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'.
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.
[Diagram: math-venn-diagram] - Available in full study mode
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.
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.
[Diagram: math-venn-diagram] - Available in full study mode