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.
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.
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, $P(A|B) + P(A'|B) = \frac{P(A\cap B) + P(A'\cap B)}{P(B)} = \frac{P(B)}{P(B)} = 1.$
Yes, in fact ${\mathbb P}(\cdot | B)$ satisfies all the axioms of probability.