Is Bayes' theorem useless if $A$ is a subset of $B$? Using Bayes' theorem, $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$
wouldn't this just be the percentage of $A$ in $B$ if $A$ is a subset of $B$?
If A is a subset of B:
$$P(B|A) = 1,$$
$$P(A|B) = \frac{P(A)}{P(B)}$$
Therefore Bayes theorem becomes $\frac{P(A)}{P(B)}$ if $A$ is a subset of $B$.
So you don't really need to use Bayes' theorem in these situations, right?
 A: Bayes' Rule is still valid when $A \subset B$, but sometimes the conditional probabilities can be deduced intuitively in these situations. For example, it makes sense that if $A \subset B$, then $P(B|A) = 1$, since if $A$ happens, we know $B$ must have happened - you don't need Bayes Rule to see that. You are also right that in this case $P(A|B)=P(A)/P(B)$, which is just a special case of Bayes Rule, so I'd still say you "need" Bayes Rule in this case.   
To make this more concrete, take an example: suppose you roll a fair die and $B$ is the event that the roll is even, while $A$ is the event the roll is a '6'. Clearly $A \subset B$. $P(B|A) = 1$, since if you rolled a '6', you clearly rolled an even number. Also
$$ P(A|B) = \frac{ P(B|A) \cdot P(A) }{ P(B) } = \frac{P(A)}{P(B)} = \frac{\frac{1}{6}}{ \frac{1}{2} } = \frac{1}{3} $$
This makes intuitive sense - conditioned on the fact that the roll is even, there are three equally likely outcomes - $\{2,4,6\}$ - therefore, the probability of rolling a '6' is $1/3$.  
