Question

Let A and B be events. The symmetric difference A?B is defined to be the set of all elements that are in A or B but not both.

In logic and engineering, this event is also called the XOR (exclusive or) of A and B.
Show that P(AUB) = P(A) + P(B)-2P(AnB), directly using the axioms of probability.

Answer 1

Correction:

P(AΔB) = P(A) + P(B) - 2P(AnB)

is what could be proven using the axioms of probability, and considering the case of symmetric difference given.

Answer:

P(AΔB) = P(A) + P(B) - 2P(AnB)

Has been shown.

Explanation:

We are required to show that

P(AUB) = P(A) + P(B) - 2P(AnB)

directly using the axioms of probability.

Note the following:

AUB = (AΔB) U (AnB)

Because (AΔB) U (AnB) is disjoint, we have:

P(AUB) = P(AΔB) + P(AnB)..................(1)

But again,

P(AUB) = P(A) + P(B) - P(AnB)...............(2)

Comparing (1) with (2), we have

P(AΔB) + P(AnB) = P(A) + P(B) - P(AnB)

P(AΔB) = P(A) + P(B) - 2P(AnB)

Where AΔB is the symmetric difference of A and B.