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**OOO****Guest**

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.

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

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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**OOO****Guest**

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

From a Venn Diagram?

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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**DC92****Guest**

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!

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,663

Thanks DC92, much appreciated.

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

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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