In a bakery there are 3 bakers, and only 2 kinds of cakes: chocolate and cheese.
Let us define events: $A$- Cheese cake, $B_i$- Bake was baked by the $i$-th baker
We are asked to find the odds that the cake was not baked by Baker_$_1$ given the fact it was not a cheese cake, meaning: $$P(\bar{B_1}|\bar{A})=?$$ So I know there's the right, long, set-theory-based answer: $$P(\bar{B_1}|\bar{A})= \frac{P(\bar{B_1}\cap\bar{A})}{P(\bar{A})}=\frac{P(\bar{A}\cap({B_2}\cup{B_3}))}{P(\bar{A})}=\frac{P(\bar{A}\cap B_2)+ P(\bar{A}\cap B_3)}{\sum_{i=1}^{3}{P({\bar A}|B_i)P(B_i)}}$$ The last phase can be justified by saying there's independence of bakers since this is how $\Omega$ is decided, and the denominator is just the complete probability theorem.
Although this is true, my question is whether I can (or why not, if not) do the following: $$P(\bar{B_1}|\bar{A})= 1-P({B_1}|\bar{A})$$ and then find $P({B_1}|\bar{A})$ with Bayes' theorem: $$P({B_1}|\bar{A}) = \frac{P(\bar{A}|{B_1})P(B_1)}{P(\bar{A})}=\frac{P(\bar{A}|{B_1})P(B_1)}{\sum_{i=1}^{3}{P({\bar A}|B_i)P(B_i)}}$$ $$\rightarrow P(\bar{B_1}|\bar{A})= 1-\frac{P(\bar{A}|{B_1})P(B_1)}{\sum_{i=1}^{3}{P({\bar A}|B_i)P(B_i)}}$$
In general I know that: $$P(\bar A|B)= 1-P(A|B)$$ But the real question I'm curious about is whether having the complementary event on both parts of the conditional probability is forcing us to use the first way I presented because some events are missed/repeat.
Any insight on that matter?
Both versions are fine, and lead to the same answer. By going through the complement of $B_1$, you avoid the lengthy individual handling of $B_2$ and $B_3$, which don't play a distinguishable role in the question at hand. This would be even more prominent if it were 10 bakers instead of only 3.
Basically, for the answer you need to consider only $B_1$ and $\bar{B_1}$. Making $\bar{B_1}$ explicit as $\bar{B_1} = B_2 \cup B_3$ and then putting down the set theoretical ramifications is lots of work for no real gain. You do have to consider all bakers in the denominator, but using the sum symbol allows you to write down just one formula instead if putting each summand separately.
You could write the result of the first approach as
$$\frac{\sum_{i=2}^3P(\bar{A}\cap B_i)}{\sum_{i=1}^{3}{P({\bar A}|B_i)P(B_i)}},$$
if you wanted to.