# Introduction

In this article, we’ll be going through some identities involving sets

# Basic Properties

$$A \cup A^C = U$$
$$A \cap A^C = \emptyset$$
$$A \cap U = A$$
$$A \cup \emptyset = A$$

# Distributive Property

$$A \cup (B \cap C) = (A\cup B) \cap (A \cup C)$$
$$A \cap (B \cup C) = (A\cap B) \cup (A \cap C)$$
$$A \cap (A \cup B) = (A\cap A) \cup (A \cap B) = A \cup (A \cap B) = A$$
$$A \cup (A \cap B) = (A\cup A) \cap (A \cup B) = A \cap (A \cup B) = A \text{ (by the latest property)}$$

## Example Proof

Prove this:
$|A| = |A \cap B| + |A \cap B^c|$
$$A = A \cap U$$
$$ = A \cap (B \cup B^c)$$
$$ = (A \cap B) \cup (A \cap B^c)$$
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
$$|A| = |A \cap B| + |A \cap B^c| - (A \cap B) \cap (A \cap B^c)$$
$$|A| = |A \cap B| + |A \cap B^c| - |(A \cap A) \cap (B \cap B^c)|$$
$$|A| = |A \cap B| + |A \cap B^c|$$

# Probability Rules with Sets

Here are the basic definitions:
$$p(\text{A or B}) = p(A \cup B)$$
$$p(\text{A and B}) = p(A \cap B)$$
What this ends up being is
$$p(A \cup B) = p(A) + p(B) - p(A \cap B)$$
If you do just $p(A) + p(B)$ then you overcount the areas they intersect. In fact, we can expand this to an arbitrary number of unions.
$$p(A \cup B \cup C) = p(A) + p(B) + p(C) \\ - P(A \cap B) - p(A \cap C) - p(B \cap C)\\ + p(A \cap B \cap C)$$
This presents another problem. If we just added up the singular probabilities and subtracted all of the two-set intersections, we would subtract the overall intersection one too many times.

This is something called the

$$p(\text{A and B}) = P(A \cap B) = p(A)p(B|A) = p(B)p(A|B)$$

This is something called the

**principle of inclusion-exclusion**. If you extend it to n-unions. You just add the single, subtract the 2's, add the 3's, subtract the 4's, ...$$p(\text{A and B}) = P(A \cap B) = p(A)p(B|A) = p(B)p(A|B)$$

# Complements

Complements initially seem annoying to work with, but given these three properties, we can solve every problem.
$$p(A^c) = 1 - p(A)$$
$$(A \cup B)^c = A^c \cap B^c$$
$$(A \cap B)^c = A^c \cup B^c$$

# Example Problem

If A and B are independent, prove A and $B^c$ are independent. Also prove these are independent: $\\$
$A^c$ and $B^c$ $\\$
$A^c$ and B

This is what independent means: $P(A \cap B) = P(A)P(B)$ $$P(A^c \cap B^c) = 1 - P(A \cup B)$$ $$ = 1 - (P(A) + P(B) - P(A \cap B))$$ $$ = 1 - (P(A) + P(B) - P(A)P(B))$$ $$ = 1 - P(A) - P(B) + P(A)P(B)$$ $$ = (1 - P(A))(1- P(B))$$ $$ = P(A^c)P(B^c)$$

This is what independent means: $P(A \cap B) = P(A)P(B)$ $$P(A^c \cap B^c) = 1 - P(A \cup B)$$ $$ = 1 - (P(A) + P(B) - P(A \cap B))$$ $$ = 1 - (P(A) + P(B) - P(A)P(B))$$ $$ = 1 - P(A) - P(B) + P(A)P(B)$$ $$ = (1 - P(A))(1- P(B))$$ $$ = P(A^c)P(B^c)$$