GCSE Sets and Venn diagrams Practice Questions

Write down what you know

There are $30$ students altogether, and the question gives $n(A) = 18$, $n(B) = 14$ and $n(A \cap B) = 7$.

Work out $n(A \cap B')$

Every one of the students who study both French and German is also one of the students who study French, so taking them away leaves exactly the students who study French but not German: $18 - 7 = 11$.

State the answer

So there are $11$ students who study French but not German.

Name the four regions of the Venn diagram

Two circles cut the universal set into exactly four regions, and every region has a name: $A' \cap B'$ — the people who have neither a car nor a bus pass, $A \cap B'$ — the people who have a car but not a bus pass, $A' \cap B$ — the people who have a bus pass but not a car and $A \cap B$ — the people who have both a car and a bus pass.

Say in symbols what the question describes

The people who have neither a car nor a bus pass are exactly the people who have neither a car nor a bus pass — one whole region of the diagram — and the symbols for that region are $A' \cap B'$.

State the answer

So $A' \cap B'$ is the set that contains exactly the people who have neither a car nor a bus pass.

Name the four regions of the Venn diagram

Two circles cut the universal set into exactly four regions, and every region has a name: $A' \cap B'$ — the people who own neither a dog nor a cat, $A \cap B'$ — the people who own a dog but not a cat, $A' \cap B$ — the people who own a cat but not a dog and $A \cap B$ — the people who own both a dog and a cat.

Say in symbols what the question describes

The people who own a cat but not a dog are exactly the people who own a cat but not a dog — one whole region of the diagram — and the symbols for that region are $A' \cap B$.

Rule out $A \cap B'$

$A \cap B'$ is the set of people who own a dog but not a cat, which is not what the question describes.

Rule out $A \cap B$

$A \cap B$ is the set of people who own both a dog and a cat, which is not what the question describes.

Rule out $A \cup B$

$A \cup B$ is the set of people who own a dog or a cat or both, which is not what the question describes.

State the answer

So $A' \cap B$ is the set that contains exactly the people who own a cat but not a dog.

Write down the size of the universal set

The universal set $\xi$ is everybody in the question: all $35$ people in a survey. Every region of the Venn diagram has to add back to that.

Write down the facts you are given

The question gives $n(A) = 18$, $n(A \cap B) = 5$ and $n(A' \cap B') = 7$. Everything else has to be worked out from the diagram.

Work out $n(A \cup B)$

Every one of the people who own neither a dog nor a cat is also one of the people altogether, so taking them away leaves exactly the people who own a dog or a cat or both: $35 - 7 = 28$.

Work out $n(A \cap B')$

Every one of the people who own both a dog and a cat is also one of the people who own a dog, so taking them away leaves exactly the people who own a dog but not a cat: $18 - 5 = 13$.

Work out $n(B)$

Every one of the people who own a dog but not a cat is also one of the people who own a dog or a cat or both, so taking them away leaves exactly the people who own a cat: $28 - 13 = 15$.

Read the completed Venn diagram

Every person sits in exactly one region of the diagram, and the completed diagram now shows all 4 of them: $7$ people who own neither a dog nor a cat, $13$ people who own a dog but not a cat, $10$ people who own a cat but not a dog and $5$ people who own both a dog and a cat.

Check the regions add back to $35$

The regions do not overlap and together they cover everyone, so they must add back to the $35$ people you started with. They do, so the diagram is consistent.

Write the probability as a fraction

A probability is the number of ways the event can happen divided by the size of the group being chosen from. Here the choice is made from all $35$ people, so the denominator is $35$.

Write the fraction in its simplest form

The highest common factor of $15$ and $35$ is $5$, so divide the top and the bottom by $5$ to get $\frac{3}{7}$.

State the answer

So the probability is $\frac{3}{7}$.

Write down what you know

There are $44$ people altogether. Every probability in the options is out of that $44$, because the person is chosen from everybody.

Work out $n(A \cap C')$

Every one of the people who own both a dog and a rabbit is also one of the people who own a dog, so taking them away leaves exactly the people who own a dog but not a rabbit: $20 - 9 = 11$.

Work out $n(A \cap B \cap C')$

Every one of the people who own all three of a dog, a cat and a rabbit is also one of the people who own both a dog and a cat, so taking them away leaves exactly the people who own a dog and a cat but not a rabbit: $10 - 6 = 4$.

Work out $n(A \cap B' \cap C')$

Every one of the people who own a dog and a cat but not a rabbit is also one of the people who own a dog but not a rabbit, so taking them away leaves exactly the people who own a dog but not a cat and a rabbit: $11 - 4 = 7$.

Complete the rest of the diagram

Filling in every region first means each statement can be tested straight from the finished diagram.

Work out the probability the true statement is about

There are $7$ people who own a dog but not a cat and a rabbit out of $44$, so that probability is $\frac{7}{44}$ — which is what the first statement claims.

Test the statement about $A \cap B \cap C$

There are $6$ people who own all three of a dog, a cat and a rabbit, so that probability is really $\frac{3}{22}$, not $\frac{1}{8}$. The statement is false.

Test the statement about $A' \cap B \cap C'$

There are $6$ people who own a cat but not a dog and a rabbit, so that probability is really $\frac{3}{22}$, not $\frac{1}{6}$. The statement is false.

Test the statement about $A \cap B$

There are $10$ people who own both a dog and a cat, so that probability is really $\frac{5}{22}$, not $\frac{1}{5}$. The statement is false.

Test the statement about $A' \cap B' \cap C'$

There are $5$ people who own none of a dog, a cat and a rabbit, so that probability is really $\frac{5}{44}$, not $\frac{1}{10}$. The statement is false.

Read the completed Venn diagram

Every person sits in exactly one region of the diagram, and the completed diagram now shows all 8 of them: $5$ people who own none of a dog, a cat and a rabbit, $7$ people who own a dog but not a cat and a rabbit, $6$ people who own a cat but not a dog and a rabbit, $4$ people who own a dog and a cat but not a rabbit, $8$ people who own a rabbit but not a dog and a cat, $3$ people who own a dog and a rabbit but not a cat, $5$ people who own a cat and a rabbit but not a dog and $6$ people who own all three of a dog, a cat and a rabbit.

Check the regions add back to $44$

The regions do not overlap and together they cover everyone, so they must add back to the $44$ people you started with. They do, so the diagram is consistent.

Note that every region count is a whole number

Each number in the diagram counts people, so it has to be a whole number between $0$ and $44$. If a subtraction ever gave a negative number you would know you had taken it from the wrong set.

Check the answer with inclusion-exclusion

Adding $n(A)$, $n(B)$ and $n(C)$ counts each overlap twice and the middle three times, so subtract the three pairwise overlaps and add the middle back once. That gives $39$, exactly what the seven regions inside the circles add to.

State the answer

So the correct statement is that the probability of choosing one of the people who own a dog but not a cat and a rabbit is $\frac{7}{44}$.