Union, intersection, complement: A Venn diagram shows sets inside the universal set U.
A ∪ B = in A or B (or both); A ∩ B = in both; A′ = not in A.
The numbers in the regions are counts (or probabilities).
IB-style question — read the notation
In a group of 30 students, 18 play football (F), 12 play basketball (B), and 7 play both.
Describe in words what F ∩ B and (F ∪ B)′ mean, and give n(F ∩ B).
Step by step
- F ∩ B = students who play both.
- (F ∪ B)′ = students who play neither sport.
Final answer
F ∩ B = play both (7 students); (F ∪ B)′ = play neither sport.
∩ = and, ∪ = or: Read ∩ as 'and' (both), ∪ as 'or' (at least one).
The dash ′ flips to 'not'.
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Start in the middle and work outward: Fill the intersection first, then subtract to get the 'only A' and 'only B' regions, then place 'neither' last so all regions add to the total.
Fill the overlap first, then 'only A' and 'only B' (each = set total − overlap), then 'neither'.
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IB-style question — fill the regions
Of 30 students, 18 play football, 12 play basketball, 7 play both.
Find how many play only football, only basketball, and neither.
Step by step
- Only football = football − both; only basketball = basketball − both.
- Neither = total − (only F + both + only B).
Final answer
Only football 11, only basketball 5, both 7, neither 7.
Don't double-count the overlap: The 18 footballers include the 7 who play both — subtract the overlap to get 'only football'.
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Region count ÷ total: Once the regions are filled, a probability is the region count ÷ total. 'Only A', 'both', 'neither' and 'A or B' all come straight from the diagram.
IB-style question — probabilities from regions
Using the filled Venn (only F 11, both 7, only B 5, neither 7; total 30), find P(only football) and P(plays neither).
Step by step
- Only football region ÷ total.
- Neither region ÷ total.
Final answer
P(only football) = 11/30; P(neither) = 7/30.
'Plays a sport' = 1 − P(neither): P(plays at least one) = P(F ∪ B) = 1 − P(neither) = 1 − 7/30 = 23/30.
Add the two, subtract the overlap: The addition rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B) — you subtract the overlap so it isn't counted twice.
If A and B are mutually exclusive (can't both happen), P(A ∩ B) = 0.
IB-style question — addition rule
P(A) = 0.5, P(B) = 0.4, P(A ∩ B) = 0.2.
Find P(A ∪ B).
Step by step
- Addition rule.
- Evaluate.
Final answer
P(A ∪ B) = 0.7.
Only drop the overlap if mutually exclusive: P(A ∪ B) = P(A) + P(B) only when the events can't both occur (P(A ∩ B) = 0).
Otherwise you must subtract the overlap.
IB-style question — complement of a union
For two events, P(A) = 0.5, P(B) = 0.4 and P(A ∩ B) = 0.2.
Find (a) P(A′ ∩ B′) and (b) P(A ∩ B′).
Step by step
- First the union (addition rule).
- (a) A′ ∩ B′ is everything OUTSIDE the union (De Morgan).
- (b) A ∩ B′ is the part of A not in B.
Final answer
(a) 0.3. (b) 0.3.
A′ ∩ B′ is the region OUTSIDE both circles (1 − P(A∪B)); A ∩ B′ is the part of A's circle not overlapping B.
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