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Probability Theory Validity Probability Theory Validity

Probability Theory Validity - PowerPoint Presentation

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Probability Theory Validity - PPT Presentation

A bar is obeying the law when it has the following property If any of the patrons are below the age of 18 then that person is not drinking alcohol Legal or Illegal Patron Age Drink Alice ID: 713117

probability heads true amp heads probability amp true tails soda flip prime possibility beer rule numbers land inductive age

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Slide1

Probability TheorySlide2

ValiditySlide3

A bar is

obeying the law

when it has the following property:

If any of the patrons are below the age of 18, then that person is not drinking alcohol.Slide4

Legal or Illegal?

Patron

Age

Drink

Alice

Over 18

Beer

Bill

Over 18

Soda

Cecil

Under 18

Beer

Doug

Under 18

SodaSlide5

Ignore People over 18

Patron

Age

Drink

Alice

Over 18

Beer

Bill

Over 18

Soda

Cecil

Under 18

Beer

Doug

Under 18

SodaSlide6

Make Sure No One Else Has Alcohol

Patron

Age

Drink

Alice

Over 18

Beer

Bill

Over 18

Soda

Cecil

Under 18

Beer

Doug

Under 18

SodaSlide7

Illegal!

Patron

Age

Drink

Alice

Over 18

Beer

Bill

Over 18

Soda

Cecil

Under 18

Beer

DougUnder 18SodaSlide8

Legal or Illegal?

Patron

Age

Drink

Evan

Over 18

Beer

Fred

Over 18

Soda

Gale

Under 18

Soda

HarrietUnder 18SodaSlide9

Ignore People over 18

Patron

Age

Drink

Evan

Over 18

Beer

Fred

Over 18

Soda

Gale

Under 18

Soda

HarrietUnder 18SodaSlide10

Make Sure No One Else Has Alcohol

Patron

Age

Drink

Evan

Over 18

Beer

Fred

Over 18

Soda

Gale

Under 18

Soda

HarrietUnder 18SodaSlide11

Legal or Illegal?

Patron

Age

Drink

Ivan

Over 18

Beer

Janet

Over 18

Soda

Kerry

Over 18

Soda

LennyOver 18SodaSlide12

Ignore Everyone

Patron

Age

Drink

Ivan

Over 18

Beer

Janet

Over 18

Soda

Kerry

Over 18

Soda

LennyOver 18SodaSlide13

Legal! No One Underage Has Alcohol

Patron

Age

Drink

Ivan

Over 18

Beer

Janet

Over 18

Soda

Kerry

Over 18

Soda

LennyOver 18

SodaSlide14

A bar is

obeying the law

when it has the following property:

If any of the patrons are below the age of 18, then that person is not drinking alcohol.

An argument is

valid

when it has the following property:

If any possibility (evaluation) makes the premises true, then the conclusion is not false.Slide15

Valid or Fails the Test?

Possibility

Premise

Conclusion

Possibility

1

True

True

Possibility

2

True

False

Possibility

3FalseTruePossibility 4FalseFalseSlide16

Ignore the Possibilities Where Premise Is Not True

Possibility

Premise

Conclusion

Possibility

1

True

True

Possibility

2

True

False

Possibility

3FalseTruePossibility 4False

FalseSlide17

Make Sure No Remaining Possibilities Make the Conclusion False

Possibility

Premise

Conclusion

Possibility

1

True

True

Possibility

2

True

False

Possibility

3FalseTruePossibility 4False

FalseSlide18

Make Sure No Remaining Possibilities Make the Conclusion False

Possibility

Premise

Conclusion

Possibility

1

True

True

Possibility

2

True

False

Possibility

3FalseTruePossibility 4False

FalseSlide19

Example

P

Q

~~P

((Q

→ P

) → Q)

T

T

T

T

T

F

TFFTFTF

F

F

FSlide20

Example

P

Q

~~P

((Q

→ P

) → Q)

T

T

T

T

T

F

TFFTFT

F

F

F

FSlide21

Example

P

Q

~~P

((Q

→ P

) → Q)

T

T

T

T

T

F

TFFTFT

F

F

F

FSlide22

RecapSlide23

Inductive Arguments

An

inductive argument

tries to

show that its conclusion is supported by its premises.

In other words,

it tries

to show that the truth of its premises makes it

more likely

that its conclusion will be true. Slide24

Inductive Strength

The quality of an inductive argument is measured by its strength – the degree to which its premises raise the probability of its conclusion.

If they don’t raise the probability very much, the argument is not very strong. If they do, the argument is strong.Slide25

Additional Evidence

Even strong inductive arguments with true premises can be shown to be bad arguments with the addition of more evidence.Slide26

Inductive Syllogism

In standard form:

1) Most bankers are rich.

2) Bill is a banker.

3) Bill is rich.

The general form of the argument I just gave is:

1) Most X’s are Y.

2) A is an X.

C) A is Y.

This type of argument is called an inductive syllogism. Slide27

Evaluating Inductive Syllogisms

The

strength of an inductive syllogism depends primarily on the strength of the generalization.

But our assessment of the argument also has to do with the amount of available evidence that has been taken into account. Slide28

Inductive Generalization

The

argument form of an inductive generalization is:

1) Most of the observed sample of X’s are Y.

C) Most X’s are Y.Slide29

Samples

Ideally, when we are trying to find out whether a large percentage of a group has a certain property, we would check every member of the group.

But for a lot of groups, that’s just not possible – there are too many to check. Instead, we look at a

sample

, or a subset of the group.Slide30

Representative Samples

The success of an inductive generalization depends on how good the match is between the sample and the entire group.

If our sample of bankers is 90% rich, but bankers on a whole are only 30% rich, our argument will not be a good one.Slide31

The Elements of Sentential LogicSlide32

Sentential Logic: Vocabulary

Sentence letters: A, B, C,…

Logical connectives: ~, &, v, →, ↔

Punctuation: ), (Slide33

Sentential Logic: Grammar

All

sentence letters are WFFs.

If φ is a WFF, then ~φ is a WFF.

If φ and ψ are WFFs, then (

φ & ψ

), (

φ v ψ

), (

φ → ψ

), (

φ ↔ ψ

) are also WFFs.

Nothing else is a WFF.Slide34

Sentential Logic: Examples

(Q & R)

((Q & R) v P)

~((Q & ~R) v P)

(S → ~((Q & ~R) v P))

~((~P ↔ S) → ~((Q & ~R) v P))Slide35

Sentential Logic: Outside Parentheses

(

Q & R

)

(

(Q & R) v P

)

~((Q & ~R) v P)

(

S → ~((Q & ~R) v P)

)

~((~P ↔ S) → ~((Q & ~R) v P))Slide36

Convention: Omit Outside Parentheses

Q & R

(Q & R) v P

~((Q & ~R) v P)

S → ~((Q & ~R) v P)

~((~P ↔ S) → ~((Q & ~R) v P))Slide37

Note on Negation

Can’t omit parentheses when negation is main connective. Example:

1. ~((Q & ~R) v P)

Is false whenever P is true. But (2) is true whenever P is true:

2. ~(Q & ~R) v PSlide38

Probability TheorySlide39

Probability Theory: Vocabulary

Sentence letters: A, B, C,…

Logical connectives: ~, &, v

Punctuation: ), (

Real numbers between 0 and 1: 0.5, 0.362,

π

/4…

Probability function symbol:

Pr

Equality sign: =Slide40
Slide41

Domain RangeSlide42

Functions

A

function

is a relation between the members of two sets X and Y, called its

domain

and its

range

.

The function takes each member of its domain and relates it to exactly one member of its range.Slide43

Numerical Functions

Most functions that you’ve learned about are numerical functions: their domains are numbers or pairs of numbers, and their range is also numbers. Addition:

+: 2,2 4

+: 16,9 25

+: 9,16 25

+: 0.5,0.25 0.75

+: 7,-18 -11Slide44

Truth-Functions

But we also learned about truth-functions in class too.

~:T F

~:F T

&:T,T T

&:T,F F

&:F,T F

&:F,F FSlide45

Pr

Pr

” is the symbol for a

probability function

. It’s a function from SL formulas to real numbers in [0, 1]. There are lots of probability functions, but they all follow these rules:

Rule 1: 0 ≤

Pr

(

φ

)

1

Rule 2: If φ is a tautology, then Pr(φ) = 1Rule 3: If φ and ψ are mutually exclusive, then Pr(φ v ψ) = Pr(φ) + Pr(ψ)Slide46

Probabilities of Complex Sentences

Just as in Sentential Logic, where you can calculate the truth-values of complex sentences if you know the truth-values of their parts, we can calculate the probabilities of complex sentences from the probabilities of their parts.Slide47

Probability of Negation

If you know the

Pr

(

φ

), you can calculate

Pr

(~

φ

):

Pr

(~

φ

) = 1 – Pr(φ)If the probability that it will rain tomorrow is 20%, then the probability that it will not rain is 1 – 20% = 80%. If the probability that the die will land 4 is 1/6, then the probability that it will not land 4 is 1 – 1/6 = 5/6. Slide48

Probability of Disjunction

There are two ways we need to use to calculate the probability of a disjunction (φ v ψ).

The first way we use if φ and ψ are mutually exclusive: they can’t both be true together. Then the rule is:

Pr

(φ v ψ) =

Pr

(φ) +

Pr

(ψ)Slide49

Examples of Mutually Exclusive Possibilities

When you roll a fair 6-sided die, the probability it will land on any one side is 1/6.

It cannot land on two sides at the same time. Slide50

Examples of Mutually Exclusive Possibilities

Landing 4 on one roll and landing 6 on the same roll are mutually exclusive possibilities.

So the probability that on one roll it will land 4

or

it will land 6 is 1/6 + 1/6 or 2/6.Slide51

Inclusive “Or”

In logic, we use inclusive “or.”

(φ v ψ) can be true when both φ and ψ are true. Thus we must be able to calculate the probabilities disjunctions of events that are not mutually exclusive.

Why doesn’t our old rule work?Slide52

Coin Flips

Suppose I flip a coin twice.

The probability that it will land heads on the

first

flip is 50%.

The probability that it will land heads on the

second

flip is 50%.

What’s the probability it will land heads on the first flip

or

the second flip?Slide53

Coin Flips

Not

this:

Pr

(F v S) =

Pr

(F) +

Pr

(S)

= 50% + 50%

= 100%Slide54

Coin Flips

First

Second

Heads

Heads

Heads

Tails

Tails

Heads

Tails

TailsSlide55

Heads on the First Flip: Two Possibilities

First

Second

Heads

Heads

Heads

Tails

Tails

Heads

Tails

TailsSlide56

Heads on the Second Flip: Two Possibilities

First

Second

Heads

Heads

Heads

Tails

Tails

Heads

Tails

TailsSlide57

Heads on First O

r

Second:

Not

2 + 2

First

Second

Heads

Heads

Heads

Tails

Tails

Heads

TailsTailsSlide58

Rule for Events that Aren’t Mutually Exclusive

In adding the probabilities of the first and second flips, we

double counted

the possibility that the coin lands heads in the first flip

and

lands heads in the second flip.

Rule:

Pr

(φ v ψ) =

Pr

(φ) +

Pr

(ψ) –

Pr(φ & ψ)Slide59

Rule for Events that Aren’t Mutually Exclusive

But how do we calculate this part?

Rule:

Pr

(φ v ψ) =

Pr

(φ) +

Pr

(ψ) –

Pr

(φ & ψ)Slide60

Dependence and Independence

Two events A and B are independent if A happening does not increase the probability that B will happen.

(This is equal to the claim that B happening does not increase the probability that A will happen.)

A and B are independent:

Pr

(A) =

Pr

(A/ B)

Pr

(B) =

Pr

(B/ A)Slide61

Conjunctions: First Rule

If A and B are independent, then

Pr

(

φ

&

ψ

) =

Pr

(

φ

) x

Pr

(ψ) Slide62

Coin Flips

Suppose I flip a coin twice.

The probability that it will land heads on the

first

flip is 50%.

The probability that it will land heads on the

second

flip is 50%.

What’s the probability it will land heads on the first flip

and

the second flip?Slide63

Probability of Independent Events

Coin flips are independent. What happens on one flip does not affect what happens on a different flip.

Thus:

Pr

(F &

S

) =

Pr

(

F

) x

Pr

(

S) = 50% x 50% = 25%Slide64

First and Second: 1 out of 4 (25%)

First

Second

Heads

Heads

Heads

Tails

Tails

Heads

Tails

TailsSlide65

Back to “Or”

Now, returning to our earlier question: What’s the probability the coin will land heads on the first flip

or

the second flip?

Pr

(φ v ψ) =

Pr

(φ) +

Pr

(ψ) –

Pr

(φ & ψ)

= Pr(F) + Pr(S) – Pr(F & S) = Pr(F) + Pr(S) – Pr(F) x Pr(S) = 50% + 50% – 25% = 75%Slide66

Heads on First O

r

Second:

3 out of 4 (75%)

First

Second

Heads

Heads

Heads

Tails

Tails

Heads

Tails

TailsSlide67

Non-Independent Events

We can’t always multiply the probabilities of the conjuncts to get the probability of a conjunction.

For example, the probability of a coin landing heads is 50%:

Pr

(H) = 50%. What’s the probability that it lands heads and lands heads on the same flip? Why, 50% of course. But

Pr

(H & H) = 50% x 50% = 25%...

Wrong!Slide68

Non-Independent Events

In general, if A happening

raises

or

lowers

the probability that B will happen, then we can’t use the standard multiplication rule.Slide69

Even Numbers

The even numbers are the numbers: 2, 4, 6, 8, 10, 12…

On a 6-sided die there are only three even numbers: 2, 4, 6.Slide70

Prime Numbers

The prime numbers are the numbers that are only divisible by themselves and one: 2, 3, 5, 7, 11, 13, 17…

On a 6 sided die, there are four prime numbers: 1, 2, 3, 5Slide71

Rolling Even

The probability of rolling an

even

number is:

Pr

((R2 v R4) v R6)

=

Pr

(R2 v R4) +

Pr

(R6)

=

Pr

(R2) + Pr(R4) + Pr(R6) = 1/6 + 1/6 + 1/6= 1/2Slide72

Rolling Prime

The probability of rolling a

prime

number is:

Pr

((R1 v R2) v (R3 v R5))

=

Pr

(R1 v R2) +

Pr

(R3 v R5)

=

Pr

(R1 v R2) + Pr(R3) + Pr(R5) = Pr(R1) + Pr(R2) + Pr(R3) + Pr(R5)= 1/6 + 1/6 + 1/6 + 1/6= 2/3Slide73

Rolling Prime and

Even

We

can’t

just multiply these probabilities together to get the right result.

Pr

(

P

&

E

) =

Pr

(

P) x Pr(E) = 2/3 x 1/2 = 2/6 = 1/3… Wrong!!!Slide74

Rolling Prime and

Even

Why? Because there is only one even prime number and there is a 1/6 chance you roll it.Slide75

Pr(Prime/ Odd) = 100%

If you roll an odd number this

raises

the probability that you rolled a prime number.

Since every odd number on the die is prime:

Pr

(P/ O) = 1

O = T

P

R1

R1

R2

R2

R3R3R4R4R5

R5

R6

R6Slide76

Pr(Prime/ Even) = 1/3 <

Pr

(Prime) = 2/3

If you roll an even number this

lowers

the probability that you rolled a prime number.

There’s a 2/3 unconditional probability that you’ll roll a prime. But if you roll an even number, there’s only a 1/3 chance it’s prime.

O = T

P

R1

R1

R2

R2

R3R3R4

R4

R5

R5

R6

R6Slide77

General Conjunction Rule

So rolling prime P and rolling even E are not independent.

The rule

Pr

(

φ

&

ψ

) =

Pr

(

φ

) x

Pr(ψ) only works for events that are independent.The general rule, for any two events, is: Pr(φ & ψ) = Pr(φ) x Pr(ψ/ φ)Slide78

Pr(Prime

and

Even)

Pr

(E & P) =

Pr

(E) x

Pr

(P/ E)

= 1/2 x

Pr

(P/E)

= 1/2 x 1/3 = 1/6Pr(P & E) = Pr(P) x Pr(E/P) = 2/3 x Pr(E/P) = 2/3 x 1/4 = 2/12 = 1/6Slide79

Conditional Probability

We’ll talk more about conditional probability next time.Slide80

Rules

Pr

(~

φ

) =

1 –

Pr

(

φ

)

Pr

(φ v ψ) =

Pr(φ) + Pr(ψ) – Pr(φ & ψ)Pr(φ v ψ) = Pr(φ) + Pr(ψ) when φ and ψ are mutually exclusivePr(φ & ψ) = Pr(φ) x Pr(ψ/ φ)Pr(φ & ψ) = Pr(φ

) x

Pr

(

ψ

)

when

φ

and

ψ

are independent

Slide81

Sample ProblemsSlide82

Imagine you have an ordinary deck of 52 playing cards, shuffled and face-down.

If you select one card from the deck, what’s the probability it will be:

The ace of spades?

The four of clubs?Slide83

If you select one card from the deck, what’s the probability it will be:

The ace of spades or the queen of diamonds?

Any ace?

Not an ace?