Basic Probability and Sample Spaces

This lesson introduces the fundamental ideas of probability as required by the AQA GCSE Mathematics specification. You will learn how to describe the likelihood of events using the probability scale, calculate simple probabilities from equally likely outcomes, list outcomes systematically, and use sample space diagrams when two events are combined.

The Probability Scale

Probability is a measure of how likely an event is to happen. It is always expressed as a number between 0 and 1 (inclusive).

Probability value	Meaning
0	The event is impossible — it can never happen
0.5	The event is even chance — equally likely to happen or not
1	The event is certain — it will definitely happen

Probabilities can be written as fractions, decimals, or percentages.

Fraction	Decimal	Percentage	Likelihood
0	0	0%	Impossible
1/4	0.25	25%	Unlikely
1/2	0.5	50%	Even chance
3/4	0.75	75%	Likely
1	1.0	100%	Certain

Exam Tip: A probability can never be less than 0 or greater than 1. If your answer falls outside this range, check your working — you have made an error.

Calculating Simple Probabilities

When all outcomes are equally likely, the probability of an event is:

P(event) = number of favourable outcomes / total number of possible outcomes

Worked Example 1

A fair six-sided die is rolled once. What is the probability of rolling a 4?

Number of favourable outcomes = 1 (just the number 4)
Total number of outcomes = 6 (the numbers 1, 2, 3, 4, 5, 6)

P(4) = 1/6

Worked Example 2

A bag contains 3 red balls, 5 blue balls, and 2 green balls. A ball is chosen at random. What is the probability that the ball is blue?

Total number of balls = 3 + 5 + 2 = 10
Number of blue balls = 5

P(blue) = 5/10 = 1/2

Exam Tip: Always simplify your fractions where possible. Examiners expect answers in their simplest form unless the question says otherwise.

Listing Outcomes

When solving probability problems, it is important to list outcomes systematically so that none are missed. This means working through possibilities in a logical order.

Worked Example 3

A coin is flipped and a die is rolled. List all the possible outcomes.

Organising by coin result first:

H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6

There are 12 possible outcomes in total.

This can be confirmed by the counting principle: 2 (coin outcomes) x 6 (die outcomes) = 12.

Sample Space Diagrams

A sample space diagram (also called a two-way table or possibility space) is a grid that shows all possible outcomes when two events are combined. It is especially useful for finding probabilities involving two dice, two spinners, or a combination of events.

Worked Example 4 — Two Dice

Two fair six-sided dice are rolled and their scores are added together. Find the probability of getting a total of 7.

	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Total number of outcomes = 6 x 6 = 36
Outcomes that give a total of 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes

P(total of 7) = 6/36 = 1/6

Worked Example 5 — Two Spinners

Spinner A has sections labelled 1, 2, 3. Spinner B has sections labelled 1, 2, 3, 4. Both spinners are spun and the results are multiplied.

	1	2	3	4
1	1	2	3	4
2	2	4	6	8
3	3	6	9	12

Total outcomes = 3 x 4 = 12

P(product is 6) = 2/12 = 1/6 (the outcomes (2,3) and (3,2))

P(product is even) = count all even products in the table, then divide by 12.

Even products: 2, 4, 2, 4, 6, 8, 6, 12 = 8 outcomes

P(product is even) = 8/12 = 2/3

Exam Tip: When drawing a sample space diagram in the exam, always label the rows and columns clearly. This makes it easy for the examiner to see your working and award method marks even if you make an arithmetic slip.

The Complement of an Event

The complement of an event A is the event "A does not happen", written as A' (or sometimes as "not A").

P(A') = 1 - P(A)

Worked Example 6

The probability that it rains tomorrow is 0.3. What is the probability that it does not rain?

P(not rain) = 1 - 0.3 = 0.7

Exam Tip: The complement rule is one of the most useful tools in probability. If it is hard to calculate P(event) directly, try calculating P(not event) and subtracting from 1 instead.

Summary

Probability is measured on a scale from 0 (impossible) to 1 (certain).
For equally likely outcomes: P(event) = favourable outcomes / total outcomes.
Always list outcomes systematically to avoid missing any.
A sample space diagram is a grid showing all outcomes for two combined events.
The complement rule states that P(not A) = 1 - P(A).
Probabilities can be expressed as fractions, decimals, or percentages.

Exam Tip: In multiple-mark questions, always show your sample space diagram or outcome list — you will earn method marks even if your final answer is incorrect.

Extended Worked Examples

Worked Example 7 — A More Complex Sample Space

Two ordinary fair dice are rolled. Find the probability that the difference between the two scores (larger minus smaller) is exactly 2.

Begin by constructing the full sample space. There are $6 \times 6 = 36$ equally likely outcomes. We record $|a - b|$ for every pair $(a, b)$ :

	1	2	3	4	5	6
1	0	1	2	3	4	5
2	1	0	1	2	3	4
3	2	1	0	1	2	3
4	3	2	1	0	1	2
5	4	3	2	1	0	1
6	5	4	3	2	1	0

Counting the cells showing 2, we find the ordered pairs $(1,3), (3,1), (2,4), (4,2), (3,5), (5,3), (4,6), (6,4)$ , giving 8 outcomes.

$P(\text{difference} = 2) = \frac{8}{36} = \frac{2}{9}$

Because the outcomes in the sample space are equally likely, this is a theoretical probability arrived at purely from counting — no experiment is required.

Worked Example 8 — Combining a Coin and a Spinner

A fair coin (H, T) is flipped and a fair four-sided spinner (1, 2, 3, 4) is spun. Produce the full sample space and find the probability that the outcome contains a head and an even number.

	1	2	3	4
H	H1	H2	H3	H4
T	T1	T2	T3	T4

There are $2 \times 4 = 8$ equally likely outcomes. The favourable outcomes are H2 and H4, giving:

$P(\text{head and even}) = \frac{2}{8} = \frac{1}{4}$

Compare this with $P(\text{head}) \cdot P(\text{even}) = \frac{1}{2} \cdot \frac{2}{4} = \frac{1}{4}$ . The two match because the coin and the spinner are independent events — a link we will formalise in a later lesson.

Worked Example 9 — Probability from a Described Bag

A bag contains sweets: 4 strawberry, 7 lemon, 5 mint, 4 cola. A sweet is selected at random. Find:

$P(\text{strawberry})$
$P(\text{mint or cola})$
$P(\text{not lemon})$

Total $= 4 + 7 + 5 + 4 = 20$ .

$P(\text{strawberry}) = \frac{4}{20} = \frac{1}{5}$

Because mint and cola cannot both be the outcome of a single selection, they are mutually exclusive:

$P(\text{mint or cola}) = \frac{5}{20} + \frac{4}{20} = \frac{9}{20}$

Using the complement rule:

$P(\text{not lemon}) = 1 - P(\text{lemon}) = 1 - \frac{7}{20} = \frac{13}{20}$

Worked Example 10 — A Subtle Listing Problem

A fair three-sided spinner (red, amber, green) is spun twice. Find the probability that the two colours are different.

The sample space contains $3 \times 3 = 9$ equally likely outcomes. Constructing it carefully:

	R	A	G
R	RR	RA	RG
A	AR	AA	AG
G	GR	GA	GG

The outcomes with the same colour are RR, AA, GG (3 outcomes). The outcomes with different colours are the remaining 6.

$P(\text{different}) = \frac{6}{9} = \frac{2}{3}$

Alternatively, $P(\text{different}) = 1 - P(\text{same}) = 1 - \frac{3}{9} = \frac{2}{3}$ . Two correct methods reach the same answer — a useful check in an exam.

Worked Example 11 — Reading a Probability Scale

A probability $p$ is marked on a scale between 0 and 1 at $p = 0.35$ . Describe the likelihood and convert to a fraction and a percentage.

As a decimal: $0.35$
As a fraction: $\frac{35}{100} = \frac{7}{20}$
As a percentage: $35\%$

Since $0.35 < 0.5$ , the event is unlikely (but not impossible). Pupils often confuse "unlikely" with "impossible" — the two are very different. Impossible means $p = 0$ ; unlikely means $0 < p < 0.5$ .

Worked Example 12 — Picking with Multiple Constraints

A deck of 20 cards contains 8 red cards and 12 blue cards. Each red card shows a letter (4 vowels, 4 consonants). Each blue card shows a digit (6 odd, 6 even). A card is chosen at random. Find $P(\text{red vowel})$ , $P(\text{red or odd})$ , and $P(\text{not blue even})$ .

The sample space has 20 equally likely outcomes. The number of red vowel cards is 4.

$P(\text{red vowel}) = \frac{4}{20} = \frac{1}{5}$

For $P(\text{red or odd})$ , the red set (8 cards) and the odd-digit set (6 cards) are mutually exclusive — no card is both red and odd-digit (red cards carry letters, blue cards carry digits). Therefore:

$P(\text{red or odd}) = \frac{8}{20} + \frac{6}{20} = \frac{14}{20} = \frac{7}{10}$

For $P(\text{not blue even})$ : blue-even cards number 6. Using the complement:

$P(\text{not blue even}) = 1 - \frac{6}{20} = \frac{14}{20} = \frac{7}{10}$

Worked Example 13 — Two-Coin Sample Space

Two fair coins are flipped. List the outcomes systematically and find $P(\text{at least one head})$ .

The sample space is $\{HH, HT, TH, TT\}$ , which is 4 equally likely outcomes. "At least one head" occurs in $HH, HT, TH$ — 3 outcomes.

$P(\text{at least one head}) = \frac{3}{4}$

Using the complement: $P(\text{at least one head}) = 1 - P(TT) = 1 - \frac{1}{4} = \frac{3}{4}$ . Both methods agree.

Answering at different grade levels

Exam-style question: A fair spinner has five equal sections labelled 1, 2, 3, 4, 5. The spinner is spun once. Find the probability that the result is an even prime number.

Grades 3–4 answer: List the outcomes on the sample space {1, 2, 3, 4, 5}. An even prime is just 2. One favourable outcome out of five equally likely outcomes gives a theoretical probability of $\frac{1}{5}$ .
Grades 5–6 answer: Identify that the events "even" and "prime" both apply only to 2 (since 2 is the only even prime — this is a key definition). Using $P(E) = \frac{\text{favourable outcomes}}{\text{total outcomes}}$ , we have $P(\text{even prime}) = \frac{1}{5} = 0.2 = 20\%$ .
Grades 7–9 answer: State that 2 is the unique even prime (any other even number has 2 as a factor, so has at least three divisors). Therefore the favourable set is $\{2\}$ , and all outcomes on a fair spinner are equally likely and mutually exclusive. Hence $P(\text{even prime}) = \frac{1}{5}$ . For a follow-up "spun 200 times" part, the expected frequency would be $\frac{1}{5} \times 200 = 40$ , calculated from the theoretical probability rather than from relative frequency.

AQA alignment: This content is aligned with AQA GCSE Mathematics (8300) specification — specifically Topic P P1 Language of probability, P2 Probability scale and relative frequency, P3 Sample space diagrams and listing outcomes, P4 Mutually exclusive and exhaustive events (introduction). Assessed on Papers 1, 2, and 3.

Basic Probability and Sample Spaces

Basic Probability and Sample Spaces

The Probability Scale

Calculating Simple Probabilities

Worked Example 1

Worked Example 2

Listing Outcomes

Worked Example 3

Sample Space Diagrams

Worked Example 4 — Two Dice

Worked Example 5 — Two Spinners

The Complement of an Event

Worked Example 6

Summary

Extended Worked Examples

Worked Example 7 — A More Complex Sample Space

Worked Example 8 — Combining a Coin and a Spinner

Worked Example 9 — Probability from a Described Bag

Worked Example 10 — A Subtle Listing Problem

Worked Example 11 — Reading a Probability Scale

Worked Example 12 — Picking with Multiple Constraints

Worked Example 13 — Two-Coin Sample Space

Answering at different grade levels

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