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Statistics · Probability

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Venn diagrams

Using a Venn diagram to organise events: the intersection A∩B ("and"), the union A∪B ("or"), and the complement A′ ("not"); filling in the regions from the numbers in a problem (starting with the overlap); reading probabilities off the diagram; and the addition formula P(A∪B) = P(A) + P(B) − P(A∩B).

Statistics · Probability

Venn diagrams

Using a Venn diagram to organise events: the intersection A∩B ("and"), the union A∪B ("or"), and the complement A′ ("not"); filling in the regions from the numbers in a problem (starting with the overlap); reading probabilities off the diagram; and the addition formula P(A∪B) = P(A) + P(B) − P(A∩B).

Why it works

A Venn diagram draws each event as a circle inside a rectangle (the universal set ξ\xi, all possible outcomes). Where circles overlap, outcomes belong to both events. This turns "and / or / not" into regions you can simply count.

The three set operations:
  • Intersection ABA \cap B — outcomes in both AA and BB (the overlap).
  • Union ABA \cup B — outcomes in AA or BB (or both) — everything inside
either circle.
  • Complement AA' — outcomes not in AA — everything outside circle AA.
Below, 6060 students study French (FF), 3535 study German (GG), and 2020 study both, out of 100100:ξ25FG402015Fill the overlap first. The numbers in a Venn diagram are the outcomes in that region only, so they add up to the total. Always start with the intersection, then work outwards:
  • both =20= 20,
  • French only =6020=40= 60 - 20 = 40,
  • German only =3520=15= 35 - 20 = 15,
  • neither (outside both) =100(40+20+15)=25= 100 - (40 + 20 + 15) = 25.
Now any probability is just a region total over 100100. For example P(neither)=25100=0.25P(\text{neither}) = \tfrac{25}{100} = 0.25, and P(FG)=40+20+15100=0.75P(F \cup G) = \tfrac{40+20+15}{100} = 0.75.

The addition formula. Adding P(F)P(F) and P(G)P(G) counts the overlap twice, so

P(AB)=P(A)+P(B)P(AB).P(A \cup B) = P(A) + P(B) - P(A \cap B).

Here P(F)+P(G)P(FG)=0.60+0.350.20=0.75P(F) + P(G) - P(F\cap G) = 0.60 + 0.35 - 0.20 = 0.75 — the same 0.750.75. Subtract the intersection exactly once to undo the double-count.

Useful regions. "Exactly one of AA, BB" is the two only parts added: P(A)+P(B)2P(AB)P(A) + P(B) - 2P(A\cap B). "Neither" is P((AB))=1P(AB)P((A\cup B)') = 1 - P(A\cup B).