# Thread: Proof of axioms of conditional probability

1. ## Proof of axioms of conditional probability

Show that P(A given B) = P(A intercept B) / P(B) implies the following axioms:

1.) P(A given B) = 1, where B is a subset of A

2.) If {Ai intersect B},i = 1 2 … are mutually exclusive, then
P(A1 ∪ A2 ∪… given B)= P(A1 given B)+ P(A2 given B)+…

3.) If B ⊃ H and B ⊃ G and P(G ) ≠ 0 , then
P(G )/P( H ) = P (G given B)/ P(H given B)

2. Originally Posted by trukka
Show that P(A given B) = P(A intercept B) / P(B) implies the following axioms:
1.) P(A given B) = 1, where B is a subset of A
2.) If {Ai intersect B},i = 1 2 … are mutually exclusive, then
P(A1 ∪ A2 ∪… given B)= P(A1 given B)+ P(A2 given B)+…
3.) If B ⊃ H and B ⊃ G and P(G ) ≠ 0 , then
P(G )/P( H ) = P (G given B)/ P(H given B)
For #1, just notice that $B \subseteq A\; \Rightarrow \;B \cap A = B$.

For #2, given that $\left( {A_k \cap B} \right)$ is a collection of pair-wise disjoints events then $P\left[ {\bigcup\limits_k {\left( {A_k \cap B} \right)} } \right] = \sum\limits_k {P\left( {A_k \cap B} \right)}$.
Then note that you have a common denominator of $P(B)$.

#3 is just algebra