Intersection between three events

Question

If I have $S_{1}=(F\cap\bar{G}\cap \bar{H})$, $S_{2}=(\bar{F}\cap G\cap \bar{H})$ and $S_{3}=(\bar{F}\cap\bar{G}\cap H)$, how can I apply the law of total probability for

$\mathbb{P}(S_{1}\cup S_{2}\cup S_{3})=\mathbb{P}(S_{1})+\mathbb{P}(S_{2})+\mathbb{P}(S_{3})-\mathbb{P}(S_{1}\cap S_{2})-\mathbb{P}(S_{1}\cap S_{3})-\mathbb{P}(S_{2}\cap S_{3})+\mathbb{P}(S_{1}\cap S_{2}\cap S_{3})$

Thanks in advance for any help!

Answer 1

Notice that if $i\ne j$, then $S_i \cap S_j = \emptyset$,

Hence $$P\left(\bigcup_{i=1}^3 S_i\right)=\sum_{i=1}^3 P(S_i)$$

We have

\begin{align}P(S_1)&= P(E \cap \bar{F}\cap \bar{G}) \\ &= P(E)-P(E \cap (F \cup G)) \tag{1} \\&=P(E)-P(E \cap F)-P(E\cap G)+P(E \cap F\cap G) \tag{2} \end{align}

To go from $(1)$ to $(2)$,

We use

$$P(E \cap (F \cup G))=P((E\cap F) \cup (E\cap G))$$ and we use inclusion exclusion principle.

Do the same thing for $P(S_2)$ and $P(S_3)$ or just do a symmetric argument and add them up.

Answer 2

Yes, I guess. He says that applying the formule below…

…we have what is written in the previous picture.