$P(A|B) =P(C|B)P(A|B∩C) +P(C'|B)P(A|B∩C')$

Question

I'm struggling with the following problem:

Show that $P(A|B) =P(C|B)P(A|B∩C) +P(C'|B)P(A|B∩C')$ assuming all conditioning events have positive probability.

I have played around with many different pieces I know such as $P(A∩B∩C)=P(A)P(B|A)P(C|A∩B)$ as well as $1=P(C|B)+P(C'|B)$ and can't seem to find anything that works.

Answer 1

$$\mathsf P(A\mid B)~ {=\dfrac{\mathsf P(A\cap B)}{\mathsf P(B)}\\=\dfrac{\mathsf P(A\cap B\cap C)+\mathsf P(A\cap B\cap C')}{\mathsf P(B)}\\\vdots\\=}$$

---

I have played around with many different pieces I know such as P(A∩B∩C)=P(A)P(B|A)P(C|A∩B) as well as 1=P(C|B)+P(C'|B) and can't seem to find anything that works.

Hint:

Hint: change the order: $\mathsf P(B\cap C\cap A)=\mathsf P(B)\mathsf P(C\mid B)\mathsf P(A\mid B\cap C)$

Answer 2

$P_{\Gamma}(A)=P(A|\Gamma)$ is probability so

$P_{\Gamma}(A|B)=\frac{P_{\Gamma}(A\cap B)}{P_{\Gamma}(B)}=\frac{P(A\cap B|\Gamma)}{P(B|\Gamma)}=\frac{\frac{P(A\cap B \cap\Gamma)}{P(\Gamma)}}{\frac{P(B \cap \Gamma)}{P(\Gamma)}}=\frac{P(A\cap B \cap\Gamma)}{P(B \cap \Gamma)}=P(A|B\cap\Gamma)$

and by this we can see that in the conditional probability if you set another condition it becomes the intersection of the conditions. And so you can prove your exercise by the use of total law of prob/ty