conditional probability of 3 events into 2 conditional probabilities

Question

Consider: $$P(A,B |D)$$

Now I consider $C_i$ such that: $\cup_i C_i = \Omega $

This leads to: $$P(A,B |D) = \sum_iP(A,B,C_i |D)= \sum_iP( A,B,C_i,D) P(D) $$

Is there any way to rewrite the right side in a product of two conditional probabilitys?

Answer 1

NB: You also require the sequence to be pairwise disjoint: $\forall i~\forall j~(i\neq j\to C_i\cap C_j=\emptyset)$

Then, by the Law of Total Probability, you will have:

$\begin{align}\mathsf P(A,B\mid D)&=\sum_i\mathsf P(A,B,C_i\mid D)\\[1ex]&=\sum_i\left(\mathsf P(A,B,C_i,D)\,\middle/\,\mathsf P(D)\right)\\[1ex]&=\sum_i\left(\mathsf P(A,B\mid C_i,D)\,\mathsf P(C_i,D)\,\middle/\,\mathsf P(D)\right)\\[1ex]&=\sum_i\mathsf P(A,B\mid C_i,D)\,\mathsf P(C_i\mid D)\end{align}$