Post on 19-Dec-2015
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Pertemuan 03
Peluang Kejadian
Matakuliah : A0392 - Statistik Ekonomi
Tahun : 2006
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Outline Materi:
• Konsep dasar peluang
• Peluang bebas dan bersyarat
• Peluang total dan kaidah Bayes
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Basic Business Statistics (9th Edition)
Basic Probability
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Konsep Dasar Peluang
• Basic Probability Concepts– Sample spaces and events, simple
probability, joint probability
• Conditional Probability– Statistical independence, marginal probability
• Bayes’ Theorem
• Counting Rules
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Sample Spaces
• Collection of All Possible Outcomes– E.g., All 6 faces of a die:
– E.g., All 52 cards of a bridge deck:
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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
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Visualizing Events
• Contingency Tables
• Tree Diagrams
Red 2 24 26
Black 2 24 26
Total 4 48 52
Ace Not Ace Total
Full Deck of Cards
Red Cards
Black Cards
Not an Ace
Ace
Ace
Not an Ace
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Contingency Table
A Deck of 52 Cards
Ace Not anAce
Total
Red
Black
Total
2 24
2 24
26
26
4 48 52
Sample Space
Red Ace
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Full Deck of Cards
Tree Diagram
Event Possibilities
Red Cards
Black Cards
Ace
Not an Ace
Ace
Not an Ace
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Probability
• Probability is the Numerical Measure of the Likelihood that an Event Will Occur
• Value is between 0 and 1
• Sum of the Probabilities of All Mutually Exclusive and Collective Exhaustive Events is 1
Certain
Impossible
.5
1
0
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(There are 2 ways to get one 6 and the other 4)E.g., P( ) = 2/36
Computing Probabilities
• The Probability of an Event E:
• Each of the Outcomes in the Sample Space is Equally Likely to Occur
number of event outcomes( )
total number of possible outcomes in the sample space
P E
X
T
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Computing Joint Probability
• The Probability of a Joint Event, A and B:
( and )
number of outcomes from both A and B
total number of possible outcomes in sample space
P A B
E.g. (Red Card and Ace)
2 Red Aces 1
52 Total Number of Cards 26
P
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P(A1 and B2) P(A1)
TotalEvent
Joint Probability Using Contingency Table
P(A2 and B1)
P(A1 and B1)
Event
Total 1
Joint Probability Marginal (Simple) Probability
A1
A2
B1 B2
P(B1) P(B2)
P(A2 and B2) P(A2)
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Computing Compound Probability
• Probability of a Compound Event, A or B:( or )
number of outcomes from either A or B or both
total number of outcomes in sample space
P A B
E.g. (Red Card or Ace)
4 Aces + 26 Red Cards - 2 Red Aces
52 total number of cards28 7
52 13
P
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P(A1)
P(B2)
P(A1 and B1)
Compound Probability (Addition Rule)
P(A1 or B1 ) = P(A1) + P(B1) - P(A1 and B1)
P(A1 and B2)
TotalEvent
P(A2 and B1)
Event
Total 1
A1
A2
B1 B2
P(B1)
P(A2 and B2) P(A2)
For Mutually Exclusive Events: P(A or B) = P(A) + P(B)
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Computing Conditional Probability
• The Probability of Event A Given that Event B Has Occurred:
( and )( | )
( )
P A BP A B
P B
E.g.
(Red Card given that it is an Ace)
2 Red Aces 1
4 Aces 2
P
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Conditional Probability Using Contingency Table
BlackColor
Type Red Total
Ace 2 2 4
Non-Ace 24 24 48
Total 26 26 52
Revised Sample Space
(Ace and Red) 2 / 52 2(Ace | Red)
(Red) 26 / 52 26
PP
P
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Conditional Probability and Statistical Independence
• Conditional Probability:
• Multiplication Rule:
( and )( | )
( )
P A BP A B
P B
( and ) ( | ) ( )
( | ) ( )
P A B P A B P B
P B A P A
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Conditional Probability and Statistical Independence
• Events A and B are Independent if
• Events A and B are Independent When the Probability of One Event, A, is Not Affected by Another Event, B
(continued)
( | ) ( )
or ( | ) ( )
or ( and ) ( ) ( )
P A B P A
P B A P B
P A B P A P B
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Bayes’ Theorem
1 1
||
| |
and
i ii
k k
i
P A B P BP B A
P A B P B P A B P B
P B A
P A
Adding up the parts of A in all the B’s
Same Event
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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?
.50 | .4 | .10P R P C R P C R
| ?P R C
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Bayes’ Theorem Using Contingency Table
||
| |
.4 .5 .2 .8
.4 .5 .1 .5 .25
P C R P RP R C
P C R P R P C R P R
(continued)
Repay
Repay
CollegeCollege 1.0.5 .5
.2
.3
.05.45
.25.75
Total
Total
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Counting Rule 1
• If any one of k different mutually exclusive and collectively exhaustive events can occur on each of the n trials, the number of possible outcomes is equal to kn.– E.g., A six-sided die is rolled 5 times, the
number of possible outcomes is 65 = 7776.
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Counting Rule 2
• If there are k1 events on the first trial, k2 events on the second trial, …, and kn events on the n th trial, then the number of possible outcomes is (k1)(k2)•••(kn).
– E.g., There are 3 choices of beverages and 2 choices of burgers. The total possible ways to choose a beverage and a burger are (3)(2) = 6.
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Counting Rule 3
• The number of ways that n objects can be arranged in order is n! = n (n - 1)•••(1).– n! is called n factorial– 0! is 1– E.g., The number of ways that 4 students can
be lined up is 4! = (4)(3)(2)(1)=24.
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Counting Rule 4: Permutations
• The number of ways of arranging X objects selected from n objects in order is
– The order is important.– E.g., The number of different ways that 5
music chairs can be occupied by 6 children are
!
!
n
n X
! 6!
720! 6 5 !
n
n X
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Counting Rule 5: Combintations
• The number of ways of selecting X objects out of n objects, irrespective of order, is equal to
– The order is irrelevant.– E.g., The number of ways that 5 children can
be selected from a group of 6 is
!
! !
n
X n X
! 6!
6! ! 5! 6 5 !
n
X n X