The Probability of the Difference of Two Events

What this lecture covers

This short recitation problem shows how to prove a probability identity purely from the axioms, without appealing to intuition. Starting from two events A and B, the goal is to show that the probability of their symmetric difference (the points in exactly one of the two events) equals P(A) + P(B) - 2P(A ∩ B).

It's a good follow-up to the lecture on probability axioms: it demonstrates the habit of naming disjoint pieces of a Venn diagram and building up a proof step by step using only the additivity axiom.

Key ideas

• Simplifying notation first: before manipulating a messy equation, name the disjoint regions involved (here called C, D, and E) to make the algebra manageable.
• Partitioning into disjoint pieces: A splits into C (in A but not B) and E (in both); B splits into D (in B but not A) and E.
• Additivity axiom as the only tool: every step in the proof rests on the fact that probabilities of disjoint sets add.
• Adding and subtracting a term: inserting +P(E) - P(E) into an expression is a standard trick for connecting a derived quantity to known probabilities.
• Result: P(C ∪ D) = P(A) + P(B) - 2P(A ∩ B), i.e. the probability that exactly one of A or B occurs.

Before you watch

• Know the three probability axioms (non-negativity, normalization, additivity), covered in the first lecture of this course.
• Be comfortable with basic set notation: union, intersection, and complement.

Check your understanding

1. Why can the additivity axiom be applied to C and D, but not directly to A and B?
2. How does writing A as the union of C and E help connect P(A) to P(C) and P(E)?
3. Why is adding and subtracting P(E) a valid algebraic step, and what does it accomplish here?
4. Can you sketch a Venn diagram and label the regions C, D, and E yourself before checking the derivation?