0

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?

brandon_ducks
  • 117
  • 7

1 Answers1

1

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.

AJKOER
  • 239