Venn Diagrams GCSE Questions, Worked Examples and Answers

The symbol $\in$ means “is an element of”. Since $7$ appears in the list for $P$,

$7\in P$

The symbol $\notin$ means “is not an element of”. Since $9$ does not appear in the list for $E$,

$9\notin E$

List the two sets first.

$M=\{3,6,9,12,15\}$

$F=\{1,2,3,4,6,12\}$

Start with the intersection: $3$, $6$ and $12$ are in both lists.

$M\cap F=\{3,6,12\}$

The $M$-only region contains $9$ and $15$. The $F$-only region contains $1$, $2$ and $4$.

Crossing off every used element leaves $5$, $7$, $8$, $10$, $11$, $13$ and $14$ outside both circles.

$M\text{ only}=\{9,15\}$

$M\cap F=\{3,6,12\}$

$F\text{ only}=\{1,2,4\}$

$\text{neither}=\{5,7,8,10,11,13,14\}$

The intersection is the overlap, so

$n(A\cap B)=4$

The union is every region inside at least one circle. Count the overlap once.

$n(A\cup B)=11+4+7=22$

$A'$ means everything not in $A$. This includes the $B$-only region and the region outside both circles.

$n(A')=7+3=10$

For $S\cup D$, count everyone in either circle, including the overlap once.

$P(S\cup D)=\frac{12+8+5}{30}=\frac{25}{30}=\frac56$

$D'$ contains everyone not in $D$: the $S$-only region and the region outside both circles.

$P(D')=\frac{12+5}{30}=\frac{17}{30}$

Start with the intersection because the $13$ pupils who study both are included in both set totals.

$\text{French only}=31-13=18$

$\text{Spanish only}=24-13=11$

There are $18+13+11=42$ pupils inside the circles.

$\text{neither}=52-42=10$

So the four regions are $18$, $13$, $11$ and $\boxed{10}$ outside the circles.

The four separate regions must total $48$.

$x+9+14+7=48$

$x+30=48$

$x=18$

Set $G$ contains its only region and the intersection.

$n(G)=18+9=\boxed{27}$

Adding the two set totals counts every person in the intersection twice: once in the cat total and once in the dog total. Subtract the intersection once to leave one copy.

$n(C\cup D)=n(C)+n(D)-n(C\cap D)$

$n(C\cup D)=45+38-21$

$n(C\cup D)=\boxed{62}$

“Exactly two” means the three pairwise overlaps, but not the centre where students do all three.

$4+5+3=12$

“At least one” means every region inside the circles.

$8+6+7+4+5+3+2=35$

Therefore $\boxed{12}$ students do exactly two activities and $\boxed{35}$ do at least one. The other $5$ do none.

Start at the centre with $5$. Each pair total includes these $5$ students, so subtract $5$ to find each pair-only region.

$B\cap C\text{ only}=14-5=9$

$B\cap P\text{ only}=12-5=7$

$C\cap P\text{ only}=11-5=6$

Now subtract the known regions from each whole-set total.

$B\text{ only}=34-(9+7+5)=13$

$C\text{ only}=31-(9+6+5)=11$

$P\text{ only}=29-(7+6+5)=11$

The number inside at least one circle is

$13+11+11+9+7+6+5=62$

Therefore the number studying none of the subjects is

$72-62=\boxed{10}$

Add the four disjoint regions and set their total equal to $40$.

$(2x+1)+(x+3)+(x-1)+5=40$

$4x+8=40$

$4x=32$

$x=8$

The three regions in the union are $17$, $11$ and $7$.

$n(A\cup B)=17+11+7=\boxed{35}$

All four disjoint region probabilities must add to $1$. The known regions total

$0.28+0.17+0.31=0.76$

The complement of the union is the region outside both circles.

$P((A\cup B)')=1-0.76=\boxed{0.24}$

The words “from set $B$” restrict the sample space to the $15$ people in $B$.

Of those $15$, the $9$ in the intersection are also in $R$.

$P(R\mid B)=\frac{n(R\cap B)}{n(B)}$

$P(R\mid B)=\frac{9}{9+6}$

$P(R\mid B)=\frac{9}{15}=\boxed{\frac35}$

The condition $T\cup S$ becomes the new denominator. Count each region that lies in $T$ or $S$ once.

$n(T\cup S)=8+6+4+5+3+2=28$

To be in $C$ as well, the student must lie in $(T\cup S)\cap C$. Those regions contain $5$, $3$ and $2$ students.

$n((T\cup S)\cap C)=5+3+2=10$

$P(C\mid T\cup S)=\frac{10}{28}=\boxed{\frac5{14}}$

First find the number in the union by removing the $12$ outside both circles.

$n(A\cup B)=70-12=58$

Use the union rule and let the intersection be the missing part.

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

$58=39+32-n(A\cap B)$

$58=71-n(A\cap B)$

$n(A\cap B)=\boxed{13}$

Let $x$ be the number with both types of training. The union rule gives

$n(F\cup R)=42+34-x=76-x$

So the number with neither is

$70-(76-x)=x-6$

The question says this is half the number with both.

$x-6=\frac{x}{2}$

$2x-12=x$

$x=12$

There are $34-12=22$ volunteers in $R$ but not $F$. The condition restricts the denominator to all $34$ people in $R$.

$P(F'\mid R)=\frac{22}{34}=\boxed{\frac{11}{17}}$

$B'$ means everything outside $B$. Intersecting this with $A$ leaves only the part of circle $A$ that is outside circle $B$.

Shade the A-only region. Do not shade the overlap.