Use Venn diagrams to find probabilities for intersecting events, apply the addition rule P(A or B) = P(A) + P(B) - P(A and B), and solve problems involving two overlapping events.
Worked examples
Reading a Venn diagram
Straightforward
Problem
A Venn diagram shows events A and B within a sample space of 20 equally likely outcomes. There are 6 outcomes in A only, 4 outcomes in B only, 3 outcomes in both A and B, and 7 outcomes in neither. Find P(A or B).
1
Identify the outcomes that make up A or B
A or B includes every outcome inside at least one of the two circles: A only, B only, and both A and B.
2
Count the favourable outcomes
n(A or B)=6+4+3=13
3
Calculate the probability
P(A or B)=n(sample space)n(A or B)=2013
Answer
P(A or B)=2013
Filling in a Venn diagram
Moderate
Problem
In a class of 30 students, 18 play sport, 14 study music, and 7 do both. A student is chosen at random. Find P(neither sport nor music).
1
Find the number in each region
Sport only: 18−7=11. Music only: 14−7=7. Both: 7.
2
Find the number in neither
Total in sport or music: 11+7+7=25.
n(neither)=30−25=5
3
Calculate the probability
P(neither)=305=61
Answer
P(neither sport nor music)=61
Using the addition rule to find P(A and B)
Challenging
Problem
For two events A and B: P(A)=53, P(B)=21, and P(A or B)=54. Find P(A and B).
1
Write down the addition rule
P(A or B)=P(A)+P(B)−P(A and B)
2
Rearrange to make P(A and B) the subject
P(A and B)=P(A)+P(B)−P(A or B)
3
Substitute the known values
P(A and B)=53+21−54
4
Convert to a common denominator and simplify
=106+105−108=103
Answer
P(A and B)=103
Practise
Q1·Straightforward
A Venn diagram shows events A and B within a sample space of 20 equally likely outcomes. There are 6 outcomes in A only, 4 outcomes in B only, 3 outcomes in both A and B, and 7 outcomes in neither. Find P(A).
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Explanation
The number of outcomes in A is n(A)=6+3=9 (A only plus both). The total number of outcomes is 20.
P(A)=209
Q2·Straightforward
A Venn diagram shows events A and B within a sample space of 20 equally likely outcomes. There are 6 outcomes in A only, 4 outcomes in B only, 3 outcomes in both A and B, and 7 outcomes in neither. Find P(A and B).
/
Explanation
The number of outcomes in both A and B is 3. The total is 20.
P(A and B)=203
Q3·Straightforward
A Venn diagram shows events C and D within a sample space of 30 equally likely outcomes. There are 10 outcomes in C only, 8 outcomes in D only, 5 outcomes in both C and D, and 7 outcomes in neither. Find P(C or D).
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Explanation
The number of outcomes in C or D (or both) is 10+8+5=23. The total is 30.
P(C or D)=3023
Q4·Straightforward
A Venn diagram shows events E and F within a sample space of 24 equally likely outcomes. There are 7 outcomes in E only, 5 outcomes in F only, 4 outcomes in both E and F, and 8 outcomes in neither. Find P(F).
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Explanation
The number of outcomes in F is n(F)=5+4=9 (F only plus both).
P(F)=249=83
Q5·Moderate
In a class of 30 students, 18 play sport, 14 study music, and 7 do both. A student is chosen at random. Find P(sport and music).
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Explanation
The number of students who play sport and study music is 7. The total is 30.
P(sport and music)=307
Q6·Moderate
In a survey of 28 people, 16 own a smartphone, 12 own a tablet, and 5 own both. A person is chosen at random. Find P(smartphone only).
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Explanation
Number who own a smartphone only: 16−5=11.
P(smartphone only)=2811
Q7·Moderate
In a class of 30 students, 18 play sport, 14 study music, and 7 do both. A student is chosen at random. Find P(neither sport nor music).
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Explanation
Number in sport only: 18−7=11. Number in music only: 14−7=7. Total accounted for: 11+7+7=25. Number in neither: 30−25=5.
P(neither)=305=61
Q8·Moderate
In a group of 40 people, 25 prefer tea, 20 prefer coffee, and 10 prefer both. A person is chosen at random. Find P(tea or coffee) using the addition rule.
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Explanation
Using the addition rule:
P(tea or coffee)=4025+4020−4010=4035=87
Q9·Challenging
In a class of 35 students, every student studies at least one of English or Maths. 22 study English and 20 study Maths. Find P(English only).
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Explanation
Since every student studies at least one subject: n(English)+n(Maths)−n(both)=35. So 22+20−n(both)=35, giving n(both)=7. Number who study English only: 22−7=15.
P(English only)=3515=73
Q10·Challenging
For two events A and B: P(A)=53, P(B)=21, and P(A or B)=54. Find P(A and B).
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Explanation
Rearranging the addition rule:
P(A and B)=P(A)+P(B)−P(A or B)=53+21−54
Converting to a common denominator of 10:
=106+105−108=103
Q11·Challenging
A Venn diagram shows events A and B within a sample space of 40 equally likely outcomes. There are 14 outcomes in A only, 11 outcomes in B only, and 9 outcomes in both A and B. Find P(A′ and B′), the probability of neither A nor B.
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Explanation
Outcomes in A only: 14. Outcomes in B only: 11. Outcomes in both: 9. Total inside the circles: 14+11+9=34. Outcomes in neither: 40−34=6.
P(A′ and B′)=406=203
Q12·Challenging
In a group of 60 people, 35 own a car, 28 own a bike, and 8 own neither. A person is chosen at random. Find P(owns both a car and a bike).
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Explanation
Number who own at least one: 60−8=52. Using inclusion–exclusion: n(car)+n(bike)−n(both)=52. So 35+28−n(both)=52, giving n(both)=11.