Lessonpart 2 lesson 3 Sample space events.pdf

Lessonpart 2 lesson 3 Sample spa

1. ةعماجفيوس ينب
Probability and Statistics for Engineers
STAT 301

2. Chapter 2: Lesson 1
Events

3. Contents
❑Statistical Experiment
❑Sample Space
❑Events

4. Is some procedure (or process) that we do
and it results in an outcome.
An Experiment
Is an experiment we do not know its exact
outcome in advancebut we know the set of
all possible outcomes.
It is also called statistical experiment
A random experiment
StatisticalExperiment:

5. Definition:
➢Thesetofallpossibleoutcomesofa
statisticalexperimentiscalledthesample
spaceandisrepresentedbythesymbolS.
➢Eachoutcome(elementormember)ofthe
samplespaceSiscalledasamplepoint.
TheSampleSpace:

6. ➢Thesamplespaceofpossibleoutcomes
whenacoinistossed,maybewritten:
S={H,T}
➢where
HandTcorrespondto"heads"and
"tails,"respectively.
TheSampleSpace(Example1):

7. Consider the experiment of tossing a die .
If we are interested in the numberthat shows
on the top face, the sample space would be :
S
1={1,2,3,4,5,6}
TheSampleSpace(Example2):
S
2= {even,odd}
If we are interested only in whether the
number is even or odd, the sample space
is

8. Example 2. illustrates the fact that :
More than one sample space can be used to
describe the outcomes of an experiment.
In this case S
1provides more information
than S
2
TheSampleSpace(Example2):
It is desirable to use a sample space that
gives the most information concerning the
outcomes of the experiment

9. An experiment consists of flipping a coin and
then flipping it a second time if a head occurs.
If a tail occurs on the first, flip, then a die is
tossed once.
TheSampleSpace(Example3):
To list the elements of the sample space, we
construct the tree diagram
S= {HH. HT. T1, T2, T3, T4, T5, T6}.

10. TheSampleSpace(Example3):

11. Sample spaces with a large or infinite number
of sample points are best described by a
statement or RuleMethod.
TheSampleSpace(Example4):
If the possible outcomes of an experiment are
the set of cities in the world with a. population
over 1 million, our sample space is written
S = {x | x is a city with a population over 1 million},
which reads "S is the set of all xsuch that xis
a city with a population over 1 million."

12. ➢An event Ais a subset of the sample
space S. That is AS.
➢We say that an event Aoccurs if the
outcome (the result) of the experiment
is an element of A.
Events:
Definition

13. ➢Sis an event
➢(is called the impossible event)
➢SSis an event
➢(Sis called the sure event)
Events:

14. Given the sample space S = { t | t > 0 },
where tis the life in years of a certain
electronic component .
The event Athat the component fails
before the end of the fifth year is the
subset A = { t | 0 < t < 5 }.
Events(Example1):

15. Experiment:Selectingaballfromabox
containing6ballsnumbered1,2,3,4,5
and6.(ortossingadie)
Events(Example2):
Thisexperimenthas6possibleoutcomes
Thesamplespaceis
S={1,2,3,4,5,6}.

16. Events(Example2):
Considerthefollowingevents:
E
1
=gettinganevennumber={2,4,6}S
E
2
=gettinganumberlessthan4={1,2,3}S
E
3
=getting1or3={1,3}S
E
4
=gettinganoddnumber={1,3,5}S
E
5
=gettinganegativenumber={}=S
E
6
= getting a number less than 10
={1,2,3,4,5,6} = SS

17. Events:
➢n(S)= no. of outcomes (elements) in S.
➢n(E)= no. of outcomes (elements) in the
event E.
Notation:

18. Experiment:
Selecting 3 items from manufacturing
process; each item is inspected and classified
as defective (D) or non-defective (N).
Events(Example3):
S={DDD,DDN,DND,DNN,NDD,NDN,NND,NNN}
This experiment has 8 possible outcomes

19. Events(Example3):

20. Events(Example3):
A={at least 2 defectives}=
{DDD,DDN,DND,NDD}S
B={at most one defective}=
{DNN,NDN,NND,NNN}S
C={3 defectives}=
{DDD}S
Consider the following events:

21. Operations on Events

22. Contents
❑Complement
❑Intersection
❑Mutually Exclusive
❑Union

23. OperationonEvents(Complement):
The complement of an event Awith respect to
Sis the subset of all elements of Sthat are not
in A.
We denote the complement, of Aby the
symbol A` or A
C
.
Definition
A
c
= {x S: xA}
A
c
occursifAdoesnot.

24. OperationonEvents(Complement):
Venn Diagram
S

25. OperationonEvents(Example1):
Let Rbe the event that a red card is
selected from an ordinary deck of 52
playing cards, and let Sbe the entire:
deck. Then R
C
is the event that the card
selected from the deck is not a red but a
blackcard.

26. OperationonEvents(Example2):
Consider the sample space
S = {book, catalyst, cigarette, precipitate,
engineer, rivet}.
Let A = {catalyst, rivet, book, cigarette}
ThenA' = {precipitate, engineer}.

27. LetAandBbetwoeventsdefinedonthe
samplespaceS.
OperationonEvents(Intersection):
Definition
The intersection of two events Aand B
denoted by the symbol A B,
Is the event containing all elements that are
common to Aand B.

28. AB= AB = {x S: xAand xB}
ABConsistsofallpointsinboth
AandB.
ABOccursifbothAandBoccur
together.
OperationonEvents(Intersection):

29. S
OperationonEvents(Intersection):
Venn Diagram

30. OperationonEvents(Example1):
Let Cbe the event that a person selected at
random in an Internet cafe is a college
student, and let Mbe the event that the
person is a male. Then CM is the event of
all malecollege students in the Internet cafe.

31. OperationonEvents(Example2):
Let
M = {a ,e,I,o,u}andN = {r, s,t}
M N = .
Mand Nhave no elements in common and,
therefore, cannot both occur simultaneously.

32. TwoeventsAandBaremutually
exclusive(ordisjoint)ifandonlyif
AB=;thatis,AandBhaveno
commonelements(theydonotoccur
together).
MutuallyExclusive:
Definition

33. MutuallyExclusive:
AB
Aand Bare not
mutually exclusive
AB= 
Aand Bare mutually
exclusive (disjoint)
Venn Diagram

34. The union of the two events Aand B,
denoted by the symbol AB, is the
event containing all the elements that
belong to Aor Bor both.
OperationonEvents(Union):
Definition

35. OperationonEvents(Union):
AB={xS:xAorxB}
ABConsistsofalloutcomesinAorin
BorinbothAandB.
ABOccursifAoccurs,orBoccurs,or
bothAandBoccur.
ThatisABOccursifatleastoneof
AandBoccurs.

36. OperationonEvents(Union):
Venn Diagram
S

37. Union(Examples):
LetA = {a,b,c} and B = {b,c,d,e}
ThenAB = {a,b,c,d,e}.
If M = {x | 3 < x < 9}andV = {y \5 < y < 12},
Then
M N = [z | 3 < z < 12}.

38. Union(Examples):
Venn Diagram

39. Union(Examples):

40. Exercises

41. Exercises

42. Exercises

43. Exercises

44. Counting
Techniques

45. Contents
❑Multiplication Rule
❑Permutations

46. CountingSamplePoints:
Therearemanycountingtechniqueswhich
canbeusedtocountthenumberpointsinthe
samplespace(orinsomeevents)without
listingeachelement.
Inmanycases,wecancomputethe
probabilityofaneventbyusingthecounting
techniques.

47. MultiplicationRule:
Ifanoperationcanbeperformedinn
1
ways,andifforeachofthesewaysa
secondoperationcanbeperformedinn
2
ways,thenthetwooperationscanbe
performedtogetherinn
1n
2ways.
Theorem

48. MultiplicationRule(Example1):
How many sample points are: there: in
the sample space when a pair of dice is
thrown once?

49. MultiplicationRule(Example1):
Thefirstdiecanlandinanyoneofn
1=6
ways.Foreachofthese6waysthe
seconddiecanalsolandinn
2=6ways.
Therefore,thepairofdicecanlandin:
n
1n
2= (6)(6) = 36 possible ways.

50. MultiplicationRule(Example1):

51. MultiplicationRule:
If an operation can be performed in n
1ways,
and if for each of these a second operation
can be performed in n
2ways, and for each of
the first two a third operation can be
performed in n
3 ways, and so forth, then the
sequence of k operations can be performed in
n
1 n
2 ……..n
kways.
Theorem

52. MultiplicationRule(Example2):
Samisgoingtoassembleacomputerby
himself.Hehasthechoiceofordering
chipsfromtwobrands,aharddrive
fromfour,memoryfromthree,andan
accessorybundlefromfivelocalstores.
Howmanydifferent,wayscanSam
ordertheparts?

53. MultiplicationRule(Example2):
Since n
1 = 2 , n
2 = 4 , n
3 = 3 and n
4 = 5
There are :
n
1×n
2 ×n
3×n
4
= 2 ×4 ×3 ×5 = 120
different ways to order the parts.
Solution:

54. MultiplicationRule(Example3):
How many even four-digit numbers can
be formed from the digits 0, 1, 2, 5, 6,
and 9if each digit can be used only once?

55. MultiplicationRule(Example3):
Since the number must be even, we have only
n
1= 3 (0,2,6) choices for the units position
Hence we consider the units position by two
parts, 0or not 0.
However, for a four-digit number the
thousands position cannot be 0.
Solution:

56. MultiplicationRule(Example3):
if units position is 0 ( i.e. n
1 = 1 ) we have
n
2 = 5: thousands position.
n
3 = 4: hundreds position.
n
4 = 3: tens position.
Therefore, in this case we have a total of
n
1×n
2 ×n
3×n
4
= 1 ×5 ×4 ×3 = 60
even four-digit numbers

57. MultiplicationRule(Example3):

58. MultiplicationRule(Example3):
if units position is not 0 ( i.e. n
1 = 2 ) we have
n
2 = 4: thousands position.
n
3 = 4: hundreds position.
n
4 = 3: tens position.
Therefore, in this case we have a total of
n
1×n
2 ×n
3×n
4
= 2 ×4 ×4 ×3 = 96
even four-digit numbers

59. MultiplicationRule(Example3):

60. MultiplicationRule(Example3):
Sincethetwocasesaremutually
exclusiveofeachother,thetotal
numberofevenfour-digitnumbers
canbecalculatedby:
60+96 = 156
even four-digit numbers

61. Permutations:
Apermutationisanarrangementofallor
partofasetofobjects.
Consider the three letters a, b, and c. The
possible permutations are:
abc, acb,bac, bca, cab, and cba.
There are 6 distinct arrangements
Definition

62. Permutations:
We can reach the same answer if we
use multiplication rule:
n
1×n
2 ×n
3
= 3 ×2 ×1= 6 permutation

63. Permutations(Factorial):
In general, ndistinct objects can be
arranged in
n(n -l)(n -2) • • • (3)(2)(1) ways.
This product is called factorialand
represents by n!
The number of permutations of nobjects
is n!.

64. Permutations:
The number of permutation of n
distinct objects taken rat a time is
Theorem()
nr
rn
n
P
rn ,,2,1,0;
!
!
=
−
=

65. Permutations:

66. Permutation(Example1):
In one year, three awards (research,
teaching, and service) will be given for a
class of 25 graduate students in a statistics
department. If each student can receive
at most one award, how many possible
selections are there?

67. Permutation(Example1):
Since the awards are distinguishable, it is a
permutation problem. The total number of
sample points is( ) !22
!25
!325
!25
325 =
−
=P !22
!22232425 
= 13800232425 ==

68. Permutations(Example2):
A president and a treasurer are to be chosen
from a student club consisting of 50 people.
How many different choices of officers are
possible if
(a) There are no restrictions;
(b) Awill serve only if he is president;
(c) Band Cwill serve together or not at all:
(d) Dand Ewill not serve together?

69. Permutations(Example2):
(a) The total number of choices of the
officers if there are no restrictions
is:( )
24504950
!250
!50
250 ==
−
=P

70. Permutations(Example2):
(b) Since A will serve only if he is the president
, we have two situations here:
(i) A is selected as the president, which yields
49possible outcomes;
President
A
Treasurer
B
C
.
AX
AB AC AD … AAX

71. Permutations(Example2):( )
23524849
!249
!49
249 ==
−
=P 2401235249 =+
(ii) Officers are selected from the remaining
49 people which has the number of choices
Therefore, the total number of choices is:

72. Permutations(Example2):( )
22564748
!248
!48
248 ==
−
=P 225822562 =+
(C) The number of selections when Band C
serve together is 2
The number of selections when both B and
C are not chosen is :
BC CB
Therefore, the total number of choices is:

73. Permutations(Example2):
(i) The number of selections when Dserves as
an officer but not Eis (2) (48) = 96
President
D
Treasurer
48
E not Exist
Treasurer
D
President
48
E not Exist
+
(ii) The number of selections when Eserves as
an officer but not Dis also (2) (48) = 96

74. Permutations(Example2):( )
22564748
!248
!48
248 ==
−
=P 244822569696 =++
(iii) The number of selections when both D
and Eare not chosen is
Therefore, the total number of choices is:
OR: Since D and E can only serve together in 2
ways, the answer is 2450 -2 = 2448.

75. Permutations:
The number of distinct permutations of
nthings of which n
1are of one kind, n
2
of a second kind,..., n
kof a kthkind is:
Theorem !..... ! !
!
21 knnn
n


76. Permutations(Example3):
How many words consisting of 3 letters that
can be construct from a x x?
axxxaxxxa=33
!2!1
!3
=

77. Permutations(Example4):
Inacollegefootballtrainingsession,the
defensivecoordinatorneedstohave10
playersstandinginarow.Amongthese
10players,thereare1freshman,2
sophomores,4juniors,and3seniors,
respectively.Howmanydifferentways
cantheybearrangedinarowifonly
theirclasslevelwillbedistinguished?

78. Permutations(Example4):
the total number of arrangements is12600
!3 !4!2!1
!10
=

79. Permutations:
The number of ways of partitioning a set
of nobjects into rcells with n
1elements
in the first cell, n
2elements in the second,
and so forth, is :
Theorem
!..... ! !
!
21321 rr nnn
n
,....,n,n,nn
n

=







 nn....nnn
r=++++
321 where

80. Permutations(Example5):
In how many ways can 7graduate
students be assigned to one triple and
two double hotel rooms during a
conference ?210
!2 !2 !3
!7
223
7
=

=








,,

81. Combinations:
Inmanyproblems,weareinterestedinthe
numberofwaysofselectingrobjectsfromn
objectswithoutregardtoorder.These
selectionsarecalledcombinations.

82. Combinations:
Thenumberofcombinationsofndistinct
objectstakenratatimeisdenotedby
andisgivenby:
Theorem







r
n ()
nr
rnr
n
r
n
,,2,1,0;
!!
!
=
−
=









83. Combinations(Notes):
is read as “ n“ choose “ r”.
Or ncombination r







r
n 1=








n
n 1
0
=







n n
n
=








1 







−
=








rn
n
r
n

84. Combinations(Notes):()
!
1.........21
r
rnnnn
r
n −−−−
=







 35
123
567
3
7
=


=









85. Ifwehave10equal–priorityoperationsand
only4operatingroomsareavailable,inhow
manywayscanwechoosethe4patientsto
beoperatedonfirst?
Combinations(Example1):

86. ( ) !6!4
!10
!410!4
!10
4
10

=
−
=







 n= 10r= 4
The number of different ways for selecting
4 patients from 10 patients is
Combinations(Example1):( )( )1234561234
12345678910


= )(210 waysdifferent=

87. 210
1234
78910
4
10
=


=







 OR
Combinations(Example1):

88. How many different letter arrangements
can be made from the letters in the word
of STATISTICS ?
Combinations(Example2):
Here we have total 10 letters, while 2 letters
(S, T) appear 3 times each, letter appears
twice, and letters A and C appear once each.50400
1! !1 !2 !3 !3
!10
1,12,33
10
=

=








,,