Probability help

1. Jun 13, 2008

ehrenfest

1. The problem statement, all variables and given/known data
Let A and B be events. Show that $A\bar{B} \cup B \bar{A}$ is the event which exactly one of A and B occurs. Moreover,

$$P(A\bar{B}) \cup B \bar{A}) = P(A)+P(B)-2P(AB)$$

2. Relevant equations

3. The attempt at a solution

First of all I think that the problem statement is ambiguous because they don't specify the order of operations. I think they mean$(A\bar{B}) \cup (B \bar{A})$

So, basically I was trying to prove this using only the property
$$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$
but that failed. I don't know what else to do.

2. Jun 13, 2008

Dick

You might notice there is a difference between A union B and (A-B) union (B-A). What is it?

3. Jun 13, 2008

ehrenfest

Well the symmetric difference (A-B) union (B-A) doesn't contain the middle of the Venn Diagram. How is that useful?

4. Jun 13, 2008

Dick

Well, it might be useful in connecting your probability formulae together. The first one is a symmetric difference. The second is a union. So why does one have P(A intersect B) with a factor of one and the other with a factor of two? Why are you asking this?

Last edited: Jun 13, 2008
5. Jun 13, 2008

Dick

I do. I see P(AB) and P(A intersect B) in your problem statement quite clearly.

6. Jun 13, 2008

ehrenfest

OK. I think I see what is going on now.

Rewrite the original equation as

$$P(AB)+P((A\bar{B}) \cup (B \bar{A})) = P(A)+P(B)-P(AB)$$

Since AB and S(A,B) are disjoint, we can add the probabilities on the RHS to obtain P(A \cup B) and on the LHS we can obtain the same thing simply using the property of probability measures:

$$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$

7. Jun 13, 2008

Dick

Good. Yes, the only difference is how many times you include P(AB) in the LHS.

Last edited: Jun 14, 2008
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