# Probability Theory. Validity 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.

## Presentation on theme: "Probability Theory. Validity 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."— Presentation transcript:

Probability Theory

Validity

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? PatronAgeDrink AliceOver 18Beer BillOver 18Soda CecilUnder 18Beer DougUnder 18Soda

Ignore People over 18 PatronAgeDrink AliceOver 18Beer BillOver 18Soda CecilUnder 18Beer DougUnder 18Soda

Make Sure No One Else Has Alcohol PatronAgeDrink AliceOver 18Beer BillOver 18Soda CecilUnder 18Beer DougUnder 18Soda

Illegal! PatronAgeDrink AliceOver 18Beer BillOver 18Soda CecilUnder 18Beer DougUnder 18Soda

Legal or Illegal? PatronAgeDrink EvanOver 18Beer FredOver 18Soda GaleUnder 18Soda HarrietUnder 18Soda

Ignore People over 18 PatronAgeDrink EvanOver 18Beer FredOver 18Soda GaleUnder 18Soda HarrietUnder 18Soda

Make Sure No One Else Has Alcohol PatronAgeDrink EvanOver 18Beer FredOver 18Soda GaleUnder 18Soda HarrietUnder 18Soda

Legal or Illegal? PatronAgeDrink IvanOver 18Beer JanetOver 18Soda KerryOver 18Soda LennyOver 18Soda

Ignore Everyone PatronAgeDrink IvanOver 18Beer JanetOver 18Soda KerryOver 18Soda LennyOver 18Soda

Legal! No One Underage Has Alcohol PatronAgeDrink IvanOver 18Beer JanetOver 18Soda KerryOver 18Soda LennyOver 18Soda

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.

Valid or Fails the Test? PossibilityPremiseConclusion Possibility 1True Possibility 2TrueFalse Possibility 3FalseTrue Possibility 4False

Ignore the Possibilities Where Premise Is Not True PossibilityPremiseConclusion Possibility 1True Possibility 2TrueFalse Possibility 3FalseTrue Possibility 4False

Make Sure No Remaining Possibilities Make the Conclusion False PossibilityPremiseConclusion Possibility 1True Possibility 2TrueFalse Possibility 3FalseTrue Possibility 4False

Make Sure No Remaining Possibilities Make the Conclusion False PossibilityPremiseConclusion Possibility 1True Possibility 2TrueFalse Possibility 3FalseTrue Possibility 4False

Example PQ~~P((Q → P) → Q) TTTT TFTF FTFT FFFF

Example PQ~~P((Q → P) → Q) TTTT TFTF FTFT FFFF

Example PQ~~P((Q → P) → Q) TTTT TFTF FTFT FFFF

Recap

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.

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.

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

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.

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.

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.

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.

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.

The Elements of Sentential Logic

Sentential Logic: Vocabulary Sentence letters: A, B, C,… Logical connectives: ~, &, v, →, ↔ Punctuation: ), (

Sentential Logic: Grammar i.All sentence letters are WFFs. ii.If φ is a WFF, then ~φ is a WFF. iii.If φ and ψ are WFFs, then (φ & ψ), (φ v ψ), (φ → ψ), (φ ↔ ψ) are also WFFs. iv.Nothing else is a WFF.

Sentential Logic: Examples (Q & R) ((Q & R) v P) ~((Q & ~R) v P) (S → ~((Q & ~R) v P)) ~((~P ↔ S) → ~((Q & ~R) v P))

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))

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))

Note on Negation Can’t omit parentheses when negation is main connective. Example: 1. ~((Q & ~R) v P) (1)Is false whenever P is true. But (2) is true whenever P is true: 2. ~(Q & ~R) v P

Probability Theory

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: =

Domain Range

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.

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 -11

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 F

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) = 1 Rule 3: If φ and ψ are mutually exclusive, then Pr(φ v ψ) = Pr(φ) + Pr(ψ)

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.

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.

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)

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.

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.

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?

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?

Coin Flips Not this: Pr(F v S) = Pr(F) + Pr(S) = 50% + 50% = 100%

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)

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)

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)

Conjunctions: First Rule If A and B are independent, then Pr(φ & ψ) = Pr(φ) x Pr(ψ)

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?

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%

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

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%

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!

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.

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.

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, 5

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 = 1/2

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 = 2/3

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!!!

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

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 = TP R1 R2 R3 R4 R5 R6

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 = TP R1 R2 R3 R4 R5 R6

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)

Pr(Prime and Even) Pr(E & P) = Pr(E) x Pr(P/ E) = 1/2 x Pr(P/E) = 1/2 x 1/3 = 1/6 Pr(P & E) = Pr(P) x Pr(E/P) = 2/3 x Pr(E/P) = 2/3 x 1/4 = 2/12 = 1/6

Conditional Probability We’ll talk more about conditional probability next time.

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 exclusive Pr(φ & ψ) = Pr(φ) x Pr(ψ/ φ) Pr(φ & ψ) = Pr(φ) x Pr(ψ) when φ and ψ are independent

Sample Problems

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?

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?

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