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
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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: =Slide40Slide41
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?