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KONSEP DASARPROBABILITAS
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
2
CHAPTER TOPICS
Basic Probability Concepts Sample spaces and events, simple probability,
joint probability
Conditional Probability Statistical independence, marginal probability
Bayes’ Theorem Counting Rules
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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:
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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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
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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VISUALIZING EVENTS
Contingency Tables
Tree Diagrams
Red 2 24 26Black 2 24 26
Total 4 48 52
Ace Not Ace Total
FullDeckof Cards
RedCards
BlackCards
Not an Ace
Ace
Ace
Not an Ace
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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SIMPLE EVENTSThe Event of a Happy Face
There are 5 happy faces in this collection of 18 objects.
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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The Event of a Happy Face AND Yellow
JOINT EVENTS
1 Happy Face which is Yellow
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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SPECIAL EVENTS
Impossible Event Impossible event E.g., Club & Diamond on 1 card
draw Complement of Event For event A, all events not in A Denoted as A’ E.g., A: Queen of Diamond
A’: All cards in a deck that are not Queen ofDiamond
Impossible Event
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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SPECIAL EVENTS
Mutually Exclusive Events Two events cannot occur together E.g., A: Queen of Diamond; B: Queen of Club Events A and B are mutually exclusive
Collectively Exhaustive Events One of the events must occur The set of events covers the whole sample space E.g., A: All the Aces; B: All the Black Cards; C: All the
Diamonds; D: All the Hearts Events A, B, C and D are collectively exhaustive Events B, C and D are also collectively exhaustive
(continued)
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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CONTINGENCY TABLEA Deck of 52 Cards
Ace Not anAce
Total
Red
Black
Total
2 24
2 24
26
26
4 48 52
Sample Space
Red Ace
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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FullDeckof Cards
TREE DIAGRAM
Event Possibilities
RedCards
BlackCards
Ace
Not an Ace
Ace
Not an Ace
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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PROBABILITY
Probability is the NumericalMeasure of the Likelihoodthat an Event Will Occur
Value is between 0 and 1 Sum of the Probabilities of
All Mutually Exclusive andCollective Exhaustive Eventsis 1
Certain
Impossible
.5
1
0
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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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 isEqually Likely to Occur
number of event outcomes( )
total number of possible outcomes in the sample spaceP E
X
T
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COMPUTING JOINTPROBABILITY
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
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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P(A1 and B2) P(A1)
TotalEvent
JOINT PROBABILITY USINGCONTINGENCY 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)
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COMPUTING COMPOUNDPROBABILITY
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
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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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)
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COMPUTING CONDITIONALPROBABILITY
The Probability of Event A Given that Event BHas 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
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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CONDITIONAL PROBABILITYUSING CONTINGENCY TABLE
BlackColor
Type Red Total
Ace 2 2 4Non-Ace 24 24 48Total 26 26 52
Revised Sample Space
(Ace and Red) 2 / 52 2(Ace | Red)
(Red) 26 / 52 26
PP
P
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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CONDITIONAL PROBABILITY ANDSTATISTICAL 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
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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CONDITIONAL PROBABILITY ANDSTATISTICAL INDEPENDENCE
Events A and B are Independent if
Events A and B are Independent When theProbability of One Event, A, is Not Affected byAnother Event, B
(continued)
( | ) ( )
or ( | ) ( )
or ( and ) ( ) ( )
P A B P A
P B A P B
P A B P A P B
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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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 upthe partsof A in allthe B’s
SameEvent
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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BAYES’ THEOREMUSING CONTINGENCY TABLE
50% of borrowers repaid their loans. Out of thosewho repaid, 40% had a college degree. 10% ofthose who defaulted had a college degree. What isthe probability that a randomly selected borrowerwho has a college degree will repay the loan?
.50 | .4 | .10P R P C R P C R
| ?P R C
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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BAYES’ THEOREMUSING CONTINGENCY TABLE
||
| |
.4 .5 .2 .8
.4 .5 .1 .5 .2 5
P C R P RP R C
P C R P R P C R P R
(continued)
Repay Repay
College
College1.0.5 .5
.2
.3
.05
.45
.25
.75
Total
Total
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COUNTING RULE 1
If any one of k different mutually exclusiveand collectively exhaustive events can occuron each of the n trials, the number ofpossible outcomes is equal to kn. E.g., A six-sided die is rolled 5 times, the number
of possible outcomes is 65 = 7776.
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COUNTING RULE 2
If there are k1 events on the first trial, k2events on the second trial, …, and kn eventson the n th trial, then the number of possibleoutcomes is (k1)(k2)•••(kn). E.g., There are 3 choices of beverages and 2
choices of burgers. The total possible ways tochoose a beverage and a burger are (3)(2) = 6.
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COUNTING RULE 3
The number of ways that n objects can bearranged 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.
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COUNTING RULE 4:PERMUTATIONS
The number of ways of arranging X objectsselected 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
STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.
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COUNTING RULE 5:COMBINTATIONS
The number of ways of selecting X objects outof 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
DistribusiBinomial,
Poisson, danHipergeometrik