Axioms of probability

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I came across this question:
http://imageshack.us/m/3/4510/axiomsq.png
which I'm confused about. I know what the axioms of probability are but how would I use them to "derive" that result? I could illustrate why P(A∪B) = P(A) + P(B) - P(A∩B) on a Venn diagram but I have no idea how to use the axioms to show it.
 

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  • #2
disregardthat
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You only have to use that P(X) + P(Y) = P(X∪Y) if X and Y are disjoint sets.

P(B) + P(A) = (P(B) + P(A-A∩B)) + P(A∩B) = P(B∪(A-A∩B)) + P(A∩B) = P(A∪B) + P(A∩B)

For what you have been given, note that it remains to prove that [tex]P(B) + P(A \cap \overline{B}) = P(A \cup B)[/tex]. Can you use the formula again to prove this?

Hint:
substitute A by A∪B in the formula above.
 
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  • #3
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I don't understand your explanation. The equation in the first spoiler confuses the hell out of me.
 
  • #4
I like Serena
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which I'm confused about. I know what the axioms of probability are but how would I use them to "derive" that result? I could illustrate why P(A∪B) = P(A) + P(B) - P(A∩B) on a Venn diagram but I have no idea how to use the axioms to show it.
Hi ampakine! :smile:

Could you start by writing down the axioms of chance?
Then we can pick up the proof starting from your definitions.
 
  • #5
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Hi ampakine! :smile:

Could you start by writing down the axioms of chance?
Then we can pick up the proof starting from your definitions.
Alright:
(Ax1) 0 ≤ P(A) ≤ 1
(Ax2) P(S) = 1
(Ax3) If events A and B are disjoint then P(A∪B) = P(A) + P(B)

I have no idea how I could use any of that to prove anything though.
 
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  • #6
I like Serena
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Alright:
(Ax1) 0 ≤ P(A) ≤ 1
(Ax2) P(S) = 1
(Ax3) If events A and B are disjoint then P(A∪B) = P(A) + P(B)

I have no idea how I could use any of that to prove anything though.
All right!
So (perhaps with the help of a Venn diagram), can you write A∪B as a union of disjoint sets?
 
  • #7
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What do you mean write them as a union? I know that A∪B looks like this:
figure%201%20VD.jpg

and that A∪B = A + B. The first equation is the same as A∪B = A + B - A∩B but I don't know if A and B are disjoint sets. If they are then A∩B = 0 which still satisfies both equations:
A∪B = A + B - 0
A = 0 + A∪E
^^ I couldn't type B bar so I used E instead.
 
  • #8
I like Serena
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and that A∪B = A + B. The first equation is the same as A∪B = A + B - A∩B but I don't know if A and B are disjoint sets. If they are then A∩B = 0 which still satisfies both equations:
A∪B = A + B - 0
A = 0 + A∪E
^^ I couldn't type B bar so I used E instead.
"∪" is the symbol for a union.
Saying A∪B = A + B is like saying that you consider "+" as being a synonym for "∪".
This does occur in practice, but only because a "+" is easier to type.
Let's avoid the "+" here, to avoid ambiguity.

And yes, you do not know if A and B are disjoint sets, so we'll assume for now that they are not disjoint.
In a Venn diagram you would show this by drawing 2 circles that overlap.
The first circle represents A, the second represents B, and the overlapping area represents the so called intersection, written as A∩B.

In set theory we also have subtraction, for which I'll be using the symbol "\", again to avoid ambiguity with the "-" symbol, which we'll use for subtracting numbers.
So "A \ B" is the set A from which all elements that belong to B are removed.

The bar that you wanted to use would represent the complement.
Let's use BC for that, which is a common notation (that we can type).


Can you name the 3 parts of A∪B, that will be disjoint?
 
  • #9
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Can you name the 3 parts of A∪B, that will be disjoint?
A∩BC, AC∩B and A∩B?
 
  • #10
I like Serena
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A∩BC, AC∩B and A∩B?
Yep! :smile:

So we have P(A∪B) = P( (A∩BC) ∪ (AC∩B) ∪ (A∩B) ).

Can you apply the axioms now, since the preconditions are met?
 
  • #11
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I still have absolutely no idea how to do it. I don't even know where to begin. I can make Venn diagrams in my head and see exactly what is going on but I have no idea how to answer a question like this. Its the question itself I don't understand. Am I suppose to just rearrange equations until I end up with A∪B = A + B - A∩B? I can't see how letting you assume that P(A) = P(A∩B) + P(A∩BC) provides any new information thats just another way of stating the original equation.
 
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  • #12
I like Serena
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I still have absolutely no idea how to do it. I don't even know where to begin. I can make Venn diagrams in my head and see exactly what is going on but I have no idea how to answer a question like this. Its the question itself I don't understand. Am I suppose to just rearrange equations until I end up with A∪B = A + B - A∩B? I can't see how letting you assume that P(A) = P(A∩B) + P(A∩BC) provides any new information thats just another way of stating the original equation.
Let's give the sets a couple of names, just to talk easier.
Define:
A1=(A∩BC)
B1=(AC∩B)

So A∪B = A1 ∪ B1 ∪ (A∩B) is a disjoint union.


What you have to show is that P(A∪B) = P(A) + P(B) - P(A∩B),
where A and B are not necessarily disjoint.


What you can use is axiom 3 that says that for any disjoint U and V holds:
P(U∪V) = P(U) + P(V)


Basically you can set up a couple of equations from the divisions of A and B we made earlier.
I think you already understand how to get these equations, so I'll storm ahead for now.

The equations are (from axiom 3):

P(A) = P(A1) + P(A∩B)
P(B) = P(B1) + P(A∩B)
P(A∪B) = P(A1) + P(B1) + P(A∩B)


Can you combine these equations to find the equation you have to show?
 
  • #13
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The equations are (from axiom 3):

P(A) = P(A1) + P(A∩B)
P(B) = P(B1) + P(A∩B)
P(A∪B) = P(A1) + P(B1) + P(A∩B)


Can you combine these equations to find the equation you have to show?
Yeah I can replace 2 of the terms in equation 3 with equation one and get:
P(A∪B) = P(A) + P(B1)
then if I replace P(B1) with P(B) then P(A∩B) will appear twice which doesn't matter for disjoint sets since P(A∩B) equals 0 for them but will be a problem for joint sets so to fix the equation you must substract a P(A∩B) from it. So is it all about axiom 3 then? Is the purpose of the question to highlight the fact that the equation P(A∪B) = P(A) + P(B) only holds for disjoint sets? I'm probably going to be asked this question in an exam coming up but I would have had no idea how to answer if I hadn't asked about it here first. Thanks a lot!
 
  • #14
I like Serena
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Yeah I can replace 2 of the terms in equation 3 with equation one and get:
P(A∪B) = P(A) + P(B1)
then if I replace P(B1) with P(B) then P(A∩B) will appear twice which doesn't matter for disjoint sets since P(A∩B) equals 0 for them but will be a problem for joint sets so to fix the equation you must substract a P(A∩B) from it. So is it all about axiom 3 then? Is the purpose of the question to highlight the fact that the equation P(A∪B) = P(A) + P(B) only holds for disjoint sets? I'm probably going to be asked this question in an exam coming up but I would have had no idea how to answer if I hadn't asked about it here first.
Well, this is a proposition that follows from the axioms (indeed only axiom 3 to be precise).
The purpose would be that you know you can use this one when you need to.
And it highlights how propositions are derived from the axioms.

Actually, now that I look at the original problem again, I see that I took you the long way around.
We did the complete proof, but in the problem they gave you a hint.
That is, they gave you an equation that you could assume.
With it, the proof becomes a bit shorter.

Thanks a lot!
I appreciate the thank you and you're welcome! :smile:
 

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