Elementary Question on Bayesian Probability

Question

$$Pr(\neg A|B) = Pr(A|\neg B)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$ $$Pr(\neg A|\neg B) = Pr(A|B)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$

Are the above two identities correct?

The situation in which I discovered them, they hold true for all problems of a particular type (the domain of problems with which I'm concerned).

I want to know if it is true in general.

Answer 1

$$\Bbb P(\neg A|B)=\frac{\Bbb P(\neg A\cap B)}{\Bbb P(B)}=\frac{\Bbb P(B|\neg A)\Bbb P(\neg A)}{\Bbb P(B)}=\frac{\Bbb P(\neg A)}{\Bbb P(B)}\cdot\Bbb P(B|\neg A)$$ So the first one is true if and only if $\Bbb P(\neg A)=\Bbb P(B)$.

$$\Bbb P(\neg A|\neg B)=\frac{\Bbb P(\neg A \cap {\neg B)}}{\Bbb P(\neg B)}=\frac{1-\Bbb P(A\cup B)}{\Bbb P(\neg B)}=\frac{1-\Bbb P(A)-\Bbb P(B)+\Bbb P(A\cap B)}{\Bbb P(\neg B)}\\=\frac{1-\Bbb P(A)-\Bbb P(B)}{\Bbb P(\neg B)}+\frac{\Bbb P(B)}{\Bbb P(\neg B)}\cdot\Bbb P(A|B)$$ So again this is only true in some circumstances where these other terms go away.

Answer 2

$p=P(A\mid B)$ and $1-p =P(\bar A\mid B)$ as well as $q=P(A\mid \bar B)$ and $1-q =P(\bar A\mid \bar B)$. All these values can be distinct.