# Sets and Probability

• Aug 30th 2008, 12:05 PM
kid funky fried
Sets 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 PM
Plato
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 PM
Chris L T521
Quote:

Originally Posted by kid funky fried
prove:

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

So I did this:

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

If it is no included inside of the probability term, $\displaystyle P\left\{(A\cap B)\right\}\cap C$ 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 PM
Plato
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 PM
kid funky fried
revsion
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 AM
Moo
Hi,
Quote:

Originally Posted by kid funky fried
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

Sorry to say, but this expression is not good either...
Again, $\displaystyle \cap$ is usually used for sets, whereas P(robabilities) are values, numbers...
You can't weigh apples and pears together when you buy fruits ^^
• Aug 31st 2008, 07:06 AM
kid funky fried
Thx
Thanks Moo, I think I understand now.

Kid
• Aug 31st 2008, 08:31 AM
kid funky fried
Have 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 AM
kid funky fried
Quote:

Originally Posted by Moo
Hi,

Sorry to say, but this expression is not good either...
Again, $\displaystyle \cap$ 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? (Rofl)
( 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.