϵ C.
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Presentation
Sets may be described by listing the elements or by describing some property held by all members. For example the set P of positive integers might be described by:
1. P={1,2,3,…}
2. P={x : x is a positive integer}
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Example
Let A be the set of all real numbers whose squares are equal to 25. Show how to describe A by
a) Listing its elements,
A={-5,+5} since x2=25 x=-5 or x=+5
b) By describing some property held by all members.
A={X : X2=25}
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Example
Determine which of the following statement are true and which are false.
1) {2}=2False (since 2 is areal number while {2} is a set which consists of
the real number 2)2) True (since every elements is assumed to be equal to itself)3) True (since no number satisfy
which is a subset of every set 9 and )2 (i.e. 42 Axxx
}:{ xxx
BAxxBxxxA then }1:{ and }9,4:{ 2
BA
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Notations and Venn Diagram
The universal set S is the set of all elements under consideration. Using Venn diagram it consists of a rectangle.
The null or empty set, denoted by Φ, is the set consisting of no elements. Thus Φ is a subset of every set.
. For any two sets A and B we will say that A is a subset of B, or A is contained in B (denoted A B), if every element in A is also in B. Remarks: Every set is a subset of itselfIf AB and BA then A=B
A
B
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Notations and Venn Diagram
SA B S
A B
The union of A and B, denoted by A U B, is the set of all elements which belong to either A or B or both. That is, the union of A and B contain all elements that are in at least one of the sets. The key word for expressing the union of two sets is or (meaning A or B or both).
A U B
The intersection of A and B, denoted by A ∩ B, is the set of all elements which belong to both A and B. The key word for expressing the intersections is and (meaning A and B simultaneously).
A ∩ B
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Notations and Venn Diagram
If A is a subset of S , then the complement of A, denoted by Ac (or A ), is the set of elements that are in S but not in A. the shaded area in S but not in A is Ac. Note that A U Ac =S. Note also, (Ac)c=A
SA
Ac
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Notations and Venn Diagram
SA B
Two sets, A and B, are said to be disjoint or mutually exclusive
if A ∩ B=Ф. That is, mutually exclusive sets have no elements in common.
Note that, it is easy to see that A and Ac are mutually exclusive sets for any set A.
SA B
The difference of A and B, denoted by A-B, is the set of all elements of A which do
not belong to B. Note that A-B=A ∩ Bc
Note also A=(A∩B)U(A∩Bc), and, B=(B∩A)U(B∩Ac)
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ExampleIf S= {1/2,0,л,5, - √ 2 , - 4 } and subsets of S are given byA={- √2, л,0}, B={ 5, ½,-√ 2,- 4} and C={ 1/2,-4 } findA ∩ B, A U B,(A U B ) ∩ C, A- B, ( B ∩ C )c, Bc U Cc and (A ∩ C ) U
( B ∩ C )
Solution:A ∩ B={- √ 2 } A U B= {1/2,0,л ,5, - √ 2 , - 4 } (A U B ) ∩ C ={ 1/2,-4 } A-B={л,0} Bc={л,0} Cc={0,л ,5, - √ 2} Bc U Cc ={0,л,5, - √ 2} = ( B ∩ C )c
(A ∩ C ) U ( B ∩ C )= U { 1/2,-4 } = { 1/2,-4 }
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Example From a survey of sixty students attending a university, it was
found that nine were living off campus, thirty-six were undergraduates and three were undergraduates and living off campus.
a) Find the number of these students who were undergraduates, or living offcampus, or both.
n (LU U)=n(L)+n(U)-n(L∩ U)=9+36-3=42
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b) Find the number of these students who were undergraduates and living on campus.
L Lc Sum
U 3 36
Uc
Sum 9 60
33
51
246 18
n(U ∩ Lc)=33
c) Find the number of these students who were graduate students living on campus.
n(Uc ∩ Lc)=18
SL U
6 33
18
3
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Example
للصم االمل معهد في يدرسون 300يوجد طالب , بالعربية أو باالنجليزية االشارة يدرسون 150لغة
, فقط بالعربية االشارة لغة 80 لغة يدرسون. فقط باالنجليزية االشارة
.I االشارة لغة يدرسون الذين عدد هو ما
والعربية؟ باالنجليزية
.II بالعربية؟ االشارة لغة يدرسون الذين عدد هو ما
.III االشارة لغة يدرسون الذين عدد هو ما
باالنجليزية؟
.IV االشارة لغة يدرسون ال الذين عدد هو ما
بالعربية؟ وال باالنجليزية
70)( AEn
E A
80 150
0
70
22070150)( An
1507080)( En
0)( AEn
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Some Theorems Involving Sets
1)A U B = B U A (commutative law for unions) 2) A ∩ B = B ∩ A (commutative law for inter sections) 3) A U ( B U C ) = (A U B ) U C = (associative law for unions) 4)A ∩ (B ∩ C)= (A∩B)∩C = A ∩ B ∩ C (associative law for inter section 5) A∩( B U C) = (A ∩ B)U (A∩C) (first distributive law) 6) A U (B ∩ C) = (A U B)∩ (A U C) (second distibutive law) 7) A -B = A ∩ Bc
8) if A B then Bc Ac
9) A U Ø = A , A ∩ Ø = Ø 10 ) A U S = S , A ∩ S = A 11) De Morgran's law a) ( A U B )c = A c ∩ B c
b) (A ∩ B ) c = A c U B c (De Morgan's laws can be generalized to any relation between
For Example A ∩ B c = ( A c U B ) c , and (A c U B) c = A ∩ B c
12 ) A = ( A ∩ B ) U ( A ∩ B c)
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Generalizations of Definitions
The union of the sets A1,A2,A3,…, denoted
A1 U A2 U A3 ,… is the set of all elements which are in at least one member of the collection.
The intersection of the sets A1,A2,A3,…, denoted,
A1∩A2 ∩A3,.. is the set of all elements which are in every member of the collection. The collection A1,A2,A3,…, is a pair wise disjoint if
A i ∩ A j = Ф whenever i ≠ j
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Generalizations of Some Theory Involving Sets 1-B ∩ (A1 U A2 U A3 …) = (B ∩ A1 ) U ( B ∩ A2) U (B ∩ A3 ) …
2-B U (A1∩A2 ∩A3 …)= ( B U A1) ∩ ( B U A2) ∩ (B U A3) ∩ … 3-De Morgan's laws
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Cartesian Products
The cartesian product of the sets A1, A2,… ,An, denoted by A1 A2 … An, is the set of all n-tuples with the ith coordinate drawn from the ith set, i=1,2,…,n, i.e.,
A1 A2 … An={(a1, a2,… ,an): ai ϵ Ai, i=1,2,…,n}
Example let A1 = {1,2,3}, , A2 = {a , b }, thenA1 × A2 = {(1,a),(1,b),(2,a),(2,b),(3,a),(3,b)}
هو الضرب لحاصل العناصر وعددn (A1 × A2 ) = n (A1) × n(A2 )The Cartesian product may be plotted in 2-dimensional space using points on
the horizontal axis for the elements of A1,, and points of the vertical axis for the elements of A2, the resulting ordered pairs represent
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Power Sets
The power set of A is the set of all subsets which may be formed from the elements of A .
A من المكونة الجزئية المجموعات جميعExample
Let A= { a , b , c} . Find the power set of A
Power set of A= { Ø , A , {a } , { b } , { c } , {a,b} ,
{a ,c } , { b ,c}}
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STAT 111
Chapter One
Combinatorics
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Combinatorics
In many experiment the number of outcomes in S is so large that a complete listing of these outcomes is too difficult. In such experiment, it is convenient to have a mathematical method of determining the total number of outcomes in the space S and in various events in S without compiling a list of all these outcomes. In the following, some of these methods will be presented.
التي الرياضية الطرق من مجموعة هي العد طرقدون ما تجربة حدوث مرات عدد معرفة علي تساعد
المجموعة هذه عناصر لكتابة الحاجة
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The Basic Principle of Couting (Multiplication Rule )
If an experiments consists of r steps , of which the first can be made in n1 ways , for each of these the second step can be made in n2 ways , and so forth , then the whole experiment can be made in
n1 × n2 × …× nr
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Example 1: A drug for the relief of asthma can be purchased from 5 different
manufacturers in liquid, tablet, or capsule form, all of which come in regular and extra strength. In how many different ways can a doctor prescribe the drug for a patient suffering from asthma
5x3x2 = 30
Example 2: In how many ways can 8 teaching assistants be assigned to 6
sections of a statistic course if no teacher is assigned to more than onesection? 8x7x6x5x4x3 = 20160
The Basic Principle of Couting (Multiplication Rule )
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Example 3:a) How many three-digit numbers can be formed
from the digits 0,1, 2,3,4,5, and 6 if each digit can be used only once ( مسموح غير (التكرار
6x 6 x 5 = 180 b) How many of these are odd numbers? 5x5x3 = 75c) How many are greater than 330? 1x3x5 + 3x6x5 = 15 + 90 = 105
The Basic Principle of Couting (Multiplication Rule )
الخانة هذه في الصفر اختيار لايتم
من الأكبر 4، 5، 6وهي 3الأعداد
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Example 4: If a multiple-choice test consists of 5 questions each with 4 possible answers of which only one is correct,
How many different ways can a student check off one answer to each question ?
4 × 4 × 4 × 4 × 4 = ( 4 )5 = 1024
How many ways can a student check off one answer to each question and get all the question wrong ?
3× 3 × 3 × 3 × 3 = ( 3 ) 5 = 243
The Basic Principle of Couting (Multiplication Rule )
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Permutations Frequently, we are interested in situations where the outcomes are the different orders
or arrangements that are possible for a group of objects. Different arrangements like these are called permutations.
Note:n!=n(n-1)(n-2)…3x2x10!=1
Definition A permutation is an ordered arrangement of a set of distinct objects. The number
of permutations of n distinct objects taken r at a time is denoted by where
=n! / (n-r)!= n(n-1) (n-2)….(n-r+1)
شرطين توفر عند االختيار طرق عدد هي التباديل.I ( ) مختلفة مراكز مختلفة جوائز مهم الترتيب.II مسموح غير nالتكرار
rP
nrP
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Examples
Example
A president, treasurer ( أمين and a secretary, all ,(صندوقdifferent, are to be chosen from a club consisting of 10 people. How many different s of officers are possible?
Example Five separate awards are to
be presented to selected Students from a class of 30 . How many different outcomes are possible if A student can receive any number of awards ;
Each student can receive at
most 1 award?
7208910!7
!10103 P
530
171007202627282930305 P
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notes
In if r= n then
which represents the different ways to arrange n objects in a line.
The number of different ways to arrange n
objects in a circle is
( n – 1 ) !
nrP !
!
!n
nn
nPP n
nn
r
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Examples1. In how many different ways can the five letters a, b, c, d
and e be arranged? 5! = 120
2. In how many ways can 3 Arabic books, 2 Mathematics books, and 1 Chemistry book be arranged on a bookshelf if;
The books can be arranged in any order; 6! = 720 the Arabic books must be together but the other books can
be arranged in any order ; 3! 4! = 144 the Mathematics books must be together and the Arabic
books must be together ? 3 ! 3 ! 2 ! = 72
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Example3. a) In how many ways can 5 persons line up to get on a bus. 5! = 120b) In how many ways can they line up, if 3 persons insist on following each
other? 3! 3! =36c) In how many ways can they line up, if 3 persons refuse to follow each other n(A) = n(S)-n(Ac) - 120 - 3! 3! = 84
4. a) In how many ways may four women and four children be seated in a row of chairs if the women and children to occupy alternate seats'?
2 (4 x 4 x 3 x 3 x 2 x 2 x l x l) = 2 (4! 4!) b) Repeat a) when they are seated at a round table. (4-1)! 4! 5. In how many ways can 7 people be seated at a round table if a) They can sit anywhere; (7-1)! = 6! = 720 b)Two particular people must not sit next to each other? 6! - 2! 5! = 480
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Permutations When Some Elements Are Alike
Given n objects of which n1 are alike, n2 are alike,..., nr are alike, then the total number of permutations is
Example 1How many different ways can we arrange the letters in the worda) Tool 4!/2!a) Statistics 10!/(3!3!2!)
Example 2How many different arrangement are there of the following set {a , α , α , β , β , γ } . 6!/(3!2!)
!!....!
!
21 rnnn
n
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CombinationsIn many problems we are interested in the number of selecting r objects from n without regard to order (the order is unimportant), these selections are called combinations.
يكون عندما االختيار طرق عدد هي التوافيقمهم غير الترتيبمسموح غير التكرار
Definition The number of combinations of r objects is
The number is also denoted by the symbol . Note that represents the number of ways we can draw a sample of size r from a set of n distinct elements without replacement and without regard to the order in which these r elements were chosen. It is defined that = 0 whenever n is a positive integer and r is a positive integer greater than n (clearly, there is no way in which we can select a subset which contains more elements than the set itself if r < 0 or r > n ) هو والتوافيق التباديل بين الفرق الترتيب
من ( شخصين اختيار تم A, B, C إذامهم الترتيب
AB , BA , AC , CA , BC , CBمهم غير الترتيبAB , AC , BC
)!(!
!
rnr
nC n
r
nrC
0nrC
2332 P
3)!1!2(
!3
!2
323
2 P
C
r
n
r
n
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Examples
Example 1In how many different ways can a committee of 4 be selected from among the
64 staff members of a hospital?
Example 2A committee of 7, consisting of 2 doctors, 2 teachers, and 3 engineers, is to be
chosen from a group of 5 doctors, 6 teachers, and 4 engineers. How many committees are possible?
635376!60!4
!64644 C
60041510
3 2 2
4 6 5
43
62
52
CCC
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Examples
Example 3From a group of 4 teachers and 5 doctors, how many committees of size 3 are
possible;a) with no restrictions;
b) with 1 teacher and 2 doctors;
c) With 2 teachers and 1 doctor if a certain teacher must be on the committees;
d) with at least 1 teacher:
e) with at most 2 doctors?
8493 C
52
41 CC
51
31
11 CCC
0 1 2
-------------------
3 2 1 50
43
51
42
52
41
D
CCCCCC
T
0 1 2
-------------------
3 2 1 50
43
51
42
52
41
D
CCCCCC
T
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Examples
From a group of 8 statisticians and 6 economists a committee consisting of 3 statisticians and 3 economists is to be formed. How many different committees are possible if
1- Two of the statisticians refuse to work together.
2- Two of the economists will only work together.
3- One statistician and one economist refuse to work together.
63
62
21
63
63
20 CCCCCC
83
41
22
83
43
20 CCCCCC
53
73
10
10
52
73
11
10
53
72
10
11 CCCCCCCCCCCC
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Theorem
for any positive integers n and r = 0, 1... n.
Result:
rn
n
r
n
nr
n
r
n
r
n
r1
1
1
1
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The Multinomial Theorem
The number of possible divisions of n distinct objects into r distinct groups elements in the first group, n2 elements in the second, and so forth, is
nnnnnnn
n
nnn
nr
rr
... where
!!...!
!
,...,, 212121
لتقسيم الممكنة الطرق إلي nجميع األشخاص أو األشياء من r علي تحتوي بحيث المجموعات من
العناصر n 1 , n2 , … ,nr من
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Multinomial
Example 1: If 12 People are to be divided into 3 committees of
respective sizes 3,4 and , 5 how many divisions are possible ?
In how many ways can a man divide 7 gifts among his 3 children if the eldest to receive 3 gifts and the other 2 each ?
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The Binomial Theorem
The number is often referred to as the binomial coefficient because it arises in the binomial theorem which may be started as follow : For any number x and y and any positive integer n ,
r
n
knkn
k
n yxk
nyx
0
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Examples
Example 1 Expand
Example 2 Expand
Example 3 Find the constant term in the expansion of
Since constant term correspond to the one with 3 k – 12 =0 , therefore , K = 4 and the constant term
3yx
62 2yx
122 1
xx
3223031221303 333
3
2
3
1
3
0
3xyxxyyyxyxyxyxyx
06251260262 26
6....2
1
62
0
62 yxyxyxyx
12312
0
121212
0
122
12
0
122
12
121121
k
k
kk
k
kk
k
xk
xxkx
xkx
x
4954
12
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Example 4 Find the constant term in the expansion of
Since constant term correspond to the one with 3 k – 12 =0 , therefore , K = 4 and the constant term
1231212
0
12121212
0
122
12
0
122
3212
32123
2123
2
kkk
k
kkkk
k
kk
k
xk
xxkx
xkx
x
51963120324
12 4124