the algebra of events (union, intersection, complement)
Because events are sets of outcomes, we can combine them the same way we combine any sets, and each set operation translates into a plain-language statement about what happens. There are three core moves. The union of A and B, written A ∪ B, is the event 'A or B (or both) occurs' — it gathers every outcome that is in A, in B, or in both. The intersection, written A ∩ B, is the event 'A and B both occur' — only the outcomes lying in both sets. The complement of A, written A^c (or 'not A'), is the event 'A does not occur' — everything in the sample space outside A.
Reading the translation table the other way is what makes this practical: the word 'or' in a probability question becomes a union, the word 'and' becomes an intersection, and the word 'not' becomes a complement. A small picture helps — imagine two overlapping circles inside a rectangle (the sample space). The whole shaded pair is A ∪ B; the lens where they overlap is A ∩ B; the part of the rectangle outside circle A is A^c. From these three you can also build differences, like 'A but not B' = A ∩ B^c, the part of A that misses B.
These operations obey the familiar laws of set algebra — they commute (A ∪ B = B ∪ A), associate, and distribute (A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)), and the complement of the complement gives you back A. This matters because it lets you rewrite a messy event into an equivalent, easier form before computing its probability. The algebra of events is the grammar; the axioms of probability, coming next, are the rules for attaching numbers to whatever sentences this grammar lets you write.
Roll one die. Let A = 'even' = {2,4,6} and B = 'at least 4' = {4,5,6}. Then A ∪ B = {2,4,5,6} (even or at least 4), A ∩ B = {4,6} (even and at least 4), and A^c = {1,3,5} (not even).
Or = union, and = intersection, not = complement — the three translations you reach for constantly.
Probabilities do not simply add across a union when events overlap: P(A or B) = P(A) + P(B) - P(A and B), because the overlap would otherwise be counted twice.