Bayes Theorem problem, from Finan #9.4: $P(A\mid B ∩ C)$ The Problem:
You are given $\Pr(A) = 2/5, \Pr(A ∪ B) = 3/5, \Pr(B\mid A) = 1/4, \Pr(C\mid B) = 1/3,$ and
$\Pr(C\mid A ∩ B) = 1/2$. Find $\Pr(A\mid B ∩ C)$.
My work:  
I know that $\Pr(A\mid B) \Pr(B) = \Pr(B\mid A) \Pr(A) = \Pr(A \cap B)$ .     
To solve the problem I need to solve   $\Pr(A\mid B ∩ C) = \frac{\Pr(A \cap B \cap C) }{\Pr(B \cap C)}$  
I worked out that $\Pr(B \cap C) = \Pr(C\mid B) \Pr(B) = 1/3 \cdot \Pr(B) = \Pr(B\mid C)\Pr(C)$
So, now I want to find either $\Pr(B)$ or $\Pr(B\mid C)$ & $\Pr(C)$.    
Then, I found that $\Pr(A \cap B \cap C) = 1/20$ by working out that $\Pr(C\mid A \cap B) = \frac{\Pr(A \cap B \cap C) }{\Pr(A \cap B)} = 1/2$  
I've been trying to go further than this, but end up with no new information. I'm not sure how knowing $\Pr(A ∪ B) = 3/5$ would help. Any suggestions?

From the given info, you can find $P(A \cap B)=\dfrac{1}{10}$ and hence $P(B)=P(A \cap B)+P(A \cup B)-P(A)=\dfrac{2}{10}$. Now you can find $P(B \cap C)$.
