Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional Probabilities Complementary Proof

  1. Dec 3, 2007 #1
    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!

    [tex]P\left(A\right|B)=1-P\left(not A\right|B)[/tex]
  2. jcsd
  3. Dec 4, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

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

    What is P(A&B) + P(~A&B) = ?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook