Chapter 6: The Mathematics of Probability Flashcards
❮ i,j ❯
Ordered Pair (Probability Calculus)
i = outcome of first roll
j = outcome of second roll.
Example: Ordered pairs ❮ 1,4 ❯ and ❮ 4,1 ❯ are __________ different ___________.
two; outcomes
“S”
Represents all possible outcomes in a random experiment.
“Sample Space”
“Sample Space” must be either _____________ or ____________ _____________.
finite; countably infinite
Event A
Subset of the Sample Space (S).
An Event can be ___ and ___, ____ or ___, or _____-___ and _____-___.
A and B; A or B; not-A and not-B
Aᶜ
Complement of Event A
The complement of Event A is all other Events in the Sample Space.
Sᶜ
Complement of Sample Space (S)
Sᶜ = ∅ (where “∅” is a null sign, meaning “empty set”).
The complement of a Sample Space (S) is an empty set.
p
p = Probability Measure
Takes an Event as it’s argument, and assigns a real number 0-1 as the probability of that event.
Probability Calculus (Set of 3 Axioms)
1) Probability of an Event is a real number between 0 and 1.
2) Probability of Sample Space (S) is 1.
3) If two Events are mutually exclusive (meaning both can’t happen simoultaneously), then the probability of one Event happening equals the probability of the first plus that of the second.
Mutually exclusive
both events can’t happen at the same time.
Probability Calculus (Formalized)
1) p(A) = 1≥p(A)≥0
2) p(S) = 1
3) If A ᑎ B = ∅, then p(A ᑌ B) = p(A) + p(B)
A ᑎ B
“A intersection B”
Only elements that are in both A and B
When A ᑎ B = ∅, that means A and B don’t have any intersecting elements, hence they’re mutually exclusive
A ᑌ B
“A union B”
Contains all elements by A or B, or both.
When representing the Probability Calculus in _______________ ___________, A and B are no longer “___________,” they become “_________________.”
propositional logic; Events; Propositions
If Propositions ____ and ___ are ______________-______________ (meaning both can’t occur at the same time such that ___ ᐱ ___ = ____, AND that both ___ and ___ don’t have any __________ in common), then only one proposition must be _________, and the other must be _________.
A; B; mutually exclusive; A; B; ∅; A; B; elements; true; false
Probability Calculus (Propositional Logic)
1) 0 ≥ p(A) ≥ 1
2) If proposition A is logically true (such that it’s impossible for A to be false), then p(A)=1.
3) If A ᐱ B = ∅, then p(A ᐯ B) = p(A) + p(B)
The difference between 1st Set of Probability Calculus and Propositional Logic set of Probability Calculus
Probability calculus (on it’s own) assigns probabilities to events, THINGS THAT EXIST IN THE EXTERNAL WORLD.
Probability Calculus, formalized in propositional logic, assings probabilities to LINGUISTIC ENTITIES, aka. propositions.
Theorem 6.1: p(A) + p(￢A) = 1
(Give steps on how to reach this answer)
p(A) + p(￢A) = 1
- A ᐯ ￢A is Logical Truth.
- Axiom 2 tells us, “If proposition A is a logical truth, then p(A) = 1
- Therefore, p(A ᐯ ￢A) = 1
- A and ￢A are Mutually Exclusive.
- Axiom 3 tells us, “If A ᐱ B = ∅, then p(A ᐯ B) = p(A) + p(B).”
-Therefore, p(A ᐯ ￢A) = p(A) + p(￢A)
Taken together, p(A) + p(￢A) = p(A ᐯ ￢A) = 1
Theorem 6.2: If A and B are logically equivalent, then p(A) = p(B)
(Give steps on how to reach this answer)
If A and B are logically equivalent, meaning they the same TRUTH VALUE,
then A and ￢B are mutually exclusive.
- Axiom 3: If A ᐱ B = ∅, then p(A ᐯ B) = p(A) + p(B).
- So it follows, that p(A ᐯ ￢B) = p(A) + p(￢B)
The compound sentence A ᐯ ￢B is true for some logical reasons (it’s Logically True).
- Axiom 2: If A is Logically True, then p(A) =1
- So it follows, that p(A ᐯ ￢B) = 1
Taken together, if p(A ᐯ ￢B) = 1, and p(A ᐯ ￢B) = p(A) + p(￢B), then p(A) + p(￢B) = 1.
Were not done yet though! Now we need to refer back to Theorem 6.1: p(A) + (￢A) = 1.
Plugging this theorem in for our sentence ￢B, we get p(B) + p(￢B) = 1
p(A) + p(￢B) = 1
p(B) + p(￢B) = 1
In both of these equations, we yield the same value for ￢B (1), so it follows that p(A) = p(B)
Theorem 6.3 i) p(A v B) = p(A) + p(B) - p(A ∧ B)
Theorem 6.3 ii) p(A → B) = p(￢A) + p(B) - p(￢A ∧ B)
Proof pt. i)
- p(A v B) and p(￢A ∧ B) v p(A ∧ ￢B) v p(A ∧ B) are logically equivalent.
- Great! Axiom 2: If proposition A is logically equivalent, then p(A)=1
- However, p(￢A ∧ B) and p(A ∧ ￢B) v p(A v B) are mutuallly exclusive.
- Okay then, Axiom 3: If A ∧ B = ∅, then p(A v B) = p(A) + p(B)
- p(A v B) = p( (p(￢A ∧ B) ) v (p(A ∧ ￢B) v p(A v B) ) = p(￢A ∧ B) + [p(A ∧ ￢B) v p(A v B)]
- *You can simplify the problem, the rightmost equations are equivalent with A.
- p(A v B) = p(￢A ∧ B) + p(A)
- *Now we need p(B) to complete equation at it’s stated in the 3rd axiom (p(A v B) = p(A) + p(B)), and we need to eliminate p(￢A ∧ B). Why don’t we try adding p(A ∧ B) to both sides!
- p(A v B) + p(A ∧ B) = p(￢A ∧ B) + p(A) + p(A ∧ B)
- *You can simplify the problem, ￢A ∧ B and A ∧ B are equivalent w/ B.
- p(A v B) + p(A ∧ B) = p(B) + p(A)
- p(A v B) = p(B) + p(A) - p(A ∧ B)
Proof pt. ii)
- Same as Proof pt. i). Just be sure to substitute A w/ ￢A such that…
p(￢A v B) = p(B) + p(￢A) - p(￢A ∧ B)
Rule for Conjunctive Probability with Independent Conjuncts
p(A Ʌ B) = p(A) × p(B)
In most decision problems, ______________ is _______________.
probability; conditional
p(A|B) (First Definition of A|B )
p(A|B) = p(A ∧ B) / p(B)
