if $\mathbb{P}(A) = \mathbb{P}(A \mid B)$, prove $\mathbb{P}(A^c) = \mathbb{P}(A^c \mid B)$ if $\mathbb{P}(A) = \mathbb{P}(A \mid B)$ prove $\mathbb{P}(A^c) = \mathbb{P}(A^c \mid B)$
my try:

$$\mathbb{P}(A) = \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$$
$$\mathbb{P}(A^c) = \mathbb{P}(A^c \mid B) = \frac{\mathbb{P}(A^c \cap B)}{\mathbb{P}(B)}$$
$$\frac{\mathbb{P}(B\setminus A)}{\mathbb{P}(A^c)} = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(A)}$$

and I'm stuck. Tried replacing $\mathbb{P}(A^c)$ with $1-\mathbb{P}(A)$ to no avail.
 A: $P(A|B)+ P(A^c|B)=1$. Thus, you take the first expression and subtract from 1 gives the second expression. Even for conditional probabilities, $P(A|B)+P(C|B) = P(A\cup C|B)+P(A\cap C|B)$. All the standard laws hold with $B$ in the conditioning. 
Other way to see $P(A|B)+ P(A^c|B)=1$ is that this is same as $P(A|B)+ P(A^c|B) = \frac{P(A\cap B)+ P(A^c \cap B)}{P(B)} = P(B)/P(B)=1$. (Since $P(A\cap B)+ P(A^c \cap B) = P(B)$ since the set union is $B$ with no intersection. )
$P(A)= P(A|B)$, and thus $1-P(A)= 1-P(A|B)$ which further gives $P(A^c)= P(A^c|B)$.
A: $$P(A^c|B)=\frac{P(A^c\cap B)}{P(B)}=\frac{P(B-A)}{P(B)}=\frac{P(B)-P(A\cap B)}{P(B)}=1-P(A|B)=1-P(A)$$
A: Characteristic for $P(A\mid B)$ is the equality:$$P(A\cap B)=P(A\mid B)P(B)$$
This shows that the following statements are equivalent: 


*

*$P(A\mid B)=P(A)$

*$P(A\cap B)=P(A)P(B)$


In words: the events $A$ and $B$ are independent.
If this is the case then: $$P(A^c\cap B)=P(B)-P(A\cap B)=P(B)-P(A)P(B)=P(A^c)P(B)$$ or (again) equivalently: $$P(A^c\mid B)=P(A^c)$$
In words: the events $A^c$ and $B$ are (also) independent.
