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?

One may write
$$
P(A\cap B)=P(A)\cdot P_{A}(B)=\frac25\times\frac14=\frac1{10}.
$$ From
$$
P(A\cup B)=P(A)+P(B)-P(A\cap B)
$$ one gets
$$
P(B)=P(A\cup B)+P(A\cap B)-P(A)=\frac35+\frac1{10}-\frac25=\frac3{10}.
$$ One has
$$
P(A\cap B \cap C)=P(A\cap B) \cdot P_{A\cap B}(C)=\frac1{10} \times \frac12=\frac1{20}
$$ and
$$
P(B\cap C)=P(B)\cdot P_{B}(C)=\frac3{10}\times\frac13=\frac1{10}.
$$ Finally,

$$
P_{B\cap C}(A)=\frac{P(A\cap B \cap C)}{P(B \cap C)}=\frac{\frac1{20}}{\frac1{10}}=\frac12.
$$

