Sets and Probability

**kid funky fried** · August 30th 2008, 12:05 PM

Prove:

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

**Plato** · August 30th 2008, 12:24 PM

The expression $P\left\{ {A \cap B} \right\} \cap C$ has no meaning.
It like adding a matrix to a number: meaningless!
So what are you working out?
What is the problem?

**Chris L T521** · August 30th 2008, 01:02 PM

Originally Posted by kid funky fried

prove:

$ P\{ (A \cap B)\} {\color{red}\cap C} = P\{ A \cap C\} + P\{ B \cap C\} - P\{ A \cap B \cap C\} $

So I did this:

$ \begin{gathered} P(A \cup B) = P(A) + P(B) - P(A \cap B)theorum \hfill \\ by substitution, \hfill \\ (P(A) + P(B) - P(A \cap B)) \cap C = \hfill \\ P(A \cap C) + P(B \cap C) - P(A \cap B \cap C) \hfill \\ \end{gathered} $

If it is no included inside of the probability term, $P\left\{(A\cap B)\right\}\cap C$ has no meaning, as Plato mentioned.

However, if it was $P\left\{(A\cap B)\cap C \right\}$ , then we can work with it...maybe.

From what I'm understanding, I think you are looking for $P\left\{(A\cup B)\cap C \right\}$

This is the same as $P\left\{(A\cap C)\cup (B\cap C) \right\}={\color{red}P(A\cap C)+P(B\cap C)-P(A\cap B\cap C)}$

I hope this helps

--Chris

**Plato** · August 30th 2008, 01:03 PM

As I have pointed out to you, $P\left( {A \cap B} \right) \cap C$ is totally meaning less.
First $P\left( {A \cap B} \right)$ is a number while $C$ 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!

**kid funky fried** · August 30th 2008, 01:36 PM

Hopefully, my revision has meaning.

I just want to make sure this is legit.

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

Kid

**Moo** · August 31st 2008, 01:00 AM

Hi,

Originally Posted by kid funky fried

Hopefully, my revision has meaning.

I just want to make sure this is legit.

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

Kid

Sorry to say, but this expression is not good either...
Again, $\cap$ is usually used for sets, whereas P(robabilities) are values, numbers...
You can't weigh apples and pears together when you buy fruits ^^

**kid funky fried** · August 31st 2008, 07:06 AM

Thanks Moo, I think I understand now.
I will revise-again.

Kid

**kid funky fried** · August 31st 2008, 08:31 AM

$ \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)

**kid funky fried** · August 31st 2008, 08:55 AM

Originally Posted by Moo

Hi,

Sorry to say, but this expression is not good either...
Again, $\cap$ is usually used for sets, whereas P(robabilities) are values, numbers...
You can't weigh apples and pears together when you buy fruits ^^

Moo, Do you think I could weigh apples and pears together if they are the same price?

( I do this sometimes, also!)lol

Thx for ur help. I appreciate it. I (obviously)did not realize that the intersection was the issue at hand.

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