Statistics · Conditional Probability · Year 2

Set Notation

Circles, overlaps and shading -- a compact language for describing exactly which outcomes an event contains.

7 min read 1 worked example

Describing events with sets

You can use set notation to describe events within a sample space. This lets you abbreviate probability statements that would otherwise need a full sentence.

The symbol $\mathscr{E}$ is used to represent the whole sample space — every possible outcome. Events $A$ and $B$ are drawn as circles inside a rectangle representing $\mathscr{E}$.

Intersection: A and B

The event "$A$ and $B$" is written $A\cap B$. The $\cap$ symbol is the symbol for intersection.

The intersection $A\cap B$ (shaded) is the region inside both circles — outcomes belonging to $A$ and $B$.

If $A$ and $B$ are independent, $\mathrm{P}(A\cap B) = \mathrm{P}(A)\times \mathrm{P}(B)$.

Notation — mutually exclusive events

If two events $A$ and $B$ are mutually exclusive, then their intersection is the empty set, $\varnothing$. You can write $A\cap B=\varnothing$ — there are no outcomes in both at once.

Union: A or B

The event "$A$ or $B$" is written $A\cup B$. The $\cup$ symbol is the symbol for union.

The union $A\cup B$ (shaded) is the region inside either circle (or both) — outcomes in $A$ or $B$.

If $A$ and $B$ are mutually exclusive, then $\mathrm{P}(A\cup B) = \mathrm{P}(A)+\mathrm{P}(B)$ — there's no overlap to double-count.

Complement: not A

The event "not $A$" is written $A'$. This is also called the complement of $A$.

The complement $A'$ (shaded) is everything outside circle $A$ — every outcome that is not $A$, including all of $B$ that lies outside $A$.

Key facts about the complement

$\mathrm{P}(A') = 1-\mathrm{P}(A)$. Events $A$ and $A'$ are always mutually exclusive — an outcome can never be both in $A$ and not in $A$.

Worked example

Reading events off a Venn diagram

Apex's Year 12 enrichment programme has 45 students, each enrolled in at least one strand. Let $C$ be the event that a randomly chosen student is enrolled in the coding strand, and $D$ the event that they're enrolled in the data-analysis strand. Given $n(C)=15$, $n(D)=9$ and $n(C\cap D)=3$, find: a $\mathrm{P}(C\cap D)$ b $\mathrm{P}(C\cup D)$ c $\mathrm{P}(C')$ d $\mathrm{P}(C'\cap D)$

The Venn diagram for the enrichment-programme example, showing the number of students in each region.

$n(C)$ is notation for the number of outcomes in $C$: $n(C)=15$ students do coding, and $n(C\cap D)=3$ of those also do data analysis, so $C$-only $=15-3=12$ and $D$-only $=9-3=6$.

MethodRead counts straight off the diagram, then divide by 45

$C\cap D$ is "does both strands": $\mathrm{P}(C\cap D)=\dfrac{3}{45}=\dfrac{1}{15}$.

3 outcomes in the intersection, 45 in total.

$C\cup D$ is "does coding, or data analysis, or both": $n(C\cup D)=12+3+6=21$, so $\mathrm{P}(C\cup D)=\dfrac{21}{45}=\dfrac{7}{15}$.

Add every region inside either circle.

$C'$ is "doesn't do coding": $\mathrm{P}(C')=\dfrac{30}{45}=\dfrac{2}{3}$.

$45-15=30$ students don't do coding.

$C'\cap D$ is "doesn't do coding, but does data analysis": $\mathrm{P}(C'\cap D)=\dfrac{6}{45}=\dfrac{2}{15}$.

The part of $D$ that lies outside $C$.

▪ solved

Examiner's eye

$C'\cap D$ is the set of outcomes not in $C$ but in $D$ — on the diagram, that's the part of circle $D$ that lies outside circle $C$. Reading intersections with a complement directly off a correctly filled Venn diagram avoids sign errors that creep in when working from the formula alone.

Check yourself

A tutor group has 10 students, labelled 1 to 10. $F$ is the event that a student plays a musical instrument, where $F=\{1,3,4,6,8,9\}$. $D$ is the event that a student is on the school's debate team, where $D=\{2,4,6,9,10\}$. List the outcomes in $F\cap D$ and $F\cup D$, and find $\mathrm{P}(F')$.

$F\cap D=\{4,6,9\}$.

$F\cup D=\{1,2,3,4,6,8,9,10\}$.

$F'=\{2,5,7,10\}$, so $\mathrm{P}(F')=\dfrac{4}{10}=\dfrac{2}{5}$.

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