Conditional Probabilities with Given Data I posted this question the other day but I believe it got deleted?
In any case, I also wasn't very specific with my notation and I apologize for that! Here is the question again. 
Given event A, B, and C, where B and C are independent events...(Note A occurs because B or C occurs...i.e. A is a power shortage and B is lightning storm and C is hurricane)
Express the following quantities in terms of the following:
$P(C),  P(B), P(A|CB), P(A|CB^c), P(A|C^cB), P(A|C^cB^c)$
(1) $P(B|AC)$
(2) $P(B|A)$
My Attempt:
(1) $P(B|AC) = \frac{P(BAC)}{P(AC)}$
where $P(BAC) = P(B)P(A|B)P(C|BA)$
and $P(A|B) = \frac{P(AB)}{P(B)}$
$P(BA) = P(B|A)P(A)$
$P(C|BA) = \frac{P(V|AC)P(C|A)}{P(B|A)}$
I wasn't sure what to do after this...
(2) $P(B|A) = \frac{P(BA)}{P(A)}$
I don't know how to change P(BA) into the information that we have. 
Thank's so much for any help! And if anyone has little cheats or mnemonics to remember how different conditional probabilities are related to other conditional probabilities, that would be really awesome!!
Thanks again for any help. 
 A: $(1)$
\begin{eqnarray*}
P(B \mid AC) &=& \dfrac{P(ABC)}{P(AC)}. \\
&& \\
P(ABC) &=& P(A \mid BC)\,P(B\mid C)\,P(C) \qquad\text{(by chain rule)} \\
&=&  P(A \mid BC)\,P(B)\,P(C) \qquad\text{(by independence)}. \\
&& \\
P(AC) &=& P(A\mid C)\,P(C) \\
&=& P(C)\left[P(A\mid CB)P(B\mid C) + P(A\mid CB^c)(1 - P(B\mid C))\right] \\
&&\qquad\qquad\text{(by law of total probability)} \\
&=& P(C)\left[P(A\mid CB)P(B) + P(A\mid CB^c)(1 - P(B))\right] \\
&&\qquad\qquad\text{(by independence)}. \\
&& \\
\therefore P(B \mid AC) &=& \dfrac{P(A \mid BC)\,P(B)}{P(A\mid CB)P(B) + P(A\mid CB^c)(1 - P(B))}. 
\end{eqnarray*}
$(2)$
\begin{eqnarray*}
P(B\vert A) &=& \dfrac{P(A\mid B)P(B)}{P(A)} \qquad\text{(by Bayes Thm).} \\
&& \\
P(A\vert B) &=& P(A\mid CB)\,P(C\mid B) + P(A\mid C^cB)\,(1 - P(C\mid B)) \qquad\text{(law of total prob.)}\\
&=& P(A\mid CB)\,P(C) +  P(A\mid C^cB)\,(1 - P(C)) \qquad\text{(by independence)}. \\
&& \\
P(A) &=& P(A\mid CB)\,P(CB) + P(A\mid C^cB)\,P(C^cB) \\
&&\qquad + P(A\mid CB^c)\,P(CB^c) + P(A\mid C^cB^c)\,P(C^cB^c) \\
&& \\
&=& P(A\vert CB)\,P(C)P(B) + P(A\mid C^cB)\,(1-P(C))P(B) \\
&& \qquad + P(A\mid CB^c)\,P(C)(1-P(B)) + P(A\mid C^cB^c)\,(1-P(C))(1-P(B)). \\
&& \\
\therefore P(B\vert A) &=& \frac{\left[ P(A\mid CB)P(C) +  P(A\mid C^cB)(1 - P(C))\right]P(B)}{\mathrm{P(A\vert CB)P(C)P(B) + P(A\vert C^cB)(1-P(C))P(B) + P(A\vert CB^c)P(C)(1-P(B)) + P(A\vert C^cB^c)(1-P(C))(1-P(B))}} 
\end{eqnarray*}
