The 3-set Venn diagram displays three overlapping circles labeled A, B, and C. Eight numbered segments represent all possible outcome combinations from three events.
The center segment (#1) shows the triple intersection where all three events occur simultaneously. Segments #2-#7 represent various two-way intersections and single-event-only regions. Segment #8 lies outside all circles.
View the pre-loaded "Demographics Study" example showing how three marginals and four constraints — P(A∩Bc)=0.4, P(A∩Cc)=0.18, P(B∩C)=0.38 and P(A∩B∩C)=0.08 — determine all eight region probabilities through systematic calculation.
Navigating Eight Regions
Click any numbered segment (1-8) to select and highlight it. The corresponding outcome displays in the middle panel with its probability value and explanation.
Segments #1-7 lie within at least one circle. Segment #8 represents Ac∩Bc∩Cc - outcomes where none of the three events occur.
Hover over segments to preview which outcome they represent. Hover over outcomes in the list to see their diagram location. This bidirectional connection clarifies complex notation like A∩Bc∩C.
Understanding the Eight Outcomes
Segment #1 (A∩B∩C): All three events occur - the center intersection.
Segments #2-4: Two events occur, one doesn't - the three two-way intersections excluding the center.
Segments #5-7: Only one event occurs - portions of single circles that don't overlap with others.
Segment #8 (Ac∩Bc∩Cc): None of the events occur - outside all three circles.
These eight mutually exclusive regions partition the entire sample space, and their probabilities sum to exactly 1.000.
Demographics Study: Eight Regions That Reproduce Every Given
The tool opens on this problem. It gives P(A)=0.5 for Woman, P(B)=0.5 for Employed and P(C)=0.62 for Academic, together with four constraints: P(A∩Bc)=0.4, P(A∩Cc)=0.18, P(B∩C)=0.38 and P(A∩B∩C)=0.08.
From those the eight regions come out as 0.08, 0.02, 0.24, 0.16, 0.30, 0.10, 0.00 and 0.10, in the order the tool lists them.
Regions 1 to 8 are 0.08, 0.02, 0.24, 0.16, 0.30, 0.10, 0.00 and 0.10. They sum to 1 and reproduce all three marginals and all four constraints. Region 7 is exactly zero: every male academic in this survey is employed.
They sum to exactly 1, and — the check worth doing every time — adding them back the other way returns every given you started from. The three marginals: 0.08+0.02+0.24+0.16=0.5, 0.08+0.02+0.30+0.10=0.5, and 0.08+0.24+0.30+0.00=0.62. The four constraints likewise: 0.24+0.16=0.4, 0.02+0.16=0.18, 0.08+0.30=0.38, and region 1 is 0.08 by itself.
Region 7 is worth a second look because it is exactly 0.00: nobody in this survey is a non-woman, not employed and academic. Every male academic here is employed. A zero region is legitimate — it says that combination simply does not occur — and it is why P(A∪B∪C)=1−0.10=0.90 rather than something larger.
The two conditionals the regions hand you show how strongly the events pull against each other. Among women, the employed fraction is 0.1/0.5=0.2; among everyone else it is 0.4/0.5=0.8. Employment and being a woman are very far from independent in this dataset, and the region sizes say so directly.
Why Four Constraints Are Needed, Not Two
Eight regions are eight unknowns, so eight independent equations are needed to pin them down. The three marginals and the requirement that everything sums to 1 supply four. Each given constraint supplies one more, so four constraints are exactly enough.
That accounting is easy to get wrong. The three marginals, the total, and just the two constraints P(A∩Bc) and P(A∩Cc) come to six equations for eight unknowns — a system with infinitely many solutions, not one.
Region 1 selected: the triple intersection
Eight regions need eight equations. Three marginals and the total give four; without four constraints the answer is not unique. With P(B n C) = 0.38 but no value for region 1, any triple intersection from 0.08 to 0.10 fits - the given 0.08 picks one out.
The triple intersection is the region that resolves it. Add P(B∩C)=0.38 and the eight unknowns collapse to a single free parameter, the value of region 1: every choice of P(A∩B∩C) between 0.08 and 0.10 yields eight non-negative regions consistent with all seven equations. Only the eighth given, P(A∩B∩C)=0.08, picks one of them out.
That range has a boundary worth noticing. At the low end, 0.08, region 7 falls to zero; push the triple intersection any lower and region 7 would have to go negative, which no probability may do. The problem sits exactly on that edge.
The practical rule: count equations before solving. If a three-set problem hands you the marginals and only one or two intersections, it does not have a unique answer, and any single set of eight numbers you produce is a guess at which member of the family was intended.
Selecting a Region
Clicking a numbered box fills it gold with an amber border and leaves the other seven white. Selection is a toggle: clicking the same box again clears it.
The still below shows region 8, Ac∩Bc∩Cc — the one region that lies inside the sample space Ω but outside all three circles.
Region 8 selected: outside all three circles
A selected box fills gold with an amber border; the other seven stay white. Region 8 holds 0.10 here despite having nothing drawn around it - the omega in the corner is the reminder that the rectangle is the sample space. Circle sizes are fixed, so area shows membership, not probability.
Region 8 is easy to overlook because nothing is drawn around it, yet it carries 0.10 here, as much as region 6 and more than region 2. The Ω in the top-left corner is the reminder: the rectangle is the sample space, and everything not enclosed by a circle still belongs to it.
The diagram deliberately carries no numbers. The circles and the eight boxes show you *where* each outcome lives; the probabilities live in the list beside them, and selecting a region is what ties the two together.
That split matters more with three sets than with two. All three circles are drawn at the same radius and in fixed positions, so the picture encodes set membership only — never probability. Region 5 takes up about 11% of the area inside the circles but holds 0.30, the largest probability of any region in the diagram; region 4 takes up twice that area, about 23%, and holds only 0.16. Region 7 is starker still: roughly 16% of the drawn interior, and a probability of exactly zero.
Using Show Calculations
Toggle "Show Calculations" to reveal step-by-step solutions for all eight regions. The tool displays the logical sequence for solving the system of probability equations.
The solution typically starts by finding pairwise intersections: P(A∩B)=P(A)−P(A∩Bc). Then it solves for the triple intersection using all available constraints.
Once the triple intersection is known, other regions follow systematically. For example, P(A∩B∩Cc)=P(A∩B)−P(A∩B∩C). Each calculation builds on previous results.
Reading Probability Solutions
Each of the eight outcomes displays its calculated probability to three decimal places. In the Demographics example, P(A∩B∩C)=0.080 means 8% of the population are women, employed, and have an academic background.
Compare region sizes to understand relationships. Here P(A∩Bc∩C)=0.240 is three times P(A∩B∩C)=0.080, so academic women in this dataset are predominantly not employed.
The probabilities reflect both the given marginals P(A),P(B),P(C) and the specific constraints. Different constraints yield completely different regional distributions.
Solving Systems of Equations
Three-set problems require solving systems of equations. Eight regions mean eight unknowns, so eight independent equations are needed: the three marginals, the fact that everything sums to 1, and four given constraints.
The general approach: (1) Use constraints to find pairwise intersections like P(A∩B). (2) Set up equations relating the triple intersection to known values. (3) Solve for the center region. (4) Calculate remaining regions systematically.
Click "Show Calculations" to see the General Solution Steps outlining this process. Individual region calculations appear when you select specific segments.
The Inclusion-Exclusion Principle
For three sets, the inclusion-exclusion principle calculates P(A∪B∪C) - the probability that at least one event occurs.
The formula adds individual probabilities, subtracts pairwise overlaps (to correct double-counting), then adds back the triple intersection (which was subtracted three times but should only be subtracted twice).
In the diagram, P(A∪B∪C) equals segments #1 through #7 - everything inside at least one circle.
Conditional Probability with Three Events
Conditional probability extends to three events. P(A∩B∣C) asks: given C occurred, what's the probability both A and B occur?
Visually, restrict to circle C (segments #1, #3, #5, #7), then find what fraction also lies in both A and B (only segment #1).
Calculate as P(A∩B∣C)=P(C)P(A∩B∩C). If P(A∩B∩C)=0.08 and P(C)=0.62, then P(A∩B∣C)=0.08/0.62≈0.129 or 12.9%.
When to Use 3-Set Diagrams
Use 3-set Venn diagrams when analyzing three distinct characteristics or categories simultaneously. Common applications include:
Survey analysis: Demographic studies with three attributes (age group, employment status, education level)
Medical research: Patients with three conditions or risk factors
Quality control: Products with three types of potential defects
Market segmentation: Customers categorized by three preferences or behaviors
For four or more events, contingency tables or tree diagrams become more practical than Venn diagrams.