Show that the postulates of probability are satisfied by conditional probabilities. In other words, show that is P(B) does not equal 0 then:

Question

I did prove all three of the postulates. I was just hoping someone could look them over to see if I proved it correctly? Thank you!

1. P(A|B) >= 0 (This is what I'm proving!)

   P(A|B) = P(AnB)/P(B)

   1>(or equal to) P(B) > 0

   P(AnB) c P(B) therefore, P(AnB) < P(B)

   Then, P(B)> P(AnB) >(or equal to) 0

   So, P(A|B) must be >= 0

2. P(B|B) = 1 (I'm pretty positive I proved this correctly so I'm going to leave the proof out)

3. P(A1 u A2 u ... |B) = P(a1|B) + P(A2|B) + ... for any sequence of mutually exclusive events A1, A2, ... (This is what I'm proving)

   P(A1 u A2 u ... |B) = P[(A1 u A2 u...) n B)]/ P(B)

   I know that P[(A1 u A2 u...)n B] is disjoint

   Then, P(A1 n B) + P(A2 n B) + ...

   so now I have [P(A1 n B) + P(A2 n B) +...] / P(B)

   I know that P(A1 n B)/ P(B) <=> P(A1|B) and P(A2 n B)/ P(B) <=> P(A2|B)

   So, P(A1 u A2 u ... |B) = P(a1|B) + P(A2|B) + ...