Mata kuliah
Tahun
: A0392 - Statistik Ekonomi
: 2010
Pertemuan 05
Sebaran Peubah Acak Diskrit
1
Outline Materi:
• Ruang Sampel, Konsep dasar peluang
dan Peubah Acak
• Nilai harapan peubah acak
• Peluang Bersyarat dan Bebas
• Peluang Total dan Kaidah Bayes
2
Konsep Dasar Peluang
• Basic Probability Concepts
– Sample spaces and events, simple
probability, joint probability
• Conditional Probability
– Statistical independence, marginal probability
• Bayes’ Theorem
• Counting Rules
3
Sample Spaces
• Collection of All Possible Outcomes
– E.g., All 6 faces of a die:
– E.g., All 52 cards of a bridge deck:
4
Kejadian (Events)
• Simple Event
– Outcome from a sample space with 1
characteristic
– E.g., a Red Card from a deck of cards
• Joint Event
– Involves 2 outcomes simultaneously
– E.g., an Ace which is also a Red Card from a
deck of cards
5
Visualizing Events
• Contingency Tables
Ace
Total
Black
Red
2
2
24
24
26
26
Total
4
48
52
• Tree Diagrams
Full
Deck
of Cards
Not Ace
Ace
Red
Cards
Black
Cards
Not an Ace
Ace
Not an Ace
6
Contingency Table
A Deck of 52 Cards
Red Ace
Ace
Not an
Ace
Total
Red
2
24
26
Black
2
24
26
Total
4
48
52
Sample Space
7
Tree Diagram
Event Possibilities
Full
Deck
of Cards
Red
Cards
Ace
Not an Ace
Ace
Black
Cards
Not an Ace
8
Probability
• Probability is the Numerical
Measure of the Likelihood
that an Event Will Occur
• Value is between 0 and 1
1
Certain
.5
• Sum of the Probabilities of
All Mutually Exclusive and
Collective Exhaustive Events
0
is 1
Impossible
9
Computing Probabilities
• The Probability of an Event E:
number of event outcomes
P( E ) 
total number of possible outcomes in the sample space
X

T
E.g., P(
) = 2/36
(There are 2 ways to get one 6 and the other 4)
• Each of the Outcomes in the Sample
Space is Equally Likely to Occur
10
Computing Joint Probability
• The Probability of a Joint Event, A and B:
P(A and B)
number of outcomes from both A and B

total number of possible outcomes in sample space
E.g. P(Red Card and Ace)
2 Red Aces
1


52 Total Number of Cards 26
11
Joint Probability Using
Contingency Table
Event
B1
Event
B2
Total
A1
P(A1 and B1) P(A1 and B2) P(A1)
A2
P(A2 and B1) P(A2 and B2) P(A2)
Total
Joint Probability
P(B1)
P(B2)
1
Marginal (Simple) Probability
12
Computing Compound
Probability
• Probability of a Compound Event, A or B:
P( A or B)
number of outcomes from either A or B or both

total number of outcomes in sample space
E.g. P (Red Card or Ace)
4 Aces + 26 Red Cards - 2 Red Aces

52 total number of cards
28 7


52 13
13
Compound Probability
(Addition Rule)
P(A1 or B1 ) = P(A1) + P(B1) - P(A1 and B1)
Event
Event
B1
B2
Total
A1
P(A1 and B1) P(A1 and B2) P(A1)
A2
P(A2 and B1) P(A2 and B2) P(A2)
Total
P(B1)
P(B2)
1
For Mutually Exclusive Events: P(A or B) = P(A) + P(B)
14
Computing Conditional
Probability
• The Probability of Event A Given that
Event B Has Occurred:
P( A and B)
P( A | B) 
P( B)
E.g.
P (Red Card given that it is an Ace)
2 Red Aces 1


4 Aces
2
15
Conditional Probability Using
Contingency Table
Color
Type
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
Revised Sample Space
P(Ace and Red) 2 / 52
2
P(Ace | Red) 


P(Red)
26 / 52 26
16
Conditional Probability and
Statistical Independence
• Conditional Probability:
P( A and B)
P( A | B) 
P( B)
• Multiplication Rule:
P( A and B)  P( A | B) P( B)
 P( B | A) P( A)
17
Conditional Probability and
Statistical Independence
• Events A and B are Independent if
(continued)
P( A | B)  P ( A)
or P ( B | A)  P ( B )
or P ( A and B )  P ( A) P ( B )
• Events A and B are Independent When
the Probability of One Event, A, is Not
Affected by Another Event, B
18
The Law of Total Probability
• Let S1 , S2 , S3 ,..., Sk be mutually exclusive
and exhaustive events (that is, one and only
one must happen). Then the probability of
another event A can be written as
P(A) = P(A  S1) + P(A  S2) + … + P(A  Sk)
= P(S1)P(A|S1) + P(S2)P(A|S2) + … + P(Sk)P(A|Sk)
19
The Law of Total Probability (Cont.)
S1
A
A  S1
S2….
A Sk
Sk
P(A) = P(A  S1) + P(A  S2) + … + P(A  Sk)
= P(S1)P(A|S1) + P(S2)P(A|S2) + … + P(Sk)P(A|Sk)
20
= Σn P(S )P(A|S )
i=1
i
i
Bayes’ Rule (Bayes’ Theorem)
Let S1 , S2 , S3 ,..., Sk be mutually exclusive and exhaustive
events with prior probabilities P(S1), P(S2),…,P(Sk). If an
event A occurs, the posterior probability of Si, given that A
occurred is
P( Si ) P( A | Si )
P( Si | A) 
for i  1, 2,...k
 P( Si ) P( A | Si )
21
Example
From a previous example, we know that 49% of the
population are female. Of the female patients, 8% are
high risk for heart attack, while 12% of the male patients
are high risk. A single person is selected at random and
found to be high risk. What is the probability that it is a
male? Define H: high risk F: female M: male
We know:
P(F) =
P(M) =
P(H|F) =
P(H|M) =
.49
.51
.08
P( M ) P( H | M )
P( M | H ) 
P( M ) P( H | M )  P( F ) P( H | F )
.51 (.12)

 .61
.51 (.12)  .49 (.08)
.12
22
Bayes’ Theorem
Using Contingency Table
50% of borrowers repaid their loans. Out of those
who repaid, 40% had a college degree. 10% of
those who defaulted had a college degree. What is
the probability that a randomly selected borrower
who has a college degree will repay the loan?
P  R   .50
P  C | R   .4
P  C | R   .10
PR | C  ?
23
Bayes’ Theorem
Using Contingency Table
(continued)
Repay
Repay
Total
College
.2
.05
.25
College
.3
.45
.75
Total
.5
.5
1.0
PR | C 
P C | R  P  R 
P C | R  P  R   P C | R  P  R 
.4 .5 

.2


 .8
.4 .5  .1.5 .25
24
Review probabilitas
(Tambahan Materi)
Sample space, sample points, events
•
Sample space,, adalah sekumpulan semua sample points,, yang mungkin;
dimana 
–
–
–
–
•
Contoh 1. Melemparkan satu buah koin:={Gambar,Angka}
Contoh 2. Menggelindingkan dadu: ={1,2,3,4,5,6}
Contoh 3. Jumlah pelanggan dalam antrian: ={0,1,2,…}
Contoh 4. Waktu pendudukan panggilan (call holding time): ={xx>0}
Events A,B,C,…   adalah himpunan bagian dari sample space
– Contoh 1. Angka genap pada sebuah dadu:A={2,4,6}
– Contoh 2. Tidak ada pelanggan yang mengantri : A={0}
– Contoh 3. Call holding time lebih dari 3 menit. A={xx>3}
•
•
Event yang pasti : sample space 
Event yang tidak mungkin : himpunan kosong ()
26
Kombinasi event
• Union (gabungan) :“A atau B” : AB={A atau
B}
• Irisan: “A dan B” : AB={A dan B}
• Komplemen : “bukan A”:Ac={A}
• Event A dan B disebut tidak beririsan (disjoint) bila :
AB=
• Sekumpulan event {B1,B2,…} merupakan partisi dari event
A jika
– (i) Bi  Bj= untuk semua ij
– (ii) iBi =A
27
Probabilitas (peluang)
Back to Six
• Probabilitas suatu event dinyatakan oleh P(A)
• P(A)[0,1]
• Sifat-sifat peluang
28
Conditional Probability
(Peluang bersyarat)
• Asumsikan bahwa P(B)>0
• Definisi : Conditional probability dari suatu event
A bila diketahui event B terjadi didefinisikan
sebagai berikut
• Dengan demikian
29
Teorema Probabilitas Total
• Bila {Bi} merupakan partisi dari sample
space 
• Lalu {ABi} merupakan partisi dari event A,
maka berdasarkan sifat probabilitas yang
ketujuh pada slide nomor 28
• Kemudian asumsikan bahwa P(Bi)>0 untuk
semua i. Maka berdasarkan uraian pada
slide nomor 29 dapat didefinisikan teorema
probabilitas total sbb
30
Teorema Bayes
• Bila {Bi} merupakan partisi dari sample space 
• Asumsikan bahwa P(A)>0 dan P(Bi)>0 untuk semua i.
Maka berdasarkan uraian pada slide nomor 29
• Kemudian, berdasarkan teorema probabilitas total, kita
peroleh
• Ini merupakan teorema Bayes
– Peluang P(Bi) disebut peluang a priori dari event Bi
– Peluang P(BiA) disebut peluang a posteriori dari event Bi (bila
diketahui event A terjadi)
31
Kesalingbebasan statistik dari event (Statistical
independence of event)
• Definisi : Event A dan B saling bebas
(independent) jika
• Dengan demikian
• Demikian pula
32
Peubah acak (random variables)
• Definisi : Peubah acak X (yang merupakan
bilangan riil [real-valued]) adalah fungsi bernilai riil
dan dapat diukur yang didefinisikan pada sample
space ;X:   
– Setiap titik sample (sample points)  dihubungkan
dengan sebuah bilangan riil X()
– Dengan kata lain : memetakan setiap titik sample ke
sebuah bilangan riil menggunakan peubah acak X
33
Contoh
• Sebuah koin dilempar tiga kali; setiap lemparan
akan menghasilkan head (H) atau tail (T)
• Sample space:
• Misalnya peubah acak X merupakan jumlah
total tail (T) dalam ketiga eksperimen
pelemparan koin tersebut, maka :
34