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#1 2008-10-04 09:16:32

OOO
Guest

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

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.

#2 2008-10-04 10:21:19

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

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

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

You may have noticed that you find the probability by adding the probabilities of the individual events, then taking away the probabilities of each combination of two events, and finally adding the probability for all three to happen. This pattern is called the inclusion-exclusion principle, and it applies to the union of any number of probabilities.

So P(A U B U C U D) =

P(A) + P(B) + P(C) + P(D) - P(A^B) - P(A^C) - P(A^D) - P(B^C) - P(B^D) - P(C^D) + P(A^B^C) + P(A^B^D) + P(A^C^D) + P(B^C^D) - P(A^B^C^D)


Why did the vector cross the road?
It wanted to be normal.

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#3 2008-10-04 11:58:02

OOO
Guest

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

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

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

Thank you again.

#4 2008-10-04 23:34:13

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

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

From a Venn Diagram? tongue
I don't think there's an algebraic way to do it.


Why did the vector cross the road?
It wanted to be normal.

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#5 2012-07-07 03:03:17

DC92
Guest

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

P(AUB)=P(A'^B)+P(A^B')+P(AUB)

You can see though that P(A)=P(A^B')+P(A^B) and P(B)=P(A'^B)+P(A^B). So you can say that P(A'^B)=P(A)-P(A^B) and that P(A^B')=P(B)-P(A^B).

Substituting....P(AUB)=P(A)-P(A^B)+P(B)-P(A^B)+P(A^B)=P(A)+P(B)-P(A^B)!

This is the algebric way! smile...where B' is B complementary and A' is A complementary! smile

#6 2012-07-07 19:35:07

MathsIsFun
Administrator
Registered: 2005-01-21
Posts: 7,535

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

Thanks DC92, much appreciated.


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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#7 2012-10-01 03:09:38

IrishEyes
Member
Registered: 2012-10-01
Posts: 1

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

How do you find P(A^B^C) without knowing P(AUBUC) and knowing the values P(A), P(B), P(C) and P(A^B), P(A^C), P(B^C)?

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