If A and B are two events such that P(A) > 0 and P(B) ≠ 1, then P(Ā / B̄) is…

Concept: For any events X, Y with P(Y) > 0, the conditional probability is P(X | Y) = P(X ∩ Y) / P(Y). De Morgan's law gives the complement of a union: (A ∪ B)′ = Ā ∩ B̄.

Application to this question:

1. Since P(B) ≠ 1, P(B̄) = 1 − P(B) > 0, so the conditioning event B̄ has positive probability and P(Ā | B̄) is well-defined.

2. By definition, P(Ā | B̄) = P(Ā ∩ B̄) / P(B̄).

3. By De Morgan's law, Ā ∩ B̄ = (A ∪ B)′, so P(Ā ∩ B̄) = 1 − P(A ∪ B).

4. Substituting: P(Ā | B̄) = (1 − P(A ∪ B)) / P(B̄).

Cross-check with numbers: let A, B be independent with P(A) = 0.3, P(B) = 0.4. Then P(A ∪ B) = 0.3 + 0.4 − 0.3×0.4 = 0.58, and P(B̄) = 0.6, so the formula gives (1 − 0.58) / 0.6 = 0.42 / 0.6 = 0.7. Directly, independence of A and B also makes Ā and B̄ independent, so P(Ā | B̄) = P(Ā) = 1 − 0.3 = 0.7 — the two routes agree.

So P(Ā / B̄) = (1 − P(A ∪ B)) / P(B̄).