Constructing Possibility Spaces and Calculating Theoretical Probabilities Revision Notes | GCSE AQA Higher Maths

1. Why Possibility Spaces Matter

You flip two coins. What is the probability of getting exactly one head? Many students say $\frac{1}{3}$ — because they list HH, HT, TT and pick the middle one. That answer is wrong, and a possibility space is exactly what reveals why.

A possibility space is a structured record of every outcome that could happen in an experiment. When outcomes are equally likely, it turns probability into a counting exercise: count the favourable outcomes, count the total, divide.

The key word is equally likely. A fair coin, a fair die, or a fair spinner all produce equally likely outcomes — so theoretical probability applies directly. Once you have a complete possibility space, every probability question becomes a fraction.

2. Single Experiments — Building the Possibility Space

For a single experiment, the possibility space is a simple list. Roll one fair six-sided die: the outcomes are $\{1, 2, 3, 4, 5, 6\}$. There are 6 equally likely outcomes, so each has probability $\frac{1}{6}$.

From this list you can answer any question. The probability of rolling a prime number means identifying the favourable outcomes — 2, 3, and 5 — and counting them: $P(\text{prime}) = \frac{3}{6} = \frac{1}{2}$.

Single experiments are the foundation. The real power of possibility spaces comes when you combine two experiments — and that is where a systematic table becomes essential.

3. Combined Experiments — The Two-Way Table

When two experiments happen together — rolling two dice, flipping two coins, spinning two spinners — the number of outcomes multiplies. Two dice give $6 \times 6 = 36$ outcomes. Trying to list these in a random order almost guarantees you miss some.

The solution is a two-way possibility space table. One experiment labels the rows; the other labels the columns. Every cell in the table is one outcome — and because the table is a grid, you cannot skip any.

This table shows the sum of two fair dice. Every cell is one equally likely outcome, and there are $6 \times 6 = 36$ cells in total. To find any probability, count the cells showing the target value and divide by 36.

Worked example: Using a Possibility Space Table — Sum of Two Dice

Two fair six-sided dice are rolled. Using the possibility space, find the probability that the sum is 7.

1. Total outcomes $= 6 \times 6 = 36$ — The table has 6 rows and 6 columns, giving 36 equally likely outcomes. [M1 — correct total]
2. Pairs that sum to 7: $(1,6),\ (2,5),\ (3,4),\ (4,3),\ (5,2),\ (6,1)$ — Read along the diagonal in the table where the cell value is 7. There are 6 such cells. [M1 — systematic listing]
3. $P(\text{sum} = 7) = \frac{6}{36}$ — Favourable outcomes over total outcomes. [A1]
4. $\frac{6}{36} = \frac{1}{6}$ — Divide numerator and denominator by 6. Always simplify unless the question asks for a specific form. [A1]

Answer: $P(\text{sum} = 7) = \frac{1}{6}$

4. Combined Experiments — Lists and Grids for Other Setups

Not every combined experiment involves dice. The same table method works for coins, spinners, or any pair of equally likely experiments. The approach is always the same: one experiment across the top, one down the side, fill every cell.

Consider flipping two fair coins. The possibility space has $2 \times 2 = 4$ outcomes: HH, HT, TH, TT. Notice that HT and TH are different outcomes — the first coin landed heads in one case and tails in the other. This is the trap that causes the $\frac{1}{3}$ error from the opening.

Worked example: Possibility Space for Two Coins — Probability of Exactly One Head

Two fair coins are flipped. Find the probability of getting exactly one head.

1. Possibility space: $\{HH,\ HT,\ TH,\ TT\}$ — Two coins, each with 2 outcomes: $2 \times 2 = 4$ equally likely outcomes total. [M1]
2. Exactly one head: $HT$ and $TH$ — HT means coin 1 is heads, coin 2 is tails. TH means coin 1 is tails, coin 2 is heads. These are two distinct outcomes.
3. $P(\text{exactly one head}) = \frac{2}{4} = \frac{1}{2}$ — 2 favourable outcomes out of 4 total. [A1] — This confirms the opening answer of $\frac{1}{3}$ was wrong because TH and HT were incorrectly merged.

Answer: $P(\text{exactly one head}) = \frac{1}{2}$

5. A Spinner and a Die — Applying the Method

Now combine two different experiments: a fair four-sided spinner (labelled 1, 2, 3, 4) and a fair six-sided die. The possibility space has $4 \times 6 = 24$ equally likely outcomes.

Setting up the table with the spinner values as rows and die values as columns, each cell records the pair (spinner result, die result). From this table, any probability question can be answered by counting.

Worked example: Spinner and Die — Probability of Equal Results

A fair four-sided spinner (1–4) and a fair six-sided die (1–6) are used together. Find the probability that the spinner and die show the same number.

1. Total outcomes $= 4 \times 6 = 24$ — 4 spinner outcomes multiplied by 6 die outcomes. [M1]
2. Same number: $(1,1),\ (2,2),\ (3,3),\ (4,4)$ — The spinner only goes up to 4, so matches are only possible for 1, 2, 3, and 4. The die showing 5 or 6 cannot match the spinner.
3. Number of favourable outcomes $= 4$ — There are exactly 4 pairs where both values are equal.
4. $P(\text{same number}) = \frac{4}{24} = \frac{1}{6}$ — Divide numerator and denominator by 4. [A1]

Answer: $P(\text{same number}) = \frac{1}{6}$

7. Exam Traps and Common Mistakes

Possibility space questions are reliable mark-earners — but only if you avoid the errors that examiners see most often. The mistakes below each cost marks in a predictable way, so recognising them in advance is worth the effort.

Constructing a complete possibility space and applying $P(\text{event}) = \frac{\text{favourable outcomes}}{\text{total outcomes}}$ covers the full higher-tier specification requirement for theoretical probability with equally likely outcomes — for both single and combined experiments.