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Combining Events: Unions, Intersections, Complements

Events are sets, so you can combine them like sets — 'or', 'and', and 'not'. Master this small grammar and you can rewrite almost any probability question into a form you can actually compute.

Three words: or, and, not

In the previous guide an event turned out to be nothing more than a subset of the sample space — a bundle of outcomes you single out to ask a question about. The quiet payoff of that definition arrives now: because events are just sets, you can combine them with exactly the same operations you already use for sets, and each operation translates into a plain English word. There are three core moves, and they cover an astonishing amount of ground.

The union of A and B, written A ∪ B, is the event 'A or B (or both) happens'. The intersection, written A ∩ B, is the event 'A and B both happen'. The complement of A, written A^c, is the event 'A does not happen' — everything in the sample space that is outside A. Picture rolling one die. Let A = 'even' = {2, 4, 6} and B = 'at least 4' = {4, 5, 6}. Then A ∪ B = {2, 4, 5, 6}, A ∩ B = {4, 6}, and A^c = {1, 3, 5}. Notice the small trap right away: 'or' in mathematics is inclusive, so A ∪ B includes the outcome 4, which is in both. This whole toolkit is the algebra of events.

The grammar has rules of its own

These three operations are not a loose collection of tricks; they obey a tidy algebra, and that is exactly what makes them useful. Union and intersection each commute (A ∪ B = B ∪ A) and associate, so the order you list events in never matters. They also distribute over one another, just like multiplication over addition in ordinary arithmetic: A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). And taking a complement twice brings you home: (A^c)^c = A. The point of these laws is leverage — a messy event can be rewritten into an equivalent, simpler one before you ever attach a number to it.

The single most useful pair of rules links 'not' with 'or'/'and': De Morgan's laws. They say (A ∪ B)^c = A^c ∩ B^c and (A ∩ B)^c = A^c ∪ B^c. Read aloud, the first is pure common sense: 'neither A nor B happened' is the same as 'A didn't happen and B didn't happen'. The second says 'A and B did not both happen' means 'A failed or B failed'. The deep idea is that a complement flips the operation — union becomes intersection, intersection becomes union — as it crosses the bar. This is the workhorse you reach for whenever an event about 'at least one' is awkward but its opposite, 'none', is easy.

(A or B)^c  =  (not A) and (not B)
(A and B)^c =  (not A) or  (not B)

Example (one die):
  A = even      = {2,4,6}
  B = at least 4= {4,5,6}
  A or B        = {2,4,5,6}
  (A or B)^c    = {1,3}
  A^c and B^c   = {1,3,5} and {1,2,3} = {1,3}   (matches)
De Morgan's laws in words and checked on a single die.

When events cannot overlap

A special and important case is when two events share no outcomes at all: A ∩ B is empty. We call such events mutually exclusive (or disjoint) — they cannot both happen on the same run. 'The die shows 2' and 'the die shows 5' are mutually exclusive; 'even' and 'at least 4' are not, because 4 and 6 sit in both. This matters enormously for the next guide, because adding probabilities is only safe when events are disjoint. The most familiar pair is an event and its own complement: A and A^c never overlap, and together they fill the whole sample space.

A close cousin is the idea of exhaustive events: a collection whose union is the whole sample space, so at least one of them is guaranteed to occur. When events are both mutually exclusive and exhaustive, they form a partition — they carve the sample space into non-overlapping pieces that leave nothing out. The faces of a die, {1}, {2}, {3}, {4}, {5}, {6}, are the simplest partition. Partitions are quietly central; they are how the law of total probability and conditioning will later let you break a hard question into easy slices.

Two warnings the pictures hide

Venn diagrams are wonderful for intuition, but they quietly bake in two assumptions worth saying out loud. First, the size of a region in a Venn diagram suggests probability, yet area is not probability — a tiny-looking sliver can carry a huge chance and a fat blob almost none, depending on how the underlying outcomes are weighted. The diagram shows which outcomes belong to which event; it does not, by itself, tell you how likely anything is. Second, the diagram makes events look like fixed blobs, but the same English sentence can map to very different sets depending on how finely you defined the sample space in guide one.

The 'at least one' versus 'none' duality is where De Morgan earns its keep in practice. Suppose three independent-looking switches each can be on, and you want the event 'at least one is on'. Computing that directly means juggling overlaps; but its complement is the clean single event 'all three are off'. So you compute the easy complement and subtract from 1 — the complement rule, P(A) = 1 - P(A^c). We have not proved that rule yet (the axioms in the next guide will), but you can already see why it will be the single most-used shortcut in the whole subject: rephrasing 'at least one' as 'one minus none'.

Counting an 'or': inclusion-exclusion

Here is the question that combining events is leading toward. If A and B overlap, how big is A ∪ B? A tempting but wrong move is to add the sizes of A and B — that double-counts the outcomes living in both, namely A ∩ B. Correcting for that overcount gives the inclusion-exclusion principle: the size of a union equals the sum of the sizes minus the size of the overlap. In probability language this becomes P(A or B) = P(A) + P(B) - P(A and B). Add the parts, then subtract the part you counted twice.

  1. Single out the events. One die: A = 'even' = {2, 4, 6}, B = 'at least 4' = {4, 5, 6}. We want P(A or B), with a fair die so each face has probability 1/6.
  2. Add the parts. P(A) = 3/6 and P(B) = 3/6, so a naive sum gives 6/6 = 1 — already suspicious, since A ∪ B does not cover face 1 or 3.
  3. Find the overlap. A ∩ B = {4, 6}, so P(A and B) = 2/6.
  4. Subtract the double-count. P(A or B) = 3/6 + 3/6 - 2/6 = 4/6 = 2/3, which matches counting A ∪ B = {2, 4, 5, 6} directly.

When A and B are mutually exclusive the overlap term vanishes, P(A and B) = 0, and the formula collapses to the simple addition P(A or B) = P(A) + P(B). That special case is exactly the additivity that the axioms will take as a starting assumption. For three or more events the pattern continues with alternating signs — add the singles, subtract the pairs, add back the triples — but the two-event version above is the one you will use constantly. None of this is yet 'probability theory' in the formal sense; it is the algebra of events, the grammar that the axioms in the next guide will breathe numbers into.