# Conditional Probabilities Complementary Proof

1. Dec 3, 2007

### rbzima

I'm having trouble seeing how this works out. It's blatantly obvious that this is true, but somehow I can't seem to get anywhere on paper with it to simplify it down to anything. Any help would be greatly appreciated!

$$P\left(A\right|B)=1-P\left(not A\right|B)$$

2. Dec 4, 2007

### EnumaElish

P(A|B) = P(A&B)/P(B)
P(~A|B) = P(~A&B)/P(B)

What is P(A&B) + P(~A&B) = ?