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If $p(S|\neg G) = 0$, then $p(\neg G \& R) > 0 \implies p(S\mid \lnot G\& R)=0$.

Could someone please prove this?

Many thanks in advance.

Graham Kemp
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Mijito
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  • If the event $(\neg G)$ prevents the event $S$ from occurring, then the combined events $(\neg G \text{and} R)$ still prevent the event $S$ from occurring. It is as if the event $G$ is a disease that is not cured by the event $R$. – user2661923 May 03 '22 at 11:27
  • An alternate way of responding is that $\neg G$ and $S$ are mutually exclusive events, which implies that $S$ is mutually exclusive with any subset of $\neg G$. – user2661923 May 03 '22 at 11:29
  • Actually, I'm looking for a formal proof of my statement. – Mijito May 03 '22 at 11:43
  • Here, I can't help, because my knowledge of formal Probability Theory is close to nil. – user2661923 May 03 '22 at 11:44

1 Answers1

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Use the definition of conditional probability.

$$\def\amp{\mathop{\&}}\begin{align}\mathsf P(S\mid\neg G\amp R) &=\dfrac{\mathsf P(S\amp(\lnot G\amp R))}{\mathsf P(\lnot G\amp R)}&&\text{by definition of conditional probability}\\&=\dfrac{\mathsf P(\lnot G\amp(S\amp R))}{\mathsf P(\lnot G\amp R)}&&\text{using commutivity and associativity of }\amp\\&~~\vdots&&\text{and so on...}\end{align}$$

Graham Kemp
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