Events are sets — combine them with ∩, ∪, and the complement, then add their probabilities the honest way.
In the last two lessons an event was a subset of the sample space Ω, and a probability was a number we counted out — favorable over total. But real questions rarely ask about one bare event. They ask about combinations: rainy and windy, a heart or a face card, at least one six in four rolls. The beautiful news is that events are sets, so the whole toolkit from Stage 19 · Set Operations — intersection ∩, union ∪, complement — walks straight back in. Once you can name a compound event as a set, two short rules turn its probability into arithmetic you already trust.
Let the trial be tomorrow's weather, and define two events on the same Ω:
Three new events come for free, just by reading them as set operations:
These are exactly the Venn pictures from Stage 19, but now each region is an event, and the area share of each region (when outcomes are equally likely) is its probability. Naming a wordy question as A ∩ B, A ∪ B, or Ā is the whole first move.
Events are sets on the sample space Ω. "and" is intersection A ∩ B; "or" is union A ∪ B; "not" is the complement Ā. Translate the English, then the set picture tells you everything.
Conclusion. Every "and / or / not" question is a set operation on Ω: name the region first, and the probability follows from its share.
Two relationships between events deserve their own names, because each one collapses a formula.
Two events are mutually exclusive when they cannot happen on the same trial — their intersection is empty:
A ∩ B = ∅
Their circles do not overlap at all. Rolling one die, "the result is even" and "the result is 5" are mutually exclusive — no single roll is both. But mutually exclusive events need not cover everything; here "even or 5" still misses 1 and 3.
Complementary events are a special, stronger case: they are mutually exclusive and together they fill Ω. The pair A and Ā is the model:
A ∪ Ā = Ω and A ∩ Ā = ∅
A and Ā split the sample space cleanly into two pieces with no overlap and no gap — a partition into two parts. "Rolls an even number" and "rolls an odd number" are complementary; "rolls an even number" and "rolls a 5" are merely mutually exclusive (they leave a gap).
Draw one card. Let A = "a heart" and B = "a spade." These are mutually exclusive (no card is both), since A ∩ B = ∅ — but not complementary, because a diamond or club is in neither. By contrast A = "a heart" and Ā = "not a heart" are complementary: every card is in exactly one of them.
Mutually exclusive is not the same as independent. Disjoint events that aren't trivial are actually the most dependent events there are: if one happens, the other cannot — knowing A occurred changes B's chance to 0. Independence (the topic of 33.4) means knowing A changes nothing. Different — even opposite — ideas; don't let the long words blur them.
Conclusion. Mutually exclusive means A ∩ B = ∅ (no overlap); complementary adds A ∪ B = Ω (no gap), so A and Ā partition the whole sample space.
How likely is A ∪ B? You might guess P(A) + P(B) — and you'd be off whenever the circles overlap, because the lens A ∩ B got counted twice, once inside A and once inside B. Subtract the double-count back out, and you have the general addition formula:
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Read it as a count of outcomes: |A ∪ B| = |A| + |B| − |A ∩ B|. The subtraction is the whole story — it repairs the overlap you counted twice.
Draw one card. Let A = "a heart" (13 cards) and B = "a face card" — J, Q, K of any suit (12 cards). The overlap A ∩ B is the heart face cards J♥, Q♥, K♥ (3 cards). So
|A ∪ B| = 13 + 12 − 3 = 22, and P(A ∪ B) = 2252 = 11/26 ≈ 0.423.
If you had forgotten to subtract the 3, you'd report 25/52 — too big, because J♥, Q♥, K♥ would each be counted twice.
When A and B are mutually exclusive, the overlap is empty, so P(A ∩ B) = 0 and the formula simplifies to the version you may have met first:
A, B disjoint ⟹ P(A ∪ B) = P(A) + P(B)
That special case is just the general rule with nothing to subtract. Always check first whether the events can overlap; if they can, you owe the subtraction.
The classic blunder is P(A ∪ B) = P(A) + P(B) for events that can overlap. That over-counts the lens A ∩ B. Only drop the subtraction when you have checked that A ∩ B = ∅. "Heart or face card" overlaps; "heart or spade" does not.
Conclusion. P(A ∪ B) = P(A) + P(B) − P(A ∩ B); the minus sign repairs the overlap, and it vanishes only when the events are mutually exclusive.
Because A and Ā partition Ω, their probabilities must add to 1: P(A) + P(Ā) = 1. Rearranged, that gives a shortcut that saves enormous work:
P(A) = 1 − P(Ā)
Whenever the complement is easier to count than the event itself, go around the back. The signature case is the phrase "at least one." The opposite of "at least one success" is the single, tidy event "none":
P(at least one) = 1 − P(none)
Roll a fair die 4 times. What is the probability of at least one six? Counting "exactly one, exactly two, …" is painful. The complement is one clean event — "no six on any roll." Each roll avoids a six with probability 56, and the rolls are independent, so
P(no six) = (5/6)4 ≈ 0.482, hence P(at least one six) = 1 − (5/6)4 ≈ 0.518.
Just over a coin-flip's chance — and one subtraction did the whole job.
The pattern generalizes: for n rolls, P(at least one six) = 1 − (5/6)n, which climbs toward 1 as you roll more. (Why "none" multiplies the per-roll chances — the independence rule — is the heart of 33.4.)
Conclusion. P(A) = 1 − P(Ā); when "at least one" is hard, count "none" instead and subtract from 1.
| English | As a set | The rule |
|---|---|---|
| A and B | A ∩ B | The overlapping lens — both occur |
| A or B | A ∪ B | P(A ∪ B) = P(A) + P(B) − P(A ∩ B) |
| not A | Ā | P(A) = 1 − P(Ā) |
| A, B can't co-occur | A ∩ B = ∅ | Mutually exclusive ⟹ P(A ∪ B) = P(A) + P(B) |
| A, Ā: no overlap, no gap | A ∪ Ā = Ω | Complementary — they partition Ω, P(A) + P(Ā) = 1 |
| at least one success | not "none" | P(at least one) = 1 − P(none) |
| mutually exclusive vs. independent | — | Different ideas — disjoint is the opposite of independent (→ 33.4) |
Let A = "the card is a club" and B = "the card is a queen," drawn from a standard 52-card deck. Describe A ∩ B and A ∪ B in words, and give |A ∩ B|.
In a class, P(plays soccer) = 0.5, P(plays tennis) = 0.3, and P(plays both) = 0.1. Find the probability a student plays at least one of the two sports.
Are "draw a red card" and "draw a king" mutually exclusive? Are they complementary? Explain.
A bag has 5 red and 3 blue marbles. Two are drawn (without replacement). Use a complement to find P(at least one blue).
Flip a fair coin 5 times. What is the probability of getting at least one head?
Events A and B satisfy P(A) = 0.6, P(B) = 0.5, and P(A ∪ B) = 0.9. Find P(A ∩ B), then decide whether A and B are mutually exclusive.
Six questions to lock it in. Tap the answer you think is right.
This lesson treats events as sets on the sample space Ω and builds the algebra of combining them: intersection A ∩ B, union A ∪ B, and complement Ā, the distinction between mutually exclusive (A ∩ B = ∅) and complementary (A ∪ Ā = Ω) events, the general addition rule P(A ∪ B) = P(A) + P(B) − P(A ∩ B), and the complement shortcut P(A) = 1 − P(Ā) behind every "at least one" problem. It supports CCSS HSS-CP.A.1 — describe events as subsets of a sample space using unions, intersections, and complements — and CCSS HSS-CP.B.7 — apply the Addition Rule, P(A or B) = P(A) + P(B) − P(A and B). A good check for understanding: ask your learner to give one pair of events that are mutually exclusive but not complementary, and to explain why "mutually exclusive" and "independent" are different ideas — that contrast returns in 33.4.