Axioms of Probability

Axioms of probability
Axiom : A proposition so self-evidently TRUE that it doesn't even need to be proven

Axiom 1

\forall A\ \subset \ S\ ,\ \ P\left(A\right)\ \ge \ 0

S: Sample Space
For every A that is a subset of S, the probability that event A happens is greater than 0.

Axiom 2

P\left(S\right)\ =\ 1

The Sample Space S means every outcome of the trial, so S always happens.
So the probability that S happens is 1.

Axiom 3

For every countable sequence of mutually exclusive events Ai

P\left(\bigcup _{i\ =\ 1}^{\infty }{A}_i\right)\ =\ \sum _{i\ =\ 1}^{\infty }P\left({A}_i\right)

If A1, A2, ..., An are all mutually exclusive (disjoint, no overlapping parts)
they're obviously equal.

Proving probability properties with the axioms

※ P(A^c) = 1 - P(A)

For the Sample Space S,
since it can be partitioned as S = A ∪ A^c,
by axiom 3, P(S) = P(A ∪ A^c) = P(A) + P(A^c).
By axiom 2, P(S) = 1, so

P(S) = 1 = P(A) + P(A^c)
P(A^c) = 1 - P(A) holds.

※ if A ⊂ B, P(A) ≤ P(B)

Letting set C be such that C ∩ A = Ø, A ∪ C = B,
by axiom 3, P(B) = P(A ∪ C) = P(A) + P(C).
By axiom 1, P(C) >= 0, so P(B) >= P(A) holds.

※ 0 ≤ P(A) ≤ 1

By axiom 1, P(A) >= 0 holds.
For the Sample Space S, since it can be Partitioned as S = A ∪ A^c,
by axiom 3, P(S) = P(A ∪ A^c) = P(A) + P(A^c), and
by axiom 2, P(S) = 1, and by axiom 1, P(A^c) >= 0, so
P(S) = 1 >= P(A) holds.

So 0 <= P(A) <= 1 holds.

※ P(A ∩ B^c) = P(A) - P(A ∩ B)

Since it can be Partitioned as A = (A ∩ B^c) ∪ (A ∩ B),
by axiom 3, P(A) = P((A ∩ B^c) ∪ (A ∩ B)) = P(A ∩ B^c) + P(A ∩ B), and
rearranging, P(A ∩ B^c) = P(A) - P(A ∩ B) holds.

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