Probability of symmetric difference in terms of union and intersection? Let $A$ and $B$ be two events of a sample space $Ω$.
Prove that $A \nabla B=(A \cup B)-(A \cap B)$
By writing the formulas, how I can prove this in an easy way?
 A: I assume that $A \nabla B$ is the 'symmetric difference' or 'exclusive-or (XOR)'
and that it is defined as  $A \nabla B \equiv AB^c \cup A^cB,$ where intersection is written as multiplication for brevity and ${}^c$ indicates a complement. 
Then
$$P(A \nabla B) = P(AB^c) + P(A^cB) = [P(AB^c) + P(AB) + P(A^cB)] - P(AB)\\
= P(A \cup B) - P(AB),$$
as required. In the above the first equality holds because the events are
disjoint, the second by arithmetic, and the third because the three quantities
inside square brackets add to the probability of the union.
Because you have not shown us what you have tried, I have no idea what
results about sums of probabilities you have already proved. So I am not
sure this is an acceptable answer at this point in your course. Please
check on that, and try to make any necessary modifications. (As @Crostul
has commented, this relationship is easily seen in a Venn Diagram,
but you have asked for a proof using formulas.)
A related equation involving the symmetric difference is that
$$P(A \nabla B) = P(A) + P(B) - 2P(AB).$$
Notes: (a) In case anyone else wants to show an alternate method, $\nabla$ is
rendered in MathJax as $\nabla$; I had to look it up. (b) Some texts use $A \Delta B$ instead
of $A \nabla B.$
