The 2-set Venn diagram displays two overlapping circles labeled A and B. Four numbered segments represent all possible outcomes: both events (intersection), A only, B only, and neither event (complement).
View the pre-loaded example problems by clicking the problem name buttons. Each problem shows event descriptions, marginal probabilities P(A) and P(B), and given constraints that help determine all four region probabilities.
The diagram automatically calculates probabilities for all four segments based on the given information. Click any numbered segment to highlight it and see detailed calculations.
Clicking and Hovering Interactions
Click any numbered segment (1-4) in the diagram to select it. Selected segments highlight in gold and display their corresponding outcome below. Click again to deselect.
Hover over segments to preview which outcome they represent. The segment highlights in yellow, and the matching outcome in the list also highlights.
Hover over outcomes in the left panel to see their location in the diagram. This two-way interaction helps connect symbolic notation like A∩B with visual diagram regions.
Understanding the Four Regions
Segment #1 (A∩B): The intersection where both circles overlap. Both events occur simultaneously. This is the most central region of probability calculations.
Segment #2 (A∩Bc): Event A occurs but B does not. This is the left portion of circle A that doesn't overlap with B.
Segment #3 (Ac∩B): Event B occurs but A does not. This is the right portion of circle B that doesn't overlap with A.
Segment #4 (Ac∩Bc): Neither event occurs. This region lies outside both circles, representing outcomes where both A and B fail to happen.
Student Survey: Four Regions That Reproduce the Marginals
The tool opens on this problem. It gives P(A)=0.6 for Student, P(B)=0.4 for Employed, and the single constraint P(A∩B)=0.15; from those the four regions come out as 0.15, 0.45, 0.25, 0.15.
They sum to exactly 1, and — the check worth doing every time — adding them back the other way returns the marginals you started from: 0.15+0.45=0.6 and 0.15+0.25=0.4.
Student Survey: P(A) = 0.6, P(B) = 0.4, P(A and B) = 0.15
Regions 1 to 4 are 0.15, 0.45, 0.25 and 0.15. They sum to 1, and adding them back the other way returns the marginals: 0.15 + 0.45 = 0.6 and 0.15 + 0.25 = 0.4.
That round trip is the whole method. Two marginals and one intersection are enough to pin down all four regions, because the four must sum to 1 and each marginal is the sum of two of them. Four unknowns, four equations.
The union follows immediately: P(A∪B)=1−P(Ac∩Bc)=1−0.15=0.85, which is also what the addition rule gives as 0.6+0.4−0.15. Reading it off the complement region is usually quicker than applying the rule.
Notice that P(A∩B)=0.15 is well below P(A)⋅P(B)=0.24. Being a student and being employed are not independent here — they pull against each other, which the region sizes show directly.
Health Screening: Why the Two Conditionals Differ
The second problem gives P(A)=0.2 for Has Condition, P(B)=0.3 for Test Positive, and the constraint P(A∩Bc)=0.05. The regions are 0.15, 0.05, 0.15, 0.65, again summing to 1 and reproducing both marginals.
The two conditional probabilities hiding in those four numbers are not the same, and the gap between them is the point of the example.
Health Screening: P(A) = 0.2, P(B) = 0.3, P(A and not B) = 0.05
Regions 0.15, 0.05, 0.15, 0.65. P(B|A) = 0.15/0.20 = 0.75 but P(A|B) = 0.15/0.30 = 0.50 - the same intersection over two different denominators.
Read down the row for A: P(B∣A)=0.200.15=0.75. The test catches three quarters of the people who have the condition — that is its sensitivity, and it sounds reassuring.
Now read down the column for B: P(A∣B)=0.300.15=0.50. Of everyone who tests positive, only half actually have the condition.
Same intersection region, two different denominators, two very different numbers — and it is the second one a patient holding a positive result actually wants. The reason they diverge is visible in the diagram: the region "no condition but tests positive" is 0.15, exactly as large as the true-positive region, because the 80% of people without the condition are numerous enough that even a small false-positive rate among them produces as many positives as the condition itself does. Reading P(B∣A) when you meant P(A∣B) is the base rate fallacy, and this problem is built to make it concrete.
Selecting a Region
Clicking a numbered box fills it gold with an amber border and leaves the other three white. The still below shows region 1, the intersection A∩B, selected.
Selection is a toggle: clicking the same box again clears it.
Region 1 selected
A selected box fills gold with an amber border; the other three stay white. Clicking it again clears the selection. The circles are fixed in size, so area shows membership, not probability.
The diagram deliberately carries no numbers. The circles and the four 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 is worth keeping in mind when reading any Venn diagram: circle areas here are fixed and identical, so they encode set membership, not probability. A region drawn large may hold very little probability — in the Health Screening problem the outside region holds 0.65, nearly two thirds of everything, while looking like empty background.
Using Show Calculations Feature
Toggle the "Show Calculations" button to reveal step-by-step solutions for each region's probability. The calculations explain how constraints and marginal probabilities determine each segment.
For segment #1, the calculation typically starts with given information like P(A∩B)=0.15. For segment #2, it uses P(A∩Bc)=P(A)−P(A∩B).
Each calculation shows the formula and the numerical result. This feature helps understand the logical sequence for solving Venn diagram problems from constraints to complete solutions.
Reading Probability Values
Each outcome displays its calculated probability on the right side, shown to three decimal places. These values always sum to exactly 1.000, representing the complete sample space.
The probabilities reflect the given constraints and marginal probabilities. In the "Student Survey" example, P(A∩B)=0.150 means 15% of the population are both students and employed.
Compare different regions to understand relative likelihoods. If P(A∩Bc)=0.450 is much larger than P(A∩B)=0.150, most students are unemployed in this scenario.
Working with Multiple Problems
Switch between example problems using the buttons at the top. Each problem has different event descriptions, marginal probabilities, and constraints.
The "Student Survey" problem examines students and employment. The "Health Screening" problem analyzes medical test accuracy with true positives and false positives.
Each problem demonstrates different constraint types. Some give P(A∩B) directly, others give P(A∩Bc), requiring different solution approaches.
Understanding Set Operations
Set operations form the mathematical foundation of Venn diagrams. The intersection A∩B represents "A and B" - outcomes where both events occur.
The union A∪B represents "A or B" - outcomes in either circle. Calculate it as P(A∪B)=P(A)+P(B)−P(A∩B), subtracting the overlap to avoid double-counting.
The complement Ac represents "not A" - everything outside circle A. By definition, P(A)+P(Ac)=1. These operations extend naturally to three or more sets.
Conditional Probability in 2-Set Diagrams
Conditional probabilityP(A∣B) asks "what's the probability of A given that B occurred?" Visually, focus only on circle B, then find what fraction of B overlaps with A.
Calculate it as P(A∣B)=P(B)P(A∩B). If P(A∩B)=0.15 and P(B)=0.40, then P(A∣B)=0.15/0.40=0.375 or 37.5%.
The diagram makes this intuitive: restrict attention to region B (segments #1 and #3), then see what portion is also in A (only segment #1).
Applying the Addition Rule
The addition rule calculates the probability of A or B occurring: P(A∪B)=P(A)+P(B)−P(A∩B). The diagram shows why subtraction is necessary.
Adding P(A) and P(B) counts the intersection twice - once in A's total and once in B's total. Subtracting P(A∩B) corrects this double-counting.
In terms of segments: P(A∪B) equals segments #1 + #2 + #3, which is P(A)+P(B)−P(A∩B) algebraically. The visual matches the formula perfectly.
Related Probability Tools
3-Set Venn Diagrams - Extend these concepts to three events with eight regions
Contingency Tables - Organize the same probability information in tabular format
Tree Diagrams - Show sequential conditional probabilities over time