Set-builder notation reads as "the set of all [variable] such that [condition]". For example, $\{x : x \text{ is a natural number}, x < 10\}$ means every natural number smaller than $10$. A set can also be defined by a rule linking two variables, e.g. $\{(x,y): y = 2x+1\}$.

Two overlapping circles split $\mathscr{E}$ into four regions: $A$ only, $B$ only, $A\cap B$ (the overlap), and outside both circles ($(A\cup B)'$). Read the value straight from whichever region the question describes, and add regions together to answer questions about unions.

Three-set diagrams must be filled from the inside out:

1. Place the value for $A \cap B \cap C$ in the very centre first.
2. Each "double-only" region is the pairwise intersection minus the centre: $n(A\cap B \text{ only}) = n(A\cap B) - n(A\cap B\cap C)$.
3. Each "single-only" region is that set's total minus everything else already placed inside its circle.
4. Everything outside all three circles is $n((A\cup B\cup C)')$.

Phrases like "given that" or "chosen from those who..." tell you to shrink your sample space to just that region, rather than the whole universal set $\mathscr{E}$.

Worked Example

In a class of $40$ students: $22$ study History ($H$), $18$ study Geography ($G$), $16$ study Economics ($E$); $8$ study $H$ and $G$; $7$ study $H$ and $E$; $6$ study $G$ and $E$; $3$ study all three; $2$ study none.

(a) Find the number of students who study History only.

(b) A student is chosen at random from those who study Geography. Find the probability they also study Economics.

1. Centre: $n(H\cap G\cap E) = 3$.
2. Double-only regions: $H\cap G$ only $=8-3=5$; $H\cap E$ only $=7-3=4$; $G\cap E$ only $=6-3=3$.
3. History only $= 22-(5+4+3) = 10$.
4. For (b), the new sample space is $n(G)=18$. Students in $G$ and $E$ $= 3\text{ (centre)}+3\text{ (double-only)}=6$.