IB Maths AI SLProbabilityPaper 1 & 2Venn diagrams~7 min read
Venn Diagrams
A Venn diagram turns a probability question into a picture. A rectangle is the sample space; circles are the events; the numbers in each region are frequencies or probabilities. Once the diagram is filled in, every answer — unions, intersections, complements, conditionals — is just a matter of reading off and adding up the right regions. The skill is filling it in correctly: work from the centre outwards.
📘 What you need to know
The structure: a rectangle labelled U (sample space) containing one circle per event. Overlapping circles share outcomes.
Numbers in regions: frequencies (must add to the total) OR probabilities (must add to 1). Decide which the question uses.
Work centre-outwards: fill the intersection (centre) FIRST, then the “only” regions, then “neither” (whatever’s left over).
The four regions of a 2-circle Venn: A only, A ∩ B, B only, neither (A‘ ∩ B‘).
Reading off: P(A) = all regions inside the A circle; P(A ∪ B) = everything in either circle; P(A‘) = everything outside A.
Conditional from a Venn: P(A|B) = (regions in both A and B) ÷ (all regions in B).
What a Venn diagram shows
The rectangle holds every possible outcome. Each circle is an event. Where circles overlap, outcomes belong to both events. Where they don’t, outcomes belong to one event or neither.
Mutually exclusive events are instantly visible — their circles don’t touch. Independence is NOT visible by eye; you have to check the probabilities with P(A ∩ B) = P(A)P(B).
Filling a Venn diagram — centre-outwards
The golden rule: start with the intersection. If you know how many are in “both”, every other region follows by subtraction. Filling the “only” regions first leads to double-counting errors.
Fill the intersection first (step 1), then subtract to get the “only” regions (step 2), then whatever’s left over goes in “neither” (step 3). Once filled: n(A) = 10 + 8 = 18, n(A ∪ B) = 10 + 8 + 7 = 25, n(neither) = 5.
If the intersection isn’t given directly, label it x and form an equation: the four regions must add up to the total.
Reading probabilities off a Venn
Once the diagram is filled, every probability is a region (or sum of regions) divided by the total:
Reading a Venn diagram
P(A) = all regions inside Atotal
P(A|B) = A ∩ B regionall regions inside B
For a conditional, the denominator shrinks to just the conditioning circle — exactly the reduced-sample-space idea from the last note. If frequencies are used, count the people; if probabilities are used, add the decimals.
Algebra rescue: if you can’t fill a region directly, call it x and use a known fact — all regions sum to the total (or to 1 for probabilities), or P(A ∩ B) = P(A)P(B) if the events are independent.
🧭 Recipe — any Venn diagram question
Draw the rectangle and circles: one circle per event, overlapping unless the events are mutually exclusive.
Fill the intersection first. If it isn’t given, label it x.
Fill the “only” regions by subtracting the intersection from each event’s total.
Fill “neither” last, so that all regions add to the grand total (or to 1).
Read off the answer: identify which regions the question wants, add them, and divide by the total (or the conditioning circle for a conditional).
Worked examples
WE 1
Fill a two-circle Venn (intersection given)
In a class of 32 students, 20 study French, 14 study Spanish, and 6 study both. Find the number of students in each region of the Venn diagram.
Step 1 — intersection (given)both French and Spanish = 6Step 2 — “only” regions by subtractionFrench only = 20 − 6 = 14Spanish only = 14 − 6 = 8Step 3 — “neither” fills to the totalneither = 32 − 14 − 6 − 8 = 4French only 14 · both 6 · Spanish only 8 · neither 4always subtract the intersection out of each event total — the “20 French” already INCLUDES the 6 who do both, so French only is 20 − 6, not 20.
WE 2
Fill a Venn with the intersection unknown (algebra)
60 people are asked whether they own a bike and/or a car. 38 own a bike, 33 own a car, and 9 own neither. How many own both?
let x = number who own BOTHwrite each region in terms of xbike only = 38 − xcar only = 33 − xall four regions sum to 60(38 − x) + x + (33 − x) + 9 = 6080 − x = 60x = 2020 people own botha quicker route: P(both) count = (bike) + (car) + (neither) − (total) = 38 + 33 + 9 − 60 = 20. Same answer — the algebra just makes the logic explicit.
WE 3
Read probabilities off a completed Venn
A Venn diagram for events A and B shows probabilities: A only = 0.25, A ∩ B = 0.15, B only = 0.35, neither = 0.25. Find (a) P(A), (b) P(A ∪ B), (c) P(A‘).
check the regions sum to 10.25 + 0.15 + 0.35 + 0.25 = 1 ✓(a) P(A) = all regions inside the A circleP(A) = 0.25 + 0.15 = 0.4(b) P(A ∪ B) = everything in either circleP(A ∪ B) = 0.25 + 0.15 + 0.35 = 0.75(c) P(A’) = everything outside AP(A’) = 1 − P(A) = 1 − 0.4 = 0.6(a) 0.4 · (b) 0.75 · (c) 0.6P(A ∪ B) is everything EXCEPT “neither” — you could also do 1 − 0.25 = 0.75. Two routes, same answer; use whichever is faster.
WE 4
Conditional probability from a Venn
A Venn diagram for 30 people shows: gym members only = 12, both gym and swimming = 8, swimmers only = 6, neither = 4. Find (a) P(swims | gym member), (b) P(gym member | swims).
(a) condition = gym member ⇒ denominator = gym circlegym circle total = 12 + 8 = 20of those, 8 also swimP(swims | gym) = 8/20 = 2/5(b) condition = swims ⇒ denominator = swim circleswim circle total = 8 + 6 = 14of those, 8 are gym membersP(gym | swims) = 8/14 = 4/7(a) 2/5 · (b) 4/7for a conditional, the denominator is the WHOLE conditioning circle (not the grand total of 30). The numerator is always the intersection (8 here).
WE 5
Three-circle Venn diagram
35 students are surveyed on which instruments they play. The Venn diagram regions are: piano only 8, guitar only 6, violin only 5, piano & guitar only 4, piano & violin only 3, guitar & violin only 2, all three 2, none 5. Find (a) how many play piano, (b) how many play exactly two instruments, (c) P(plays guitar), (d) P(plays all three).
(a) piano = every region inside the piano circle= 8 + 4 + 3 + 2 = 17(b) exactly two = the three “pair only” regions= 4 + 3 + 2 = 9(c) P(guitar): guitar circle total / 35guitar = 6 + 4 + 2 + 2 = 14P(guitar) = 14/35 = 2/5(d) P(all three) = centre region / totalP(all three) = 2/35(a) 17 · (b) 9 · (c) 2/5 · (d) 2/35a 3-circle Venn has 8 regions. “Exactly two” excludes the centre (all three). “Plays guitar” includes the centre. Read the wording carefully: “exactly” vs “at least”.
WE 6
Fill a Venn using independence
Events A and B are independent, with P(A) = 0.4 and P(B) = 0.5. Fill in the four regions of the Venn diagram with probabilities.
Step 1 — intersection, using independenceP(A ∩ B) = P(A) × P(B) = 0.4 × 0.5 = 0.2Step 2 — “only” regions by subtractionA only = P(A) − P(A ∩ B) = 0.4 − 0.2 = 0.2B only = P(B) − P(A ∩ B) = 0.5 − 0.2 = 0.3Step 3 — “neither” fills to 1neither = 1 − 0.2 − 0.2 − 0.3 = 0.3A only 0.2 · both 0.2 · B only 0.3 · neither 0.3the word “independent” is what unlocks the intersection: P(A ∩ B) = P(A)P(B). Without it you couldn’t fill the centre. After that it’s the standard centre-outwards method.
💡 Top tips
Always fill the intersection first — everything else follows by subtraction. Working outside-in causes double-counting.
Subtract the overlap from each event’s total: “20 study French” includes those who also study Spanish.
Check the regions add up: frequencies to the grand total, probabilities to exactly 1.
For conditionals, the denominator is the whole conditioning circle, NOT the grand total.
Stuck? Use algebra: label an unknown region x and form an equation from “all regions sum to the total” or independence.
⚠ Common mistakes
Putting the event total in the “only” region: writing 20 in “French only” when 20 is the WHOLE French circle. Subtract the overlap first.
Forgetting the “neither” region: the rectangle has space outside all circles — that region is real and often non-zero.
Dividing a conditional by the grand total: P(A|B) divides by the B circle, not by everyone.
Confusing “exactly two” and “at least two” on a 3-circle Venn: “exactly two” excludes the centre; “at least two” includes it.
Assuming overlapping circles means dependent events: overlap just means they CAN both happen. Independence is a numerical property, checked separately.
Next up: Tree Diagrams — the final note of the Probability chapter. Where Venn diagrams are best for “AND/OR/NOT” on a single experiment, tree diagrams shine for sequences of events: draw a marble, then another; flip, then flip again. You multiply along the branches and add between them — and the conditional probabilities from this note sit right on the second set of branches.
Need help with AI SL Probability?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
