# Law of total probability with intersections

So in my intro course to stats, we ecnountered the law of total probability. The definition is $$P(A) = \sum^n_{j=1}P(A\mid H_j)P(H_j)$$ However, the definition says

$$\bigcup^n_{j=1}H_j = S \quad \text{(Union of all Events form Sample Space)}\\ \bigcap^n_{j=1}H_j=\emptyset \quad \text{Events are Pairwise Disjoint}$$

Now I was wondering what happens, when they tell me (in the exam) to calculate the total probability of an event $$A$$, even though an intersection of H_1 and H_2 exist.

So I constructed the following example and wanted to ask for your help on whether that is correct.

So I constructed two Events $$A,B$$ with $$P(A) = 0.5, P(B) = 0.7 \text{ and } P(A \cap B) = 0.2$$

I then introduce an event $$M,$$ which can happen according to the following probabilites. $$P(M\mid A \setminus B)=P(M\mid A \cap B)=P(M\mid B \setminus A) = \frac{1}{3}$$

This introduces the next drawing.

And thus I ultimately calculated $$P(M) = P(A \setminus B\mid M) P(A \setminus B) +P(A \cap B \mid M)P(A \cap B) + P(B \setminus A \mid M) P(B \setminus A) = \frac{1}{3}\cdot(0.3+0.2+0.5)= \frac{1}{3}$$

Is that correct? Is thus the tric to redefine all the events of the subspace as pairwise disjoint? Or is there maybe another way/trick or something one should pay attention to?

This is indeed correct. The law is true if $$H_i$$ provide a partition of the probability space. Here, you created a new partition with the underlying distributions, but it's still a partition, so the law applies.
The law of total probability applies when you condition on a partition (i.e., a class of disjoint events that cover the sample space). Given an event $$\mathcal{A}$$ and a partition $$\mathscr{H} = \{ \mathcal{H}_1, \mathcal{H}_2, ..., \mathcal{H}_n \}$$ (where the sets $$\mathcal{H}_1, \mathcal{H}_2, ..., \mathcal{H}_n$$ are disjoint sets with $$\cup_{i=1}^n \mathcal{H}_i = \Omega$$), the law is derived as follows:
\begin{aligned} \mathbb{P}(\mathcal{A}) = \mathbb{P}(\mathcal{A} \cap \Omega) &= \mathbb{P} \Bigg( \mathcal{A} \cap \bigcup_{i=1}^n \mathcal{H}_i \Bigg) \\[6pt] &= \mathbb{P} \Bigg( \bigcup_{i=1}^n (\mathcal{A} \cap \mathcal{H}_i) \Bigg) \\[6pt] &= \sum_{i=1}^n \mathbb{P} (\mathcal{A} \cap \mathcal{H}_i) \\[6pt] &= \sum_{i=1}^n \frac{\mathbb{P} (\mathcal{A} \cap \mathcal{H}_i)}{\mathbb{P} (\mathcal{H}_i)} \cdot \mathbb{P}(\mathcal{H}_i) \\[6pt] &= \sum_{i=1}^n \mathbb{P}(\mathcal{A} | \mathcal{H}_i) \cdot \mathbb{P}(\mathcal{H}_i). \\[6pt] \end{aligned}
The third step follows from the additivity axiom of probability since $$(\mathcal{A} \cap \mathcal{H}_i) \cap (\mathcal{A} \cap \mathcal{H}_j) = \varnothing$$ for all $$i \neq j$$. (For simplicity we have assumed here that all the events in the partition have positive marginal probability. If a partition event has zero probability, the conditional probability is undefined, but since it is multiplied by zero anyway, we can define it arbitrarily.)