Why does the addition of $P(A^c | B^c)$ and $P(A | B^c)$ result in 1 (Why are they complements)?

Question

While watching Harvard's Statistics 110 on YouTube, in the lecture on conditional probability, The professor relates $P(A^c | B^c)$ to $P(A | B^c)$. It is given that $P(A^c | B^c) = 0.95$. The professor says, without explanation, that $P(A|B^C) = 1 - 0.95 = 0.05.$ So I think I'm missing something fundamental. I did some trials to prove it but, unfortunately, came up with nothing.

$$P(A^c | B^c) = \frac{P(A^c \cap B^c)}{P(B^c)} = \frac{P(A^c \cap B^c)}{1- P(B)} = \frac{1 - P(A \cup B)}{1 - P(B)}$$

Not sure, to be honest, about the benefit of those steps, or how can I come up with a relation between the two probabilities.

Answer 1

If $\mathbb P(B^\complement)>0$ then $$ \mathbb P(A^\complement\vert B^\complement)=\frac{\mathbb P(A^\complement\cap B^\complement)}{\mathbb P(B^\complement)}=\frac{\mathbb P(B^\complement)-\mathbb P(A\cap B^\complement)}{\mathbb P(B^\complement)}=1-\frac{\mathbb P(A\cap B^\complement)}{\mathbb P(B^\complement)}=1-\mathbb P(A\vert B^\complement). $$

For the second equality I used that $B^\complement$ is the disjoint union of $A^\complement\cap B^\complement$ and $A\cap B^\complement$, hence $\mathbb P(B^\complement)=\mathbb P(A^\complement\cap B^\complement)+\mathbb P(A\cap B^\complement)$.

Answer 2

These probabilities are conditioned on $B^c$. In that subset of events, the two events are mutually exclusive, and their union is the whole (sub)set of events.

Answer 3

If you know the event $B^c$ then you have either $A$ or $A^c$. Hence the sum of the two probabilities is equal to one.