Prove:
$\displaystyle
P\{ (A \cup B) \cap C\} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\}
$
If it is no included inside of the probability term, $\displaystyle P\left\{(A\cap B)\right\}\cap C$ has no meaning, as Plato mentioned.
However, if it was $\displaystyle P\left\{(A\cap B)\cap C \right\}$, then we can work with it...maybe.
From what I'm understanding, I think you are looking for $\displaystyle P\left\{(A\cup B)\cap C \right\}$
This is the same as $\displaystyle P\left\{(A\cap C)\cup (B\cap C) \right\}={\color{red}P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)}$
I hope this helps
--Chris
As I have pointed out to you, $\displaystyle P\left( {A \cap B} \right) \cap C$ is totally meaning less.
First $\displaystyle P\left( {A \cap B} \right)$ is a number while $\displaystyle C$ is a set in a probability space.
We do not intersect numbers with sets: it is meaningless.
I think that I have done this for you, did I not?
You cannot pove what you have written!
$\displaystyle
\begin{gathered}
P\{ (A \cup B) \cap C\} = P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) \hfill \\
Dist. \hfill \\
P\{ (A \cup B) \cap C\} = P\{ (A \cap C) \cup (B \cap C)\} \hfill \\
Theor. \hfill \\
P(A \cup B) = P(A) + P(B) - P(A \cap B) \hfill \\
I\_got, \hfill \\
P(A \cap C) + P(B \cap C) - P\{ (A \cap C) \cap (B \cap C)\} = \hfill \\
P(A \cap C) + P(B \cap C) - P\{ (A \cap B \cap C). \hfill \\
\end{gathered}
$
Now I understand Plato's disdain!
(I think)