GCSE Possibility spaces Practice Questions

Work out how many equally likely outcomes there are

The spinner has $6$ equal sections, so the sample space is the list of those $6$ equally likely outcomes. Every probability in this question is out of that total of $6$.

Find the outcomes for which the spinner lands on an even number

The favourable outcomes are 2, 4, 6. Each one is favourable because it divides exactly by $2$.

Write the probability as a fraction and simplify it

$3$ favourable outcomes out of $6$ equally likely outcomes. The highest common factor of $3$ and $6$ is $3$, so dividing top and bottom by $3$ gives $\frac{1}{2}$.

Work out how many equally likely outcomes there are

The first number can be any of $5$ values and the second can be any of $5$ values, so there are $5 \times 5 = 25$ equally likely ordered pairs. Each one is one cell of the possibility space grid. Every probability in this question is out of that total of $25$.

Find the outcomes for which the total is $6$

The favourable outcomes are (1, 5), (2, 4), (3, 3), (4, 2), (5, 1). Each one is favourable because it is exactly $6$.

Write the probability as a fraction and simplify it

$5$ favourable outcomes out of $25$ equally likely outcomes. The highest common factor of $5$ and $25$ is $5$, so dividing top and bottom by $5$ gives $\frac{1}{5}$.

Work out how many equally likely outcomes there are

The first number can be any of $6$ values and the second can be any of $6$ values, so there are $6 \times 6 = 36$ equally likely ordered pairs. Each one is one cell of the possibility space grid. Every probability in this question is out of that total of $36$.

Complete the possibility space grid

Each cell holds the total of the number at the top of its column and the number at the side of its row. Filling in every cell writes out the whole sample space, with nothing missed and nothing counted twice.

Search the grid systematically, one column at a time

Go along the columns in order and, within each column, down the rows. Working in a fixed order is what stops a pair being missed or written down twice — which is exactly what a possibility space is for.

Find the outcomes for which the total is $5$

The favourable outcomes are (1, 4), (2, 3), (3, 2), (4, 1). Each one is favourable because it is exactly $5$.

Rule out the other four lists

None of the other four is the complete list: the list with $2$ pairs lists each pair only once, as if $(a, b)$ and $(b, a)$ were the same outcome. They are not: they are different cells of the grid. the list with $3$ pairs has missed one of the favourable cells. the list with $3$ pairs has missed the first favourable cell. the list with $5$ pairs includes $(1, 1)$, which is an outcome of the experiment but does not make the event happen.

State the answer

The complete list of ordered pairs for which the total is $5$ is (1, 4), (2, 3), (3, 2), (4, 1) — $4$ of the $36$ equally likely outcomes.

Work out how many equally likely outcomes there are

The first number can be any of $6$ values and the second can be any of $6$ values, so there are $6 \times 6 = 36$ equally likely ordered pairs. Each one is one cell of the possibility space grid. Every probability in this question is out of that total of $36$.

Complete the possibility space grid

Each cell holds the total of the number at the top of its column and the number at the side of its row. Filling in every cell writes out the whole sample space, with nothing missed and nothing counted twice.

Count the cells that give a total of $6$

$5$ of the $36$ cells of the grid give a total of $6$, so $P = \frac{5}{36}$.

Count the cells that give a total of $10$

Only $3$ of the $36$ cells give a total of $10$, so $P = \frac{3}{36}$.

Compare the two probabilities

The two fractions have the same denominator — both are counts out of the same $36$ equally likely outcomes — so the bigger numerator wins. $5 > 3$, so a total of $6$ is more likely.

Rule out "a total of $10$ is more likely"

A total of $10$ fills only $3$ cells against $5$, so it is the less likely of the two, not the more likely.

Rule out "the two totals are equally likely"

Both totals are possible, but "possible" is not "equally likely". They come from different numbers of cells ($5$ against $3$), so their probabilities differ. This is the classic mistake in this topic.

Rule out the statement that claims $7$ and $2$ cells

It picks the right winner but miscounts the grid, so the statement as a whole is false. A reason that does not survive a recount is not a reason.

Rule out the statement that swaps the two counts

That statement attaches the count $5$ to a total of $10$. The grid says the opposite, so it is false twice over.

State the answer

A total of $6$ is more likely: $5$ of the $36$ equally likely outcomes give it, against only $3$ for a total of $10$.

Work out how many equally likely outcomes there are

The first number can be any of $6$ values and the second can be any of $6$ values, so there are $6 \times 6 = 36$ equally likely ordered pairs. Each one is one cell of the possibility space grid. Every probability in this question is out of that total of $36$.

Complete the possibility space grid

Each cell holds the difference of the number at the top of its column and the number at the side of its row. Filling in every cell writes out the whole sample space, with nothing missed and nothing counted twice.

Count the cells that give a difference of $1$

$10$ of the $36$ cells of the grid give a difference of $1$, so $P = \frac{10}{36}$.

Count the cells that give a difference of $3$

Only $6$ of the $36$ cells give a difference of $3$, so $P = \frac{6}{36}$.

Compare the two probabilities

The two fractions have the same denominator — both are counts out of the same $36$ equally likely outcomes — so the bigger numerator wins. $10 > 6$, so a difference of $1$ is more likely.

Rule out "a difference of $3$ is more likely"

A difference of $3$ fills only $6$ cells against $10$, so it is the less likely of the two, not the more likely.

Rule out "the two differences are equally likely"

Both differences are possible, but "possible" is not "equally likely". They come from different numbers of cells ($10$ against $6$), so their probabilities differ. This is the classic mistake in this topic.

Rule out the statement that claims $12$ and $5$ cells

It picks the right winner but miscounts the grid, so the statement as a whole is false. A reason that does not survive a recount is not a reason.

Rule out the statement that swaps the two counts

That statement attaches the count $10$ to a difference of $3$. The grid says the opposite, so it is false twice over.

Check that the outcomes really are equally likely

Everything in the question is fair, and every cell of the grid describes exactly one way the experiment can turn out, so each of the $36$ outcomes has probability $\frac{1}{36}$. Without that, counting cells would prove nothing.

Count the outcomes for which it does NOT happen

$26$ of the $36$ outcomes are not favourable, and $10 + 26 = 36$, so nothing has been missed or counted twice.

Work out the probability that it does not happen

An event and its opposite have probabilities that add to $1$, so the probability that it does not happen is $\frac{13}{18}$. Adding the two back together is a quick check.

Write the probability as a decimal

The same number as a decimal. The question asked for a fraction, because the fraction is exact, but the decimal shows how big the probability is.

Write the probability as a percentage

The event happens in about 27.778\% of all the possible ways the experiment can turn out.

State the answer

A difference of $1$ is more likely: $10$ of the $36$ equally likely outcomes give it, against only $6$ for a difference of $3$.