Probability: Showing A\bar{B} \cup B \bar{A} is one event

[ehrenfest]

Homework Statement

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)

Homework Equations

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.

 

[Dick]

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

 

[ehrenfest]

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

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

 

[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?

 

[Dick]

Homework Statement

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)

Homework Equations

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.

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

 

[ehrenfest]

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

Rewrite the original equation as

<br /> P(AB)+P((A\bar{B}) \cup (B \bar{A})) = P(A)+P(B)-P(AB)<br />

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:

<br /> P(A \cup B) = P(A)+P(B)-P(A \cap B)<br />

 

[Dick]

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

Rewrite the original equation as

<br /> P(AB)+P((A\bar{B}) \cup (B \bar{A})) = P(A)+P(B)-P(AB)<br />

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:

<br /> P(A \cup B) = P(A)+P(B)-P(A \cap B)<br />

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