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?
$\endgroup$2 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$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.
