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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)

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# Conditional Probability (confusing)

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