Probabilities of conditionals

P(B|A) = P(not (A and not B))

Short answer: not generally true.

• By definition, [ P(B\mid A)=\frac{P(A\cap B)}{P(A)}=1-\frac{P(A\cap B^{c})}{P(A)}. ] Your right-hand side is (P(\neg(A\land \neg B))=1-P(A\cap B^{c})) without dividing by (P(A)).

• Counterexample (independent (A,B) with (P(A)=P(B)=0.5)): [ P(B\mid A)=0.5,\quad 1-P(A\cap B^{c})=1-(0.5\times 0.5)=0.75. ] Not equal.

• When would your equality hold?

  • If (P(A)=1); or
  • If (A\subseteq B) a.s. (so (P(B\mid A)=1) and (P(\neg(A\land \neg B))=1)).

• Likely fix: [ P(B\mid A)=P\big(\neg(A\land \neg B)\mid A\big)=1-\frac{P(A\cap B^{c})}{P(A)}. ]

Think of what this means for material conditionals

Great prompt. Your RHS, [ P(\neg(A\land \neg B)), ] is exactly (P(A\to B)) if “(\to)” is the material conditional (true except when (A) and not (B)). That gives a clean link:

[ P(A\to B)=1-P(A\land \neg B) =1-P(A)+P(A)P(B\mid A)\quad (P(A)>0). ]

What this means

• Material conditional (\neq) conditional probability. Generally (P(B\mid A)\ne P(A\to B)). They coincide only in special cases (e.g., (P(A)=1) or (A\subseteq B) a.s.).

• Rare-antecedent effect. If (P(A)) is small, then (P(A\to B)\approx 1) no matter how small (P(B\mid A)) is. Example: (P(A)=0.01,\ P(B\mid A)=0.10) ⇒ (P(A\to B)=1-0.01+0.01\cdot0.10=0.991) while (P(B\mid A)=0.10). So material “if” is almost always true when (A) is rare—unlike our intuitive reading of “if”.

• Right “fix” for your equation. The conditional-probability version is: [ P(B\mid A)=P\big(\neg(A\land \neg B)\mid A\big)=P(A\to B\mid A). ]

• Deeper punchline (Lewis/Adams). Trying to treat some proposition (C) so that (P(C)=P(B\mid A)) in general runs into Lewis’s triviality results; there isn’t a standard event whose probability is always the conditional probability. A popular workaround (Adams’ Thesis) is: the assertability of “If (A) then (B)” tracks (P(B\mid A)), not its truth value as a material conditional.

Takeaway: Material “if” gives (P(A\to B)=1-P(A)+P(A)P(B\mid A)), which can be high for trivial reasons. If you want the intuitive strength of “if (A), (B)”, use (P(B\mid A)), not (P(A\to B)).