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To confirm the formula for probabilities, given that an event has occured, I wonder if it is true that:

$\mathbb{P}(A \mid B)=1-\mathbb{P}(A^{C} \mid B)$

where $\mathbb{P}(A)+\mathbb{P}(A^{C})=1$.

$A$ and $B$ are events.

2 Answers 2

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By definition, as long as $Pr(B)\neq 0$, $$P(A|B) = \frac{P(A\cap B)}{P(B)}$$ and $$P(A'|B) = \frac{P(A'\cap B)}{P(B)}.$$ Adding them together, $$P(A|B) + P(A'|B) = \frac{P(A\cap B) + P(A'\cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1.$$

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    Indeed, one could argue that this _must_ hold, for the system to be logically consistent, even if $P(B)=0$ as long as $P(A|B)$ is defined at all. After all, $P(A|B)+P(A^C|B)=P(A\lor A^C|B)=P(\top|B)=P(\top)=1$, where $\top$ denotes a tautology.2012-12-23
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    @IlmariKaronen How does one define $P(A|B)$ when $P(B) = 0$? Is the Wikipedia entry (http://en.wikipedia.org/wiki/Conditional_probability#Definition_with_.CF.83-algebra) what you have in mind?2012-12-23
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    For example. My point was that, however you define it, it had better satisfy that property if you want it to make any sense, since it would be absurd for $P(A \lor A^C|B)$ to be anything other than $1$ if it's defined at all.2012-12-23
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Yes, in fact ${\mathbb P}(\cdot | B)$ satisfies all the axioms of probability.