Let A, B, C ⊆ Ω such that 0 < P(B ∩ C) < P(B). Prove that 0 < P(B ∩ C^c) < P(B) and that: P(A|B) = P(A|B ∩ C)P(C|B) + P(
Posted: Mon Jul 11, 2022 11:43 am
Let A, B, C ⊆ Ω such that 0 < P(B ∩ C) < P(B). Provethat 0 < P(B ∩ C^c) < P(B) and that:
P(A|B) = P(A|B ∩ C)P(C|B) + P(A|B ∩ C^c) P(C^c|B)
P(A|B) = P(A|B ∩ C)P(C|B) + P(A|B ∩ C^c) P(C^c|B)