Thread: 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)

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 .

For #2, given that is a collection of pair-wise disjoints events then .
Then note that you have a common denominator of .

#3 is just algebra