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KONSEP DASAR PROBABILITAS

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Page 1: KONSEP DASAR PROBABILITAS fileKONSEP DASAR PROBABILITAS. STATISTIK & PROBABILITAS ... STATISTIK & PROBABILITAS Copyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT. 14 COMPUTING

KONSEP DASARPROBABILITAS

Page 2: KONSEP DASAR PROBABILITAS fileKONSEP DASAR PROBABILITAS. STATISTIK & PROBABILITAS ... STATISTIK & PROBABILITAS Copyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT. 14 COMPUTING

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

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STATISTIK & PROBABILITASCopyright © 2017 By. Ir. Arthur Daniel Limantara, MM, MT.

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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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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 26Black 2 24 26

Total 4 48 52

Ace Not Ace Total

FullDeckof Cards

RedCards

BlackCards

Not an Ace

Ace

Ace

Not an Ace

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SIMPLE EVENTSThe Event of a Happy Face

There are 5 happy faces in this collection of 18 objects.

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The Event of a Happy Face AND Yellow

JOINT EVENTS

1 Happy Face which is Yellow

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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

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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)

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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

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FullDeckof Cards

TREE DIAGRAM

Event Possibilities

RedCards

BlackCards

Ace

Not an Ace

Ace

Not an Ace

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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

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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

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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

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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)

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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

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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 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

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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

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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

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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

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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

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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

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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

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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.

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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.

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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.

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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

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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

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DistribusiBinomial,

Poisson, danHipergeometrik