A partition of the sample space $\Omega$ is a collection of disjoint events $S_{1}, \ldots, S_{n}$ such that $\Omega=\cup_{i=1}^{n} S_{i}$
(a) Show that for any event $A$, we have
$$
\mathbf{P}(A)=\sum_{i=1}^{n} \mathbf{P}\left(A \cap S_{i}\right)
$$
(b) Use part (a) to show that for any events $A, B$, and $C$, we have
$$
\mathbf{P}(A)=\mathbf{P}(A \cap B)+\mathbf{P}(A \cap C)+\mathbf{P}\left(A \cap B^{c} \cap C^{c}\right)-\mathbf{P}(A \cap B \cap C)
$$
(a) Since $\Omega=\cup_{i=1}^{n} S_{i}$, we have
$$
A=\bigcup_{i=1}^{n}\left(A \cap S_{i}\right)
$$
while the sets $A \cap S_{i}$ are disjoint. The result follows by using the additivity axiom.
(b) The events $B \cap C^{c}, B^{c} \cap C, B \cap C$, and $B^{c} \cap C^{c}$ form a partition of $\Omega$, so by part
(a), we have
$$
\mathbf{P}(A)=\mathbf{P}\left(A \cap B \cap C^{c}\right)+\mathbf{P}\left(A \cap B^{c} \cap C\right)+\mathbf{P}(A \cap B \cap C)+\mathbf{P}\left(A \cap B^{c} \cap C^{c}\right)
$$
The event $A \cap B$ can be written as the union of two disjoint events as follows:
$$
A \cap B=(A \cap B \cap C) \cup\left(A \cap B \cap C^{c}\right)
$$
so that
$$
\mathbf{P}(A \cap B)=\mathbf{P}(A \cap B \cap C)+\mathbf{P}\left(A \cap B \cap C^{c}\right)
$$
Similarly,
$$
\mathbf{P}(A \cap C)=\mathbf{P}(A \cap B \cap C)+\mathbf{P}\left(A \cap B^{c} \cap C\right)
$$
Combining Eqs. (1)-(3), we obtain the desired result.