Let A and B be two independent events. Show that $p(C|A)=p(B)\cdot p[C|(A\cap B)]+p(\overline{B})\cdot p[C|(A\cap\overline{B})]$ 
Let A and B be two independent events. With $p(A)>0$ and $p(B)>0$,
  show that 
$$p(C|A)=p(B)\cdot p[C|(A\cap B)]+p(\overline{B})\cdot
 p[C|(A\cap\overline{B})]$$

I tried:
$$\frac{p(C\cap A)}{p(A)}=p(B)\cdot \frac{p(C\cap (A\cap B))}{p(A\cap B)}+p(\overline{B})(\frac{p(C\cap(A\cap \overline{B}))}{p(A\cap\overline{B})}) \Leftrightarrow \\ \Leftrightarrow \\
\frac{p(C\cap A)}{p(A)} = \frac{p(C\cap (A\cap B))}{p(A)}+p(\overline{B})\frac{p(C\cap(A\cap \overline{B}))}{p(A\cap \overline{B})} \Leftrightarrow \\
\frac{p(C\cap A)}{p(A)}=\frac{p(C\cap (A\cap B))}{p(A)}+p(\overline{B})\frac{p(C\cap(A\cap \overline{B}))}{p(A\cap \overline{B})} \Leftrightarrow \\
\frac{p(C\cap A)}{p(A)}=\frac{p(C\cap (A\cap B))}{p(A)}+p(\overline{B})\frac{p(C\cap (A \cap \overline{B}))}{p(A|\overline{B})\cdot p(\overline{B})}\Leftrightarrow \\
\frac{p(C\cap A)}{p(A)}=\frac{p(C\cap (A\cap B))}{p(A)}+\frac{p(C\cap (A \cap \overline{B}))}{p(A|\overline{B})}\Leftrightarrow \\ ???$$
What do i do next?
 A: Notice that $P(CA) = P(CAB)+P(CA\bar B)$ by the law of total probability and that $P(A|\bar B) = P(A)$ by independence. Then in your last line
\begin{align*}
\frac{P(C\cap (A\cap B))}{P(A)}+\frac{P(C\cap (A\cap \bar B))}{P(A|\bar B)} 
&= \frac{P(C\cap (A\cap B))}{P(A)}+\frac{P(C\cap (A\cap \bar B))}{P(A)} \\
&= \frac{P(C\cap (A\cap B))+P(C\cap (A\cap \bar B))}{P(A)} \\
&= \frac{P(C\cap A)}{P(A)} \\
&=\frac{P(C|A)P(A)}{P(A)}\\
&= P(C|A).
\end{align*}
A: HINT: If $A,B$ indepedent so are $A,\bar{B}$ and $\bar{A}, B$. Also for any event $E$ and $F$ you can write $\Pr(E)=\Pr(E\cap F)+\Pr(E\cap \bar{F})$
A: One thing that comes to mind is to use the fact that if A and B are independent events, then A and $\bar{B}$ are also independent. See $A = (A\cap B) \cup (A \cap \bar{B})$ and of course $A \cap B$ is disjoint to $A \cap \bar{B}$ which means that 
\begin{eqnarray}
\mathbb{P}[A] & = & \mathbb{P}[A \cap B] + \mathbb{P}[ A \cap \bar{B}] \Rightarrow \\
\mathbb{P}[A] &= & \mathbb{P}[A]\mathbb{P}[B] + \mathbb{P}[A \cap \bar{B}] \Rightarrow \\
\mathbb{P}[A](1 - \mathbb{P}[B])& = &\mathbb{P}[A \cap \bar{B}] \Rightarrow\\
\mathbb{P}[A \cap \bar{B}] & = & \mathbb{P}[A]\mathbb{P}[\bar{B}]
\end{eqnarray}
