Unions and Intersections
Every question with a fully worked solution.
- Unions and Intersections - Video - Basics of Probability: Unions, Intersections, and Complements Watch
Theory
A Venn diagram sorts every outcome into regions, so probabilities can be read straight from it. This Year 10 maths page covers the intersection \(A\cap B\), the union \(A\cup B\) and the complement \(A'\), finding the overlap from totals, the addition rule, unknowns in regions, three-set diagrams, two-way tables and probabilities given a condition.
The rectangle is the universal set \(\xi\) (the whole sample space). Each circle is an event or set, a collection of distinct objects. \(n(A)\) is the number of elements in \(A\).
\(A\cap B\), the intersection, is "\(A\) and \(B\)": the overlap. \(A\cup B\), the union, is "\(A\) or \(B\) or both": everything inside either circle. \(A'\) (also \(\overline{A}\) or \(A^c\)), the complement, is everything not in \(A\).
Combinations describe single regions: \(A\cap B'\) is "\(A\) only", \(A'\cap B'\) is "neither", and \((A\cup B)'\) is the same region as \(A'\cap B'\). "Exactly one" means \(A\) only or \(B\) only. "At least one" means \(A\cup B\).
Events are mutually exclusive when they cannot happen together, so the circles do not overlap and \(A\cap B\) is empty. A probability "given" a condition uses only the region named in the condition as the total.
The addition rule stops the overlap being counted twice:
| In words | Notation | Regions |
|---|---|---|
| \(A\) and \(B\) | \(A\cap B\) | the overlap |
| \(A\) or \(B\) (or both) | \(A\cup B\) | all three regions inside the circles |
| \(A\) only | \(A\cap B'\) | \(n(A)-n(A\cap B)\) |
| neither | \(A'\cap B'=(A\cup B)'\) | total \(-\ n(A\cup B)\) |
| \(A\) or not \(B\) | \(A\cup B'\) | everything except \(B\) only |
| \(B\) given \(A\) | \(P(B\mid A)\) | \(\dfrac{n(A\cap B)}{n(A)}\) |
How to solve a Venn diagram problem
- Draw the rectangle and one circle for each event.
- Fill the overlap first. If it is not given, find it: overlap \(=n(A)+n(B)-n(A\cup B)\), where \(n(A\cup B)\) is the total minus "neither".
- Fill the "only" regions by subtracting the overlap from each circle total, then the outside region.
- Check that all the regions add to the total (or to \(1\) for probabilities).
- Read the probability: the count in the required regions over the total. For "given that", divide by the count in the given region only.
Other question types
- Unknowns in regions: add every region, set the sum equal to the total and solve (Example 3). With probabilities the sum is \(1\), and a quadratic may need a negative solution rejected.
- Described in words: "French only is twice both" means let both \(=b\), French only \(=2b\).
- Explain a mistake: if \(P(A)+P(B)\gt 1\), the events must overlap, and adding them counts the overlap twice.
- Checking independence: compare \(P(B\mid A)\) with \(P(B)\). If they are equal, knowing \(A\) does not change the chance of \(B\).
(i) \(P(\text{exactly one sport})\)
(ii) the probability that a basketball player also plays soccer.
Students in at least one circle: \(60-9=51\). Adding \(32+25=57\) counts the overlap twice, so the overlap is \(57-51=6\).
| \(B \text{ only}\) | \(=\) | \(32-6=26\) |
| \(S \text{ only}\) | \(=\) | \(25-6=19\) |
(i) Exactly one sport means \(B\) only or \(S\) only.
| \(P\) | \(=\) | \(\dfrac{26+19}{60}=\boldsymbol{\dfrac{3}{4}}\) |
(ii) Only the \(32\) basketball players count, and \(6\) of them play soccer.
| \(P(S\mid B)\) | \(=\) | \(\dfrac{6}{32}=\boldsymbol{\dfrac{3}{16}}\) |
(i) \(n(A\cup B)\)
(ii) \(P(A'\cap B)\)
(iii) \(P\big((A\cup B)'\big)\)
List the sets: \(A=\{3, 6, 9, 12, 15, 18\}\), \(B=\{1, 2, 3, 6, 9, 18\}\), so \(A\cap B=\{3, 6, 9, 18\}\).
| \(n(A\cup B)\) | \(=\) | \(n(A)+n(B)-n(A\cap B)\) |
| \(\) | \(=\) | \(6+6-4=\boldsymbol{8}\) |
(ii) \(A'\cap B\) is in \(B\) but not in \(A\): \(\{1, 2\}\). So \(P(A'\cap B)=\dfrac{2}{20}=\boldsymbol{\dfrac{1}{10}}\).
(iii) \((A\cup B)'\) is outside both circles: \(20-8=12\) numbers. So the probability is \(\dfrac{12}{20}=\boldsymbol{\dfrac{3}{5}}\).
(i) \(P(A)=0.5\), \(P(B)=0.35\) and \(P(A\cap B)=0.15\). Find \(P(A\cup B)\), \(P(\text{neither})\) and \(P(A\cap B')\).
(ii) A Venn diagram of \(50\) people has regions \(A\) only \(=3x\), both \(=x+2\), \(B\) only \(=2x-1\) and neither \(=x\). Find \(x\) and \(P(A\mid B)\).
(i) Use the addition rule, then the complement.
| \(P(A\cup B)\) | \(=\) | \(0.5+0.35-0.15=\boldsymbol{0.7}\) |
| \(P(\text{neither})\) | \(=\) | \(1-0.7=\boldsymbol{0.3}\) |
| \(P(A\cap B')\) | \(=\) | \(0.5-0.15=\boldsymbol{0.35}\) |
(ii) The four regions add to \(50\).
| \(3x+(x+2)+(2x-1)+x\) | \(=\) | \(50\) |
| \(7x+1\) | \(=\) | \(50\) |
| \(x\) | \(=\) | \(\boldsymbol{7}\) |
The regions are \(21\), \(9\), \(13\) and \(7\). Given \(B\), only the \(9+13=22\) people in \(B\) count: \(P(A\mid B)=\boldsymbol{\dfrac{9}{22}}\).
(i) How many study none?
(ii) Find \(P(\text{exactly two})\).
(iii) A drama student is chosen. Find the probability they also study art.
Start at the centre with \(3\). Pairs only: \(8-3=5\), \(6-3=3\), \(5-3=2\).
| \(\text{music only}\) | \(=\) | \(20-5-3-3=9\) |
| \(\text{art only}\) | \(=\) | \(17-5-2-3=7\) |
| \(\text{drama only}\) | \(=\) | \(14-3-2-3=6\) |
(i) Inside the circles: \(9+7+6+5+3+2+3=35\), so \(\boldsymbol{5}\) study none.
(ii) Exactly two: \(5+3+2=10\), so \(P=\dfrac{10}{40}=\boldsymbol{\dfrac{1}{4}}\).
(iii) Drama has \(14\) students. Those also in art: \(2+3=5\). So \(P=\boldsymbol{\dfrac{5}{14}}\).
Common pitfalls
Frequently asked questions
What does the intersection of two sets mean?
The intersection, \(A\cap B\), is the set of elements in both \(A\) and \(B\). On a Venn diagram it is the overlap of the two circles.
What is the difference between union and intersection?
The union, \(A\cup B\), contains everything in \(A\) or \(B\) or both: all of both circles. The intersection contains only what is in both: the overlap.
How do you find the number in both groups?
Add the two group totals and subtract the number in at least one group, which is the total minus those in neither. The overlap was counted twice in the sum, so the difference is the overlap.
What does A complement mean?
\(A'\) is everything in the universal set that is not in \(A\). Its probability is \(1-P(A)\).
How do you find a probability from a Venn diagram given a condition?
Use only the region described by the condition as the total. For \(P(B\mid A)\), divide the number in the overlap by the number in \(A\).