Every probability question in this course starts the same way: name the sample space, then describe an event inside it. An event is just a subset of that sample space, so the tools for combining subsets carry over directly. Union, intersection, and complement are already familiar; here they become the vocabulary for combining outcomes instead of abstract elements.
Picture the sample space as a full box of marbles, one per outcome. Event A is a handful pulled out and set on a red cloth. Event B is a different handful set on a blue cloth, and some marbles can sit on both cloths at once. The union is every marble touched by either cloth. The intersection is only the marbles sitting on both cloths together. The complement of A is every marble still left in the box, untouched by the red cloth at all.
For a sample space S and events A, B \subseteq S, three operations combine them into new events. The union A \cup B is the event "A occurs or B occurs (or both)." The intersection A \cap B is the event "A occurs and B occurs." The complement \overline{A} = S \setminus A is the event "A does not occur." Each of these three results is itself a subset of S, so it can be combined again using the same three operations.
For sets A and B: A \cup B = \{\text{everything in } A,\text{ in } B,\text{ or in both}\} A \cap B = \{\text{everything in both } A \text{ and } B\}
Quick review
Practice from earlier coursework, pulled in for this topic.
From Discrete Mathematics I — Logic, Proof, and Structure
Let A = {2,4,6,8,10} and B = {4,8,12,16}. How many elements does A union B have?
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Why
Combined without duplicates: 2,4,6,8,10,12,16 -- seven distinct elements (4 and 8 are shared, counted once).
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Why
From Discrete Mathematics I — Logic, Proof, and Structure
Which logical connective does set intersection mirror?
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Why
Intersection requires membership in BOTH sets, exactly matching AND's requirement that both sides be true.
Build an event from a die roll
A single die is rolled. Click outcomes to build event A. Watch how A compares against the fixed event B = 'the roll is prime.'
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Worked example
Union and intersection of two events
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1
A die is rolled. Let A = "the roll is even" = \{2,4,6\} and B = "the roll is prime" = \{2,3,5\}. Find A \cup B and A \cap B.
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List every outcome belonging to A or B (or both) for the union; list only outcomes belonging to both for the intersection.
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3\boxed{A \cup B = \{2,3,4,5,6\}, \qquad A \cap B = \{2\}}
Worked example
The complement of an event
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1
Using the same die roll, let C = "the roll is at least 4" = \{4,5,6\}. Find \overline{C}.
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\overline{C} contains every outcome in S = \{1,2,3,4,5,6\} that is not in C.
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3\boxed{\overline{C} = \{1,2,3\}}
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Where this goes next
Chapter 1 (Probability as a measure)
Nothing here is new mathematics; it is Chapter 6's set vocabulary, reused for events instead of abstract sets. The next section attaches a number to each of these sets. Every rule for combining those numbers, including the addition rule, inclusion-exclusion, and the complement rule, is the arithmetic shadow of a set operation practiced here. A shaky grip on union versus intersection now turns into a sign error later.
Check your understanding (counts toward your progress)
A card is drawn from a standard 52-card deck. Let A = 'the card is a heart' (13 cards) and B = 'the card is a face card' (12 cards, 3 per suit). How many cards belong to A ∩ B?
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Why
A ∩ B is hearts that are also face cards — exactly the 3 face cards (J, Q, K) in the heart suit.
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List which of the 13 hearts also happen to be a jack, queen, or king.
Which of these is guaranteed true for ANY two events A and B?
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Not guaranteed — if B ⊆ A, the union equals A exactly, with no increase.
A ∩ B can absolutely be empty — that's exactly what 'mutually exclusive' means.
Two events sharing no outcomes at all (mutually exclusive) is a normal, common case.
Why
The intersection is always a subset of the union (every outcome in both A and B is also in A or B), so it can never have more outcomes.
Test each choice against a mutually exclusive pair of events first — that's the case most likely to break a false claim.
A die is rolled. Let A = \{2,4,6\} and B = \{1,2,3\}. What is A \cap B?
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That's the union A ∪ B, not the intersection — the intersection keeps only outcomes in BOTH sets.
A and B do share an outcome (2), so the intersection isn't empty.
Those are the outcomes in B but not A — check which outcomes appear in both sets, not just one.
Why
The intersection keeps only outcomes belonging to both A and B. Only 2 appears in both lists.
Let A = \{1,2,3,4,5\} and B = \{3,4,5,6,7\}. How many elements are in A - B (elements of A that are NOT in B)?
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Why
A - B keeps only elements of A that don't also appear in B. Removing 3, 4, 5 (shared with B) from A leaves {1,2} — 2 elements.
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Why
Check your understanding (counts toward your progress)
A die is rolled. Let C = \{1,2\}. What is \overline{C}?
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That's C itself, not its complement — the complement is everything NOT in C.
The complement isn't empty here; most outcomes in S aren't in C.
That's the whole sample space S, not the complement of C specifically.
Why
The complement of C is every outcome in S = {1,...,6} that isn't in C.
A standard 52-card deck is the sample space. Let H = 'the card is a heart' (13 cards). How many cards are in \overline{H}?
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Why
The complement contains every card that isn't a heart: 52 - 13 = 39.
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Why
The complement is every outcome in the sample space that ISN'T in H.
Events A and B are mutually exclusive. What must be true?
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Mutually exclusive events with positive probability are actually DEPENDENT, not independent — knowing A happened tells you B definitely didn't.
Mutually exclusive only means A and B share no outcomes — it says nothing about covering the whole sample space.
Mutually exclusive events typically aren't equal at all; in fact if A = B and both are nonempty, they'd share every outcome.
Why
Mutually exclusive means A and B share no outcomes, so their intersection is the empty event, with probability 0.
Start from the definition: mutually exclusive means A and B share zero outcomes.
Practice: Union, intersection, and difference (4 problems · doesn’t affect your progress)
A store tracks two groups of customers: R = 'redeemed a coupon' and M = 'joined the mailing list.' Which set describes customers who did BOTH?
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R ∪ M is customers who did at least one of the two — that includes people who only redeemed a coupon or only joined the list, not just the ones who did both.
R - M is customers who redeemed a coupon but did NOT join the mailing list — that's not 'both.'
The complement of R is customers who did NOT redeem a coupon at all — unrelated to the mailing list.
Why
Doing both things at once is exactly what intersection means: R ∩ M keeps only the customers in both groups.
R ∩ M
Let P = \{2,3,5,7,11\} and Q = \{1,3,5,7,9\}. How many elements are in P \cup Q?
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Why
List every element that appears in P or Q at least once, with no repeats: {1,2,3,5,7,9,11} — 7 elements total.
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Why
Using the same sets P = \{2,3,5,7,11\} and Q = \{1,3,5,7,9\}, how many elements are in P \cap Q?
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Why
Only 3, 5, and 7 appear in both P and Q, so the intersection has 3 elements.
Correct answer
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Why
Look for elements written in BOTH lists, not just one.
Using the same sets P = \{2,3,5,7,11\} and Q = \{1,3,5,7,9\}, how many elements are in Q - P (elements of Q not in P)?
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Why
Q - P keeps only Q's elements that aren't shared with P. Removing 3, 5, 7 (shared) from Q leaves {1,9}, so 2 elements.
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Why
Start from Q's list, then cross off anything that also shows up in P.
Practice: The complement and the universal set (5 problems · doesn’t affect your progress)
The complement of an event A answers which question?
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That's the intersection of A with some other event B, not the complement.
That's the union of A with some other event, not the complement.
That's just A's size (a count), not a description of a different event.
Why
The complement of A is everything in the sample space S that is NOT in A — a new event built entirely from what A leaves out.
What's NOT in A, within the sample space?
A die is rolled, so S = \{1,2,3,4,5,6\}. Let D = \{2,4,6\} (even rolls). How many elements are in \overline{D}?
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Why
S has 6 outcomes; D has 3 of them, so \overline{D} has the remaining 6 - 3 = 3 outcomes: {1, 3, 5}.
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Why
The complement is every outcome in S that isn't already in D.
A card is drawn from a standard 52-card deck, so |S| = 52. Let F = 'the card is a face card' with |F| = 12. How many elements are in \overline{F}?
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Why
|\overline{F}| = |S| - |F| = 52 - 12 = 40.
Correct answer
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Why
Subtract the size of F from the size of the whole sample space.
What is the complement of the complement of an event A, i.e. \overline{\overline{A}}?
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The empty set is what A and its complement have in common (their intersection), not the double complement.
S is what A and its complement combine to form (their union), not the double complement.
It does simplify — taking the complement twice always lands back where you started.
Why
Complementing removes A from S; complementing again removes what's left, restoring exactly A.
A itself
The sample space is S=\{1,2,\ldots,12\}. Let P = \{2,3,5,7,11\} (primes in S) and E = \{2,4,6,8,10,12\} (evens in S). How many elements are in \overline{P \cup E} (elements of S in NEITHER set)?
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Why
P ∪ E has 10 distinct elements (2 is counted once even though it's in both P and E). S has 12 elements total, so the complement — elements in neither set — has 12 - 10 = 2 elements: {1, 9}.
Correct answer
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Why
First find how many elements are in P ∪ E (careful — 2 is in both, don't double count it), then subtract from the size of S.
The complement of a union is everything left over in S once every element that's in P OR E is removed.