# Exclusive-Or probability calculation

For independent, non-mutually exclusive events, calculate the probability that $A$ or $B$ will be true, but not both. That is, $P(A \oplus B)$.

I thought of two ways to compute this probability, but they give different results. Both methods take the union and "remove" the intersection, but they do it in different ways.

The first (subtraction): $$P(A \oplus B) = P(A \cup B) - P(A \cap B) \\= P(A) + P(B) - 2P(A)P(B)$$

The second (set operators): $$P(A \oplus B) = P( (A \cup B) \cap (A \cap B)' \\=[P(A)+P(B)-P(A)P(B)]\cdot [1-P(A)P(B)]$$

But take, as an example, $P(A) = 0.3$ and $P(B) = 0.7$. The first method gives 0.58, while the second gives 0.6241.

Which method is wrong, and why?

The first calculation is correct.

In the second calculation, you assumed that $A\cup B$ and $(A\cap B)'$ are independent. They are in general not independent, even if $A$ and $B$ are.

• That makes sense. Would that have something to do with the "area outside of the Venn Diagrams"?
– baum
Jan 9, 2015 at 22:11
• The area outside the Venn diagram, if by that you mean the area outside $A\cup B$, is determined once we know $\Pr(A)$, $\Pr(B)$, and that $A$ and $B$ are independent. I do not see the connection with the non-independence of $A\cup B$ and $(A\cap B)'$. Informally, one way to see the non-independence is to note that if for example $\Pr((A\cap B)')$ is small, then $\Pr(A\cap B)$ is big, forcing $\Pr(A\cup B)$ to be big. Jan 9, 2015 at 22:20

As André noted, your wrong approach assumed independence of non-independent events.

Here’s another approach. The fact that $A$ and $B$ are independent implies that $A$ and $B'$ are also independent, and that $A'$ and $B$ are independent. Furthermore, $A \oplus B$ is the union of the two exclusive possibilities $A \land B'$ and $B \land A'$. Putting this together,

$P(A \oplus B) \\= P\left((A \land B')\lor (B \land A')\right)\\=P(A \land B')+ P(B \land A')=P(A)P(B')+ P(B)P(A')\\=P(A)(1-P(B))+ P(B)(1-P(A))\\=0.3(1-0.7)+0.7(1-0.3)\\=0.58$.

A Venn diagram is also useful.

• Ahh, and from your third line I see where my $P(A)+P(B)-2P(A)P(B)$ comes from.
– baum
Jan 9, 2015 at 22:37