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

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IrishEyes
2012-10-02 01:09:38

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)?

MathsIsFun
2012-07-08 17:35:07

Thanks DC92, much appreciated.

DC92
2012-07-08 01:03:17

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! ...where B' is B complementary and A' is A complementary!

mathsyperson
2008-10-05 22:34:13

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

OOO
2008-10-05 10:58:02

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.

mathsyperson
2008-10-05 09:21:19

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)

OOO
2008-10-05 08:16:32

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.