# User:Saul/probability

## Notations

Ω - The outcome space.
ω - An outcome.
- A non event.
- Union.
- Intersection.
- Is in.
- Is not in.
- Subset of.
- Subset or equal to.

Note: there is a small difference between & but sometimes they get used interchangeably.

- Is not a subset of.
- Superset of.
- Is not a superset of.
ℙ(A) - A set function of set A - Probability of A.
Ac - A compliment - The event A does not occur.
ω∈A - The outcome ω is in the event A
A∪B - The union of A and B - The set containing all the elements from A and B without duplicates.
A∩B - The intersection of A and B - The set containing all the common elements from A and B.
A∩B = ∅ - The sets A and B are disjoint.
ℙ(A∪B) = ℙ(A) + ℙ(B) - The sets A and B are disjoint.
#A - The number of elements A contains if A is finite.
ℙ(S) = 1 - The set S is a exhaustive set as it contains all the outcomes.
Possible outcome - may have a probability of 0.

## Axioms

ℙ(A) ≥ 0 for all events A.
ℙ(Ω) = 1
(Countable additivity) For an infinite sequence of mutually exclusive events{A1,A2,A3,...}

### Deducible Properties

ℙ(∅) = 0 Since Ω ∪ ∅ ∪ ∅ ∪ ... = Ω
ℙ(Ac) = 1 - ℙ(A) Since A ∪ Ac = Ω
If A ⊆ B Then ℙ(A) ≤ ℙ(B) Since A∪(Ac ∩ B) = B
ℙ(A) ≤ 1 Since A ⊆ Ω
ℙ(A ∪ B) = ℙ(A) + ℙ(B) - ℙ(A ∩ B)

## Conditional Probability

ℙ(A | H) - Given H occurs what is the probability of A.
ℙ(A | H) = ℙ(A ∩ H) / ℙ(H)

If ℙ(H) > 0

For example:
Two dice are rolled the outcomes being i and j.
let:

A = {(i, j) : |i - j| ≤ 1}
Note {} is the set, (i, j) is all the possible outcomes and the ":" means the set of those outcomes that satisfy the following condition.

and:

H = {(i, j) : i + j = 7}

We could start by listing all the possible outcomes that are in set 'H:

i j
1 6
2 5
3 4
4 3
5 2
6 1

And all the outcomes in set A

i j
1 1
1 2
2 1
2 2
2 3
3 2
3 3
3 4
4 3
4 4
4 5
5 4
5 5
5 6
6 5
6 6

We can then find all the outcomes in A ∩ H (We could have just applied set H's condition straight to A):

i j
3 4
4 3

We can say there are 6 * 6 = 36 total possible outcomes (Ω).

Set H has 6 / 36 of these outcomes.
Set A has 16 / 36 of these outcomes.
Set A ∩ H has 2 / 36 of these outcomes.

We can now calculate the probability of A happening given H has happened.

ℙ(A | H) = ℙ(2/36) / ℙ(6/36) = 1 / 3

And the probability of H happening given A has happened.

ℙ(H | A) = ℙ(2/36) / ℙ(16/36) = 1 / 8

### Positive/Negative Relationships

A positive' relationship means that given H has happened the chance of A has increased.
If this is true it also means the reverse is true: given A, H has an increased chance of happening.

ℙ(A | H) > ℙ(A)

Which also means:

ℙ(H | A) > ℙ(H)

The same goes for negative relationships, however it is flipped: given H, A has an decreased chance of happening.

ℙ(A | H) < ℙ(A)

Which also means:

ℙ(H | A) < ℙ(H)

### Independent Events

If event H happening does not effect the chance of A then the two sets are said to be independent.
If:

ℙ(A | H) = ℙ(A)

And/Or:

ℙ(A ∩ H) = ℙ(A) * ℙ(H)

Then the two events(A and H) are independent.