If the $P(A \mid B) = 1$ does the $P(B^{c} \mid A^{c}) = 1$? If I think about this as B as a subset of A this makes sense?
But i'm not sure how to prove this.
Any tips to get started would be appreciated.
 A: 
If I think about this as B as a subset of A this makes sense? 

Yes. At least when $\mathsf P(A)>0$, then $\mathsf P(B\mid A)=1\iff \mathsf P(B\cap A)=\mathsf P(A)$ .
Also, use the rule of total probability$$\mathsf P(A)=\mathsf P(A\cap B)+\mathsf P(A\cap B^\complement)\\\mathsf P(B^\complement)=\mathsf P(A^\complement\cap B^\complement)+\mathsf P(A\cap B^\complement)$$
Now, assuming $\mathsf P(B^\complement)>0$, evaluate $\mathsf P(A^\complement\mid B^\complement)$ ...
A: Defining $P(A\mid B)$ implies that $P(B) \neq 0$, so we have $ P( A \cap B ) = P(A \mid B) P(B) = P(B) $. Then
$$ \begin{align}
P(A^c \cap B^c) &= 1-P((A^c \cap B^c)^c) \\
&= 1-P(A \cup B) \\
&= 1- P(A) - P(B) + P(A \cap B) \\
&= 1-P(A) = P(A^c) ,
\end{align} $$
and hence, provided that $P(A^c) \neq 0$, $P(B^c \mid A^c) = P(A^c \cap B^c)/P(A^c) = 1$.
A: $\mathbb P(A|B) = \frac{\mathbb P(A \cap B)}{\mathbb P(B)}$.
If it is equal to $1$, then $\mathbb P(A \cap B) = \mathbb P(B)$.
Then $\mathbb P(B^c | A^c) = \frac{1 - \mathbb P(A \cup B)}{1-\mathbb P(A)}$.
But $\mathbb P(A \cup B) = \mathbb P(A) + \mathbb P(B) - \mathbb P(A \cap B) = \mathbb P(A)$.
So yep, provided that $\mathbb P(B), \mathbb P(A^c) > 0$, then our result holds
