I'm given:$$x = .95 = P(A|B) = P(A^C|B^C)$$ $$P(B)=.05 \space \text{and} \space P(B^C)=.95$$
I want to find $\bf{P(B^C|A)}$
I know that: $$P(B^C|A) = \frac{P(B^C\cap A)}{P(A)}$$
I can find $P(A)$, but not sure what I can do do get me $P(B^C\cap A)$.
Is $P(A|B^C)=1-P(A^C|B^C)$ a valid argument?
Yes, apparently the complement rule holds for conditional probabilities as was previously noted on this very forum by Graham Kemp, to quote the proof:
$$\begin{align}\Pr(B) & = \Pr((A\cap B) \cup (A^\prime\cap B)) & \text{by total probability law} \\ & = \Pr(A\cap B)+\Pr(A^\prime\cap B) & \text{because of mutual exclusion} \\ \implies \Pr(A\cap B) & = \Pr(B)-\Pr(A^\prime\cap B) & \text{by rearangement}\\\therefore \Pr(A\mid B)&=1 - \Pr(A^\prime\mid B) & \text{by division by }\Pr(B)\end{align}$$
where the first step was also noted above by Michael.
