Hard GCSE Possibility spaces Questions

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 $3$.

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 $7$

$6$ of the $36$ cells of the grid give a total of $7$, so $P = \frac{6}{36}$.

Count the cells that give a total of $4$

Only $3$ of the $36$ cells give a total of $4$, 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. $6 > 3$, so a total of $7$ is more likely.

Rule out "a total of $4$ is more likely"

A total of $4$ fills only $3$ cells against $6$, 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 ($6$ against $3$), so their probabilities differ. This is the classic mistake in this topic.

Rule out the statement that claims $8$ 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 $6$ to a total of $4$. 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

$30$ of the $36$ outcomes are not favourable, and $6 + 30 = 36$, 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 $\frac{5}{6}$. 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 16.667\% of all the possible ways the experiment can turn out.

State the answer

A total of $7$ is more likely: $6$ of the $36$ equally likely outcomes give it, against only $3$ for a total of $4$.

Work out how many equally likely outcomes there are

The first number can be any of $4$ values and the second can be any of $4$ values, so there are $4 \times 4 = 16$ 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 $16$.

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 how many outcomes give each total

Read the completed grid: a total of $2$ appears in $1$ cell, a total of $8$ in $1$, and so on. These counts add to $16$, the whole sample space.

Compare the two extreme totals

A total of $5$ comes from $4$ cells of the grid; a total of $2$ comes from only $1$. The outcomes are equally likely — the totals are not.

Conclude that the totals are not equally likely

Because $4 \ne 1$, the two probabilities are different, so Priya is wrong. It is the $16$ ordered pairs that are equally likely, never the $7$ totals they produce.

Rule out "there are 7 possible totals, so each has probability 1/7"

Counting the $7$ possible totals as the sample space is exactly the mistake. If it were right, a total of $5$ would have probability $\frac{1}{7}$, but it really has $\frac{4}{16}$.

Rule out "everything is fair, so every total is equally likely"

Fairness makes the OUTCOMES equally likely, not the values they add up to. Some totals can be made in many ways and some in only one, which is what the grid shows at a glance.

Rule out the statement that says $2$ outcomes give a total of $5$

The verdict is right but the count is not: the grid shows $4$ cells with a total of $5$, not $2$, so that statement is still false.

Rule out the statement that says $3$ outcomes give a total of $2$

Again the verdict is right but the count is wrong: only $1$ cell of the grid give a total of $2$.

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 $16$ outcomes has probability $\frac{1}{16}$. Without that, counting cells would prove nothing.

Count the outcomes for which it does NOT happen

$12$ of the $16$ outcomes are not favourable, and $4 + 12 = 16$, 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 $\frac{3}{4}$. 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 25\% of all the possible ways the experiment can turn out.

State the answer

Priya is wrong. A total of $5$ comes from $4$ of the $16$ equally likely outcomes and a total of $2$ from only $1$, so the totals are not equally likely.

Work out how many equally likely outcomes there are

The first number can be any of $3$ values and the second can be any of $4$ values, so there are $3 \times 4 = 12$ 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 $12$.

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 $7$

The favourable outcomes are (1, 6), (3, 4). Each one is favourable because it is exactly $7$.

Count the favourable outcomes

There are $2$ of them, out of the $12$ equally likely outcomes in the sample space. Counting cells of the grid — not values — is the whole method.

Write the probability as a fraction and simplify it

$2$ favourable outcomes out of $12$ equally likely outcomes. The highest common factor of $2$ and $12$ is $2$, so dividing top and bottom by $2$ gives $\frac{1}{6}$.

Rule out the list with 2 pairs

That list has written every pair the wrong way round, so most of these are not even outcomes of this experiment.

Rule out the list with 1 pair

That list has missed one of the favourable cells.

Rule out the list with 1 pair

That list has missed the first favourable cell.

Rule out the list with 3 pairs

That list includes $(1, 2)$, which is an outcome of the experiment but does not make the event happen.

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 $12$ outcomes has probability $\frac{1}{12}$. Without that, counting cells would prove nothing.

Count the outcomes for which it does NOT happen

$10$ of the $12$ outcomes are not favourable, and $2 + 10 = 12$, 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 $\frac{5}{6}$. 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.

State the answer

The complete list of ordered pairs for which the total is $7$ is (1, 6), (3, 4) — $2$ of the $12$ equally likely outcomes.

Work out how many equally likely outcomes there are

The first card can be any of $6$ cards. The first card is NOT put back, so the second card can be any of the $5$ cards that are left. That gives $6 \times 5 = 30$ equally likely ordered pairs — the $6$ pairs with the same card twice cannot happen, and are crossed out in the grid. Every probability in this question is out of that total of $30$.

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. The cells on the diagonal are crossed out: the same card cannot be taken twice.

Find the outcomes for which the total is an even number

Going through the grid one column at a time, the favourable cells are $2$ with a first number of $1$, $2$ with a first number of $2$, $2$ with a first number of $3$, $2$ with a first number of $4$, $2$ with a first number of $5$ and $2$ with a first number of $6$. Each one is favourable because it divides exactly by $2$.

Count the favourable outcomes

There are $12$ of them, out of the $30$ equally likely outcomes in the sample space. Counting cells of the grid — not values — is the whole method.

Put the numbers into the rule

$12$ favourable outcomes out of $30$ equally likely outcomes gives $\frac{12}{30}$.

Write the fraction in its simplest form

The highest common factor of $12$ and $30$ is $6$, so divide the top and the bottom by $6$ to get $\frac{2}{5}$.

Test the statement that gives $\frac{4}{9}$

That statement counts the $9$ possible totals as if they were equally likely. They are not: they come from different numbers of cells of the grid, so the denominator must be the $30$ ordered pairs, never the $9$ totals. So it is false.

Test the statement that gives $\frac{13}{30}$

That statement has the right sample space but has counted $13$ favourable cells instead of $12$ — one cell too many. So it is false.

Test the statement that gives $\frac{11}{30}$

That statement has the right sample space but has counted $11$ favourable cells instead of $12$ — one cell has been missed. So it is false.

Test the statement that gives $\frac{4}{5}$

That statement has double-counted, giving $24$ favourable cells instead of $12$. So it is false.

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 $30$ outcomes has probability $\frac{1}{30}$. Without that, counting cells would prove nothing.

Count the outcomes for which it does NOT happen

$18$ of the $30$ outcomes are not favourable, and $12 + 18 = 30$, 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 $\frac{3}{5}$. 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.

State the answer

Only the first statement survives: the probability that the total is an even number is $\frac{2}{5}$, from $12$ favourable cells out of $30$ equally likely ordered pairs.