Circular Bayes Theorem?

Question

I came across a seemingly simple Bayes Theorem question, but I am going in circles currently to try to figure it out.

I am given only that $$P(A) = .78$$ $$P(B|A) = .66$$ $$ P(B'|A') = .45$$

I am trying to find P(A'|B'). Obviously the first step is to find P(B).

So far I have shown that $$P(B|A) = \frac{P(B)*P(A|B)}{P(A)} \rightarrow P(B)*P(A|B) = .4752$$ So to find P(B) we need to find P(A|B)

$$P(A|B) = \frac{P(A)*P(B|A)}{P(B)} = \frac{.4752}{P(B)}$$ so this doesnt help at all.

The other way to find P(B) is by $$P(B) = P(A)*P(B|A) + P(A')*P(B|A') $$ but we are not given P(B|A') or P(B'|A) so we can't find either.

How would you go about solving this?

Answer 1

Hint: $P(B'|A) = 1 - P(B|A)$.

Answer 2

Imagine 10000 such occurrences. P(A)= 0.78 so in 7800 of those A occurs. P(B|A)= 0.66 so in .66(780)= 5148 B also occurs. In the other 7800- 5148= 2652 B does not occur.

In 10000- 7800= 2200 A does not occur. P(B',A')= 0.45 so in 2200(0.45)= 900 B also does not occur. In the other 2200- 990= 1232 B does occur.

So in these 10000 occurrences, B does not occur 2652+ 990= 3642 times. Of those A did not occur 2200 times. P(A'|B')= 2200/3642= 0.6040... or about 0.60.