A conditionally independent problem

Let event A be there is no homework for Paul and even B be the weather is nice. Assume A and B are independent.

Now let event C be play baseball. And Paul only plays baseball when there is no homework and the weather is nice.

1. How to express C in term of A and B?
2. Given C, what is the probability of B?
3. Given $\bar{C}$, what is the probability of B?
4. Are A and B conditionally independent given C?
5. Are A and B conditionally independent given $\bar{C}$

Here is my thinking.

1. C should be $A \times B$
2. It should be 1 but I am not sure how to calculate it.
3. $P(B|\bar{C}) = \dfrac{P(B)-P(A)P(B)}{1-P(A)P(B)}$
4. Yes.
5. No.
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In question 1 I think the expected answer is is $A\cap B$, at least in the standard formalism where events are subsets of the underlying sample space. Which formalism would make $A\times B$ a meaningful answer? – Henning Makholm Oct 20 '11 at 0:01
You seem to be thinking that Paul plays baseball precisely when there is no homework and the weather is nice. That is not what "only ... when" means – Henry Oct 20 '11 at 0:10
@Henry: I don’t like the wording, but I think that the when and only when interpretation must be intended: otherwise parts of the question are impossible. – Brian M. Scott Oct 20 '11 at 0:41

1. Presumably you are expected to say $C=A\cap B$, as Henning Makholm suggested
2. Since, whenever $C$ happens, $B$ happens, you can say $\Pr(B|C)=1$.
3. $\Pr(B|\bar{C}) = \dfrac{\Pr(B)-\Pr(A)\Pr(B)}{1-\Pr(A)\Pr(B)}$ is certainly one possible way of writing the probability of $B$ but not $C$ divided by the probability of not $C$, i.e. the probability of $B$ but not $A$ divided by the probability of not $A$ and $B$ together).
4. $\Pr(A|B,C)=\Pr(A|C)=1$. Similarly $\Pr(B|A,C)=\Pr(B|C)=1$. So yes.
5. $\Pr(B|A,\bar{C}) = 0$ which is not the same as the answer to 3, so no.