Prove:

$\displaystyle

P\{ (A \cup B) \cap C\} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\}

$

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- Aug 30th 2008, 12:05 PMkid funky friedSets and Probability
Prove:

$\displaystyle

P\{ (A \cup B) \cap C\} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\}

$ - Aug 30th 2008, 12:24 PMPlato
The expression $\displaystyle P\left\{ {A \cap B} \right\} \cap C$ has no meaning.

It like adding a matrix to a number: meaningless!

So what are you working out?

What is the problem? - Aug 30th 2008, 01:02 PMChris L T521
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 - Aug 30th 2008, 01:03 PMPlato
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! - Aug 30th 2008, 01:36 PMkid funky friedrevsion
Hopefully, my revision has meaning.

I just want to make sure this is legit.

$\displaystyle

P(A \cap B) \cap P(C) = P(A \cap B \cap C)

$

Kid - Aug 31st 2008, 01:00 AMMoo
- Aug 31st 2008, 07:06 AMkid funky friedThx
Thanks Moo, I think I understand now.

I will revise-again.(Headbang)

Kid - Aug 31st 2008, 08:31 AMkid funky friedHave I seen the light?
$\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! (Nod)

(I think) - Aug 31st 2008, 08:55 AMkid funky fried