Why does this solution for computing $P(B^c | A)$ work?

Question

I'm scratching my head as to why this solution is correct.

We want to compute $P(B^c|A)$ given the following:

$P(B) = 0.08$

$P(A | B) = P(A^c | B^c) = p$

$p = 0.9$

I figured I'd start with

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

2. $P(A \cap B^c) = P(A|B^c) * P(B^c)$

So I was stuck here. I tried to play around with Bayes and some other identities but I didn't really get anywhere. I looked up the solution and I really don't understand their next step (Continuation of the above):

3. I don't see how they got to this one: $P(B^c \cap A) = P(A | B^c) * P(B^c) = 0.92(1-p)$

4. $P(B \cap A) = P(A | B) * P(B) = 0.08p$

5. then they somehow got to this: $P(A) = 0.92 - 0.84p$

6. which then yields: $P(B^c|A) = \frac {0.92(1-p)}{0.92-0.84p} = 0.561$ (for $p$ = 0.9)

I would really appreciate it if someone could explain why the above mentioned steps work?

Thanks in advance!

Answer 1

So the problems were steps $3$ and $5$?

Step $3$: We have $P(A|B^c)=1-P(A^c|B^c)=1-p$ and $P(B^c)=1-P(B)=0.92$. Hence the product of these probabilities is $0.92(1-p)$.

Step $5$: We already know the probabilities $P(A\cap B)$ and $P(A\cap B^c)$ from the previous steps, now use the equality $P(A)=P(A\cap B)+P(A\cap B^c)$.