How to proof P(A U B U C) without using Venn Diagram

[ooooo]

Do you know how to proof

P(A U B U C) = P(A) + P(B) + P(C) - P(A^B) - P(B^C) - P(C^A) + P(A^B^C)


^ is intersection.

Do you know how to find P(A U B U C U D)

Thank you very much.

 

[statdad]

Let

$$D = B \cup C$$

and note that

$$A \cup B \cup C = A \cup D$$

then

$$\begin{align*}<br /> \Pr(A \cup B \cup C) & = \Pr(A \cup D)\\<br /> & = \Pr(A) + \Pr(D) - \Pr(A \cap D) \\<br /> & = \Pr(A) + \Pr(B \cup C) - \Pr(A \cap D)\\<br /> & = \Pr(A) + \Pr(B) + \Pr(C) - \Pr(B \cap C) - \Pr(A \cap D)<br /> \end{align*}$$

The rest of the proof comes from realizing that

$$\Pr(A \cap D) = \Pr(A \cap \left(B \cup C\right)) = \Pr((A \cap B) \cup (A \cap C)),$$

using the Addition Rule for probability to expand the final term, and being very careful with positive and negative signs.

 

[ooooo]

Thank you so much Statdad. I would like to ask another question.

How to proof P(A U B) = P(A) + P(B) - P(A ^ B) ?

Thank you again.

 

[statdad]

This proof isn't needed for the problem you posted above - is there a reason you need it here?

 

[ooooo]

Sorry. I'm just curious. :)

 

[statdad]

No - I was interrupted by someone at the door.
Here is one method - there are others.
First, note that

$$A \cup B = (A-B) \cup (A \cap B) \cup (B - A)$$

and the three sets on the right are pair-wise disjoint. Now

$$\begin{align*}<br /> \Pr(A \cup B) & = \Pr(A-B) + \Pr(A \cap B) + \Pr(B - A)\\<br /> & = \left(\Pr(A-B) + \Pr(A \cap B) \right) + \left(\Pr(B-A) + \Pr(A \cap B)\right) - \Pr(A \cap B) \\<br /> & = \Pr(A) + \Pr(B) - \Pr(A \cap B)<br /> \end{align*}$$

Again, sorry for the abrupt end to my previous post - I'm getting really tired of our election season.