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On the Probability chapter of a 1995 mathematical statistical book I am reviewing I have found the following exercise:

Let A and B be arbitrary events. Let C be the event that either A occurs or B occurs, but not both. Express C in terms of A and B using any of the basic operations of union, intersection and complement.

Now, the book suggested answer is to describe the entire sample space as:

$$\Omega=(A\cap B)^{C}\cap(A\cup B)$$

I think the correct answer is:

$$\Omega=(A\cap B)^{C}\cup(A\cap B)$$

and

$$C=(A\cap B)^{C}$$

Where is the error that I have made?

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$C$ (the symmetric difference of $A$ and $B$) is obtained by overlaying (intersecting) $A\cup B$ and $(A\cap B)^c$, whence $C = (A\cup B) \cap (A\cap B)^c$:

Venn diagram

Another expression frequently used is $C = (A\cap B^c) \cup (B\cap A^c)$. The left-hand term is the pure red lune in the figure while the right-hand term is the pure blue lune; together, they form $C$.

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Actually, you both got it wrong!

You're right in thinking that

$$\Omega=(A\cap B)^{C}\cup(A\cap B)$$

since it is true that, for any set $D$ in $\Omega$, $D^C \cup D=\Omega$.

However, $C$ is the part of $A\cup B$ such that only one of $A$ and $B$ occurs. In other words, you need both the event $(A\cup B)$ and the event $(A\cap B)^{C}$ to occur. Thus

$$C=(A\cap B)^{C}\cap(A\cup B)$$

which is what the book claimed was $\Omega$.

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    $\begingroup$ That makes sense even thought it is not so obvious. Thanks, Luca $\endgroup$ – Luca Feb 17 '12 at 8:27
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As MånsT has pointed out, what the book claims is $\Omega$ is actually $C$, the event that exactly one of $A$ and $B$ occur. An alternative expression for $C$ is $$C = (A\cap B^c) \cup (A^c \cap B)$$ which is the disjoint union of the events "$A$ occurs and $B$ does not" and "$A$ does not occur and $B$ does". Thus, we have that $$P(C) = P(A\cap B^c) + P(A^c \cap B)$$ and since $$\begin{align*} P(A \cup B) &= P(A\cap B^c) + (A^c \cap B) + P(A \cap B)\\ &= P(A) + P(B) - P(A\cap B) \end{align*}$$ we can write $$\begin{align*}P(C) &= P(A \cup B) - P(A \cap B)\\ &= P(A) + P(B) - 2P(A\cap B) \end{align*}$$

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  • $\begingroup$ Thank you for the alternative way to express C and the extra info $\endgroup$ – Luca Feb 17 '12 at 8:30

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