How to simplify $\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$?

Question

How to simplify the following probability

$\operatorname{P} ( { C |B,A} )P( { B |A} )P( A ) + P( { {\bar B} |A} )P( A )$

$ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$

Can $P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$ be further simplify ?

Also, are there any systematic way to do the simplification instead of following heuristic. I see that there are some link between Boolean algebra and the simplification of these problem.

Particularly, my thinking is that $ = P\left( {A,B,C} \right) + P\left( {A,\bar B} \right)$ may be somehow related to $F\left( {A,B,C} \right) = \left( {ABC} \right)\,OR\,\left( {A\bar B} \right)$ but I do not know if my guts feeling is correct or not.

Please help me clear out these doubt, thank you for your enthusiasm !

Answer 1

$$\begin{align}&~~~~~~\mathsf P(C\mid B,A)\,\mathsf P(B\mid A)\,\mathsf P(A)+\mathsf P(\overline B\mid A)\,\mathsf P(A)\\&=\big(\mathsf P(B,C\mid A)+\mathsf P(\overline B\mid A)\big)\,\mathsf P(A)&&\small\text{by definition of conditional probability}\\&=\mathsf P\big((B\cap C)\cup\overline B\mid A\big)\,\mathsf P(A)&&\small\text{by sigma additivity of disjoint events}\\&=\mathsf P(C\cup\overline B\mid A)\,\mathsf P(A)&&\small\text{by absorption}\\&=\mathsf P(A\cap(C\cup\overline B) )\end{align}$$

Answer 2

Hint: $ P(C|B,A)P(B|A)P(A)=P(A,B,C)$.