01 · Pairwise or pairwise-only
How many members belong to A and B, whether or not they also belong to C?
Hint
The wording does not exclude C.
Worked solution
17: the 10 in A and B only plus the 7 in all three.
Understand · explore · practise
Three-event Venn diagrams
Fill a three-set Venn diagram from the centre out, distinguish pairwise totals from pairwise-only regions, and calculate exactly-one and at-least-two probabilities.
Before you startTwo-event Venn diagrams, intersections, unions and complements.
01 / Eight disjoint regions
Three events allow eight membership patterns.
Start with all three. Then fill exactly two, then exactly one, then none.
A pairwise intersection such as A ∩ B includes people who also belong to C.
Each member either belongs or does not belong to each of the three sets, producing 2×2×2=8 membership patterns. Use counts to calculate probabilities, not the areas of the drawn regions.
Build the region ledgerExplore
120 members: A total 46, B total 49, C total 41; A∩B 17, A∩C 13, B∩C 15; all three 7. The pairwise totals include all three.
Start with the triple intersection: 7 members belong to all three sets.
The areas are illustrative. Each listed count belongs to one disjoint region.
02 / Subtract the centre once
Pairwise-only is smaller than the pairwise total.
Watch: pairwise totals contain the centre
Pause, replay or seek freely. The notes explain the same idea and stay in view.
The pairwise totals are 17, 13 and 15; the triple total is 7.Worked example
A and B only = 17−7 = 10
This region is A ∩ B ∩ C′.
A and C only = 13−7 = 6
This region excludes B.
B and C only = 15−7 = 8
This region excludes A.
02 · Exactly two
How many belong to exactly two sets?
Hint
Add only the three pairwise-only regions.
Worked solution
10+6+8 = 24. The triple intersection is excluded.
03 / Recover each single-only region
Subtract the disjoint regions already inside that set.
Set totals are A 46, B 49 and C 41.Worked example
A only = 46−10−6−7 = 23
The already-filled regions inside A are disjoint.
B only = 49−10−8−7 = 24
Do not subtract the triple region twice.
C only = 41−6−8−7 = 20
Check the total within each circle afterwards.
03 · A-only shortcut
Explain 46−17−13+7 = 23.
Hint
Both pairwise totals subtracted the triple.
Worked solution
Subtracting A∩B and A∩C removes the triple twice. Add 7 back once. This gives A only 23.
04 · Exactly one
Find the probability that a uniformly chosen member belongs to exactly one set.
Hint
Add the three single-only regions.
Worked solution
(23+24+20)/120 = 67/120.
04 / Find the complement last
Every member must appear in exactly one region.
05 · Union
Find the union count and probability.
Hint
Add the seven regions inside at least one circle.
Worked solution
23+24+20+10+6+8+7 = 98. Probability = 98/120 = 49/60.
06 · None
How many belong to none of the sets?
Hint
Subtract the union count from 120.
Worked solution
120−98 = 22; probability 22/120 = 11/60.
07 · Check all eight
Check the completed region ledger.
Hint
Include the outside region.
Worked solution
23+24+20+10+6+8+7+22 = 120. The individual set totals recover 46, 49, 41.
05 / Translate the wording into regions
At least two includes the triple.
08 · At least two
Find the probability of membership of at least two sets.
Hint
Include exactly two and all three.
Worked solution
(10+6+8+7)/120 = 31/120.
09 · A but not B
Find P(A ∩ B′).
Hint
C may be present or absent.
Worked solution
A only and A∩C only qualify: (23+6)/120 = 29/120.
10 · A or B, but not C
Find P((A ∪ B) ∩ C′).
Hint
Exclude every region inside C.
Worked solution
A only, B only and A∩B only qualify: (23+24+10)/120 = 57/120 = 19/40.
11 · At most one
Find the probability of membership of at most one set.
Hint
Include none as well as exactly one.
Worked solution
(22+23+24+20)/120 = 89/120. It complements the at-least-two event.
06 / Understand inclusion–exclusion
The triple region needs one copy in the final union.
n(A ∪ B ∪ C) = n(A)+n(B)+n(C)
− n(A ∩ B)−n(A ∩ C)−n(B ∩ C)
+ n(A ∩ B ∩ C).
The same identity holds with probabilities instead of counts.
A member in all three is counted three times by the set totals and subtracted three times by the pairwise totals. Add the triple once so that member is counted once overall.
12 · Formula check
Use inclusion–exclusion for this population.
Hint
Insert the original inclusive totals.
Worked solution
46+49+41−17−13−15+7 = 98, agreeing with the disjoint region sum.
07 / A diagram must have possible counts
Check each region, not only the final total.
13 · Negative pair region
Can the triple count be 9 if n(A∩C)=6?
Hint
The triple intersection is contained in A∩C.
Worked solution
No. The A∩C-only count would be 6−9=−3, which is impossible.
14 · Sum only the set totals
Why is (46+49+41)/120 not a valid union probability?
Hint
Some members appear in two or three sets.
Worked solution
The numerator 136 double- or triple-counts some members and produces a value above 1. Use disjoint regions or inclusion–exclusion.
08 / Build from the centre out
Disjoint regions make compound events manageable.
Mark the triple first. Subtract it from the pairwise totals, then use the individual set totals to fill the single-only regions. Find none last. For every event, list the qualifying disjoint regions before adding their counts.
Section 1 of 8 · Eight disjoint regions