Set Operations in Probability — Section 2: Set Theory & Combinatorics

Because events are sets, every operation on sets has a corresponding probability rule. The four you'll use over and over:

Union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . Subtracting the intersection avoids double-counting the overlap.

Intersection: for independent events, $P(A \cap B) = P(A)\, P(B)$ . For dependent events, $P(A \cap B) = P(A \mid B)\, P(B)$ .

Complement: $P(A^c) = 1 - P(A)$ . Often the fastest path to a hard probability is through the complement (e.g. "at least one" $= 1 -$ "none").

De Morgan's laws: $(A \cup B)^c = A^c \cap B^c$ and $(A \cap B)^c = A^c \cup B^c$ . They let you swap unions for intersections under negation.

For three or more events, inclusion–exclusion generalizes:

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)

The pattern continues for any number of events: alternate sum of intersections of all sizes.

Because events are sets, every operation on sets has a corresponding probability rule. The four you'll use over and over:

Union: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ . Subtracting the intersection avoids double-counting the overlap.

Intersection: for independent events, $P(A \cap B) = P(A)\, P(B)$ . For dependent events, $P(A \cap B) = P(A \mid B)\, P(B)$ .

Complement: $P(A^c) = 1 - P(A)$ . Often the fastest path to a hard probability is through the complement (e.g. "at least one" $= 1 -$ "none").

De Morgan's laws: $(A \cup B)^c = A^c \cap B^c$ and $(A \cap B)^c = A^c \cup B^c$ . They let you swap unions for intersections under negation.

For three or more events, inclusion–exclusion generalizes:

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)

The pattern continues for any number of events: alternate sum of intersections of all sizes.