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Conditional Probability (confusing)

  1. Sep 11, 2008 #1
    Dear all,

    P (A |B) + P (A c|B) = 1 [A c] denotes complement of set A and of course P (B)>0

    Is the above statement true?

    How about the following two:
    P (A |B) + P (A |B c) = 1

    P (C ∪ D|B) = P (C |B) + P (D|B) − P (C ∩ D|B)
  2. jcsd
  3. Sep 11, 2008 #2
    What steps have you taken in proving/disproving these?
  4. Sep 11, 2008 #3
    first statement is true

    P (A |B) + P (A |B c) = 1 is not true

    if P(A) = x and A is independent of B, then P(A|B) = P(A|Bc) = x

    i think the third statement is true
  5. Sep 12, 2008 #4
    [tex]Pr(A|B)=\frac{Pr(A \cap B)}{Pr(B)} \quad Pr(B)>0[/tex]
    [tex]Pr(A \cup B)=Pr(A)+Pr(B)-Pr(A \cap B)[/tex]
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