Axioms of Probability: Deriving P(A∪B) = P(A) + P(B) - P(A∩B)

[ampakine]

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.

 

[disregardthat]

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

Spoiler

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 $$P(B) + P(A \cap \overline{B}) = P(A \cup B)$$. Can you use the formula again to prove this?

Hint:

Spoiler

substitute A by A∪B in the formula above.

 

[ampakine]

I don't understand your explanation. The equation in the first spoiler confuses the hell out of me.

 

[I like Serena]

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!

Could you start by writing down the axioms of chance?
Then we can pick up the proof starting from your definitions.

 

[ampakine]

Hi ampakine!

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.

 

[I like Serena]

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?

 

[ampakine]

What do you mean write them as a union? I know that A∪B looks like this:

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.

 

[I like Serena]

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 B^C for that, which is a common notation (that we can type).


Can you name the 3 parts of A∪B, that will be disjoint?

 

[ampakine]

Can you name the 3 parts of A∪B, that will be disjoint?

A∩B^C, A^C∩B and A∩B?

 

[I like Serena]

A∩B^C, A^C∩B and A∩B?

Yep!

So we have P(A∪B) = P( (A∩B^C) ∪ (A^C∩B) ∪ (A∩B) ).

Can you apply the axioms now, since the preconditions are met?

 

[ampakine]

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∩B^C) provides any new information that's just another way of stating the original equation.

 

[I like Serena]

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∩B^C) provides any new information that's just another way of stating the original equation.

Let's give the sets a couple of names, just to talk easier.
Define:
A1=(A∩B^C)
B1=(A^C∩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?

 

[ampakine]

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!

 

[I like Serena]

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!