Venn Diagrams and Probability

A Venn diagram organises events by their membership relationships. The rectangle represents the sample space. Each circle represents an event, and overlaps show outcomes belonging to more than one event.

The numbers placed inside the regions can be counts or probabilities, but they must be interpreted consistently. A schematic circle’s area is not automatically proportional to its event’s probability.

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Fill the overlap first

Use the synthetic roster of 40 club members introduced in the addition-rule lesson. Art membership A has 24 members, board-games membership B has 18, and ten belong to both.

Place 10 in the overlap first. Art only then has $24-10=14$ , and board games only has $18-10=8$ . The union contains $14+10+8=32$ , leaving $40-32=8$ outside both circles.

The circle areas are schematic; use the region counts

Synthetic roster with uniform member selection. Circle sizes do not represent event probabilities.

Diagram explanation and exact data

The four disjoint regions total 40. Art contains 14+10=24 members; games contains 10+8=18. Exactly one means the two exclusive circle portions, not the overlap.

The circle areas are schematic; use the region counts: exact reference data
Region	Count	Probability
Art only	14	7/20
Both	10	1/4
Games only	8	1/5
Neither	8	1/5

The four displayed regions are disjoint and sum to 40. In contrast, the headline event counts 24 and 18 overlap and should not be added without adjustment. The word “both” normally gives an inclusive intersection count: for two sets, it is exactly the shared region.

Read the complete event, not just one region

The event A includes art only and the overlap, so $P(A)=(14+10)/40=3/5$ . The event B includes board games only and the overlap, giving $18/40=9/20$ .

For a uniformly selected member,

P(A\cap B)=\frac{10}{40}=\frac14,\qquad P(A\cup B)=\frac{32}{40}=\frac45.

The complement $A^c$ includes board games only and neither. Its count is $8+8=16$ , so $P(A^c)=2/5$ . The complement of the union, neither event, contains only the outer region and has probability $8/40=1/5$ .

Translate less familiar set expressions

Set notation can describe a region more precisely than a long phrase:

Translate less familiar set expressions: reference table
Expression	Meaning	Count
$A\cap B^c$	Art only	14
$A^c\cap B$	Board games only	8
$A^c\cap B^c$	Neither	8
$(A\cap B)^c$	Not both	30
$(A\cap B^c)\cup(A^c\cap B)$	Exactly one	22

“Not both” includes three regions, while “neither” includes only one. The distinction is visible once the complement operation is applied to the entire expression inside the parentheses.

The full sample-space boundary matters. An outcome outside both circles is still part of the experiment; it is not an unrecorded or impossible outcome.

A condition changes the denominator

To find $P(A\mid B)$ , restrict attention to the B circle. Ten of its eighteen members are also in A:

P(A\mid B)=\frac{10}{18}=\frac59.

For $P(A\mid B^c)$ , the condition is outside B. That region includes 14 art-only and 8 neither members, so

P(A\mid B^c)=\frac{14}{22}=\frac7{11}.

Using 40 in either denominator would calculate a joint probability instead of the requested conditional. A circle diagram does not change the rule: identify the condition’s complete region, then use its total as the reference group.

Recover unknown region counts

In a separate group of 60, suppose $|A|=35$ , $|B|=28$ and $|A\cup B|=50$ . The overlap is

|A\cap B|=35+28-50=13.

Therefore A only has 22, B only has 15, and neither has 10. The probability of exactly one is $(22+15)/60=37/60$ .

This reconstructed diagram also checks feasibility. Every region must be non-negative. An overlap larger than either entire event or a union larger than the sample space would reveal inconsistent information.

Three-set diagrams need inclusive counts handled carefully

In a three-circle diagram, fill the triple intersection before the pair-only regions. If $|A\cap B|=8$ and $|A\cap B\cap C|=3$ , the part in A and B but not C contains $8-3=5$ .

For the 50-person example with event counts 20, 18 and 15, pairwise counts 8, 6 and 5, and triple count 3, the pair-only regions are 5, 3 and 2. The A-only count is $20-5-3-3=9$ ; B only is 8 and C only is 7. Their seven inside-region counts total 37, leaving 13 outside all circles.

A pairwise intersection includes the triple overlap unless the wording says “exactly these two”. Subtracting inclusive intersections without first identifying their meaning is a common source of inconsistent region totals.

Counts, probabilities and drawn areas

A count diagram can be converted to probabilities by dividing every disjoint region by the same total. A probability diagram should have disjoint region weights adding to one. Mixing counts in one region with percentages in another makes the diagram unusable.

The circles here are schematic. Their boundaries show set membership, not measured area. In particular, the visual overlap cannot establish independence. That requires checking $P(A\cap B)=P(A)P(B)$ using the numbers. For this roster, $1/4\ne(3/5)(9/20)=27/100$ , so the two membership events are not independent under uniform selection from the roster.

Practice questions

In the 40-member example, how many belong to A only?
What is $P(A\mid B^c)$ in that example?
If a 60-person group has $|A|=35$ , $|B|=28$ and union count 50, find the overlap and neither count.

Check your answers

$24-10=14$ . The event total includes the overlap, which must be removed to isolate A only.
The condition not-B contains 22 members, 14 in A. The probability is $14/22=7/11$ .
The overlap is $35+28-50=13$ . Neither contains $60-50=10$ people.