Hard GCSE Sets and Venn diagrams Questions

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 $\frac{7}{44}$.

List the elements of set $A$

$A$ is the set of even numbers in $\xi$. Working through $\xi$ gives $A = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$.

List the elements of set $B$

$B$ is the set of multiples of $3$ in $\xi$, so $B = \{3, 6, 9, 12, 15, 18\}$.

List the elements of set $C$

$C$ is the set of numbers greater than $10$ in $\xi$, so $C = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$.

Put every element into the Venn diagram

Each number goes in exactly one region, decided by which sets it is in. Elements in none of the sets go outside the circles.

Test $n(A \cap B \cap C) = 2$

$A \cap B \cap C$ has $2$ elements, so the claim that it has $2$ is true.

Test $9 \in A$

$A = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20\}$, so $9$ is not an element of $A$. The statement is false.

Test $B \subset C$

$B \subset C$ says every element of $B$ is also in $C$. Comparing $\{3, 6, 9, 12, 15, 18\}$ with $\{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$, that is false.

Test $A \cap B = \{6, 12\}$

$A \cap B$ is really $\{6, 12, 18\}$, so the statement is false.

Test $12 \notin C$

$C = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$, so $12$ is an element of $C$. The statement is false.

Check every element has been used exactly once

The regions do not overlap and together they cover $\xi$, so the region sizes must add back to $n(\xi) = 20$.

Write down the elements of $A \cap B$

The intersection holds the numbers that are in BOTH sets — the numbers that appear in the two lists.

Write down the elements of $A \cup B$

The union holds the numbers that are in at least one of the sets. Each number is written once, however many sets it is in.

Write down the elements of $A'$

$A'$ is everything in $\xi$ that is NOT in $A$.

Check the answer with inclusion-exclusion

Adding $n(A) = 10$ and $n(B) = 6$ counts the $3$ elements of $A \cap B$ twice, so subtract them once. That gives $13$, which is exactly how many numbers are written inside the circles.

State the answer

So the true statement is $n(A \cap B \cap C) = 2$. Every other statement fails when it is checked against the elements themselves.

Count the cells of the grid

Spinner $P$ can land on any of $6$ numbers and, for each of those, spinner $Q$ can land on any of $6$. Pairing every number on $P$ with every number on $Q$ gives $6 \times 6 = 36$ equally likely outcomes — one for each cell of the grid.

Work out the total in every cell

Each cell holds the total of the number on $P$ and the number on $Q$. Filling in the whole grid lists every outcome, with none missed and none counted twice.

Count the cells where the total is greater than $9$

Go through the grid and count the cells in which the total is greater than $9$. There are $6$ of them out of $36$.

Write the probability as a fraction

Every cell of the grid is equally likely, so the probability is the number of favourable cells divided by the total number of cells.

Write the fraction in its simplest form

The highest common factor of $6$ and $36$ is $6$, so divide top and bottom by $6$.

Write down the set of possible totals

The different totals form a SET: each one is written once, in order, however many cells it comes from. There are $11$ of them, not $36$.

Find the smallest and largest possible total

The smallest total comes from the smallest number on each spinner and the largest from the largest, so every total lies between $2$ and $12$. Anything outside that range has probability $0$.

Find the most likely total

A total of $7$ fills $6$ of the $36$ cells, more than any other, so it is the most likely total. Different totals are NOT equally likely — only the cells are.

Work out the probability of an even total

$18$ of the $36$ cells hold an even total, so the probability is $\frac{1}{2}$.

Note the mistake to avoid

It is tempting to divide by $11$, the number of different totals. That is wrong: the equally likely outcomes are the $36$ CELLS, so the denominator of every probability here is $36$.

Check the probabilities add to $1$

Every cell gives exactly one total, so the probabilities of the different totals must add to $1$. They do.

Note why a grid is used at all

A grid (a sample space diagram) lists every combination of the two spinners without missing any or writing any twice. That is what makes the outcomes countable — and countable is what "equally likely" needs.

Summarise the method

Fill in every cell, count the cells that match the event, and divide by the total number of cells.

Work out the probability of the smallest total

A total of $2$ can only happen one way for each cell that shows it, and $1$ of the $36$ cells do, so its probability is $\frac{1}{36}$.

State the answer

So the probability that the total is greater than $9$ is $\frac{1}{6}$.

Count the cells of the grid

Spinner $P$ can land on any of $4$ numbers and, for each of those, spinner $Q$ can land on any of $4$. Pairing every number on $P$ with every number on $Q$ gives $4 \times 4 = 16$ equally likely outcomes — one for each cell of the grid.

Work out the product in every cell

Each cell holds the product of the number on $P$ and the number on $Q$. Filling in the whole grid lists every outcome, with none missed and none counted twice.

Count the cells where the product is $4$

Go through the grid and count the cells in which the product is $4$. There are $3$ of them out of $16$.

Write the probability as a fraction

Every cell of the grid is equally likely, so the probability is the number of favourable cells divided by the total number of cells.

Check the fraction is in its simplest form

$3$ and $16$ have no common factor bigger than $1$, so $\frac{3}{16}$ is already as simple as it gets.

Write down the set of possible products

The different products form a SET: each one is written once, in order, however many cells it comes from. There are $9$ of them, not $16$.

Find the smallest and largest possible product

The smallest product comes from the smallest number on each spinner and the largest from the largest, so every product lies between $1$ and $16$. Anything outside that range has probability $0$.

Find the most likely product

A product of $4$ fills $3$ of the $16$ cells, more than any other, so it is the most likely product. Different products are NOT equally likely — only the cells are.

Work out the probability of an even product

$12$ of the $16$ cells hold an even product, so the probability is $\frac{3}{4}$.

Note the mistake to avoid

It is tempting to divide by $9$, the number of different products. That is wrong: the equally likely outcomes are the $16$ CELLS, so the denominator of every probability here is $16$.

Check the probabilities add to $1$

Every cell gives exactly one product, so the probabilities of the different products must add to $1$. They do.

Note why a grid is used at all

A grid (a sample space diagram) lists every combination of the two spinners without missing any or writing any twice. That is what makes the outcomes countable — and countable is what "equally likely" needs.

Summarise the method

Fill in every cell, count the cells that match the event, and divide by the total number of cells.

Work out the probability of the smallest product

A product of $1$ can only happen one way for each cell that shows it, and $1$ of the $16$ cells do, so its probability is $\frac{1}{16}$.

State the answer

So the probability that the product is $4$ is $\frac{3}{16}$.

List the elements of set $A$

$A$ is the set of multiples of $3$ in $\xi$. Working through $\xi$ gives $A = \{3, 6, 9, 12, 15\}$.

List the elements of set $B$

$B$ is the set of odd numbers in $\xi$, so $B = \{1, 3, 5, 7, 9, 11, 13, 15\}$.

Put every element into the Venn diagram

Each number goes in exactly one region, decided by which sets it is in. Elements in none of the sets go outside the circles.

Find the elements of $A' \cap B$

$A' \cap B$ is the set of numbers in $\xi$ that are odd but not multiples of $3$. Reading them off the diagram gives $\{1, 5, 7, 11, 13\}$.

Check every element has been used exactly once

The regions do not overlap and together they cover $\xi$, so the region sizes must add back to $n(\xi) = 15$.

Write down the elements of $A \cap B$

The intersection holds the numbers that are in BOTH sets — the numbers that appear in the two lists.

Write down the elements of $A \cup B$

The union holds the numbers that are in at least one of the sets. Each number is written once, however many sets it is in.

Write down the elements of $A'$

$A'$ is everything in $\xi$ that is NOT in $A$.

Check the answer with inclusion-exclusion

Adding $n(A) = 5$ and $n(B) = 8$ counts the $3$ elements of $A \cap B$ twice, so subtract them once. That gives $10$, which is exactly how many numbers are written inside the circles.

Note the mistake to avoid

Simply adding $n(A)$ and $n(B)$ would double-count the $3$ numbers in both sets. A number is listed only ONCE in a union, however many sets it belongs to.

Note what $\in$ and $\subset$ mean

$\in$ means "is an element of" — it goes between a NUMBER and a set. $\subset$ means "is a subset of" — it goes between two SETS. Every set here is a subset of $\xi$.

Summarise the method

List each set from its definition, place every element of $\xi$ in the right region of the diagram, and then read the answer straight off the diagram.

Work out the probability of landing in $A \cap B$

If one of the $15$ numbers in $\xi$ were picked at random, the chance of picking one of the $3$ elements of $A \cap B$ would be $\frac{1}{5}$.

Check the two circles against the universal set

Everything inside the circles plus everything outside them is the whole of $\xi$, so this must come back to $15$. It does, so nothing has been missed or written twice.

State the answer

So $A' \cap B = \{1, 5, 7, 11, 13\}$. The elements are written in order and none is repeated.