Basic Probability and Sample Spaces

Probability is the branch of mathematics that measures how likely something is to happen. Whether you are deciding if you need an umbrella, working out the chance of winning a raffle, or judging whether a game is fair, you are using probability. This lesson builds the foundations for the whole OCR GCSE Mathematics (J560) probability strand: the probability scale, equally likely outcomes, the rule for working out the probability of an event, and how to list every possible result using a sample space.

This material is assessed in Assessment Objective 1 (AO1) — accurately recalling and using the probability formula — and in AO2, where you must reason about whether outcomes are equally likely and justify your answers. Foundation-tier questions usually ask you to write down or work out a single probability, while the listing skills here underpin the harder Higher-tier work later in the course.

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Key Vocabulary

Term	Definition
Outcome	A single possible result of an experiment, such as rolling a 4.
Event	An outcome or group of outcomes you are interested in, such as "rolling an even number".
Sample space	The set of all possible outcomes of an experiment, often written in curly brackets.
Equally likely	Outcomes that all have the same chance of happening.
Fair	A dice, coin or spinner for which every outcome is equally likely.
At random	Chosen so that every item has an equal chance of being picked.
Trial	One performance of an experiment, e.g. one spin of a spinner.

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The Probability Scale

Every probability is a number from 0 to 1. Nothing can be less likely than impossible, and nothing can be more likely than certain, so no probability is ever negative or greater than 1.

Probability	Word	Meaning
$0$0	Impossible	The event cannot happen.
Between $0$0 and $\frac{1}{2}$21​	Unlikely	The event probably will not happen.
$\frac{1}{2}$21​	Even chance	Just as likely to happen as not.
Between $\frac{1}{2}$21​ and $1$1	Likely	The event probably will happen.
$1$1	Certain	The event must happen.

A probability can be written as a fraction, a decimal or a percentage, and you should be comfortable moving between all three. For example, $\dfrac{1}{4} = 0.25 = 25\%$41​=0.25=25%. On a non-calculator paper you must convert these without a calculator, so learn the common ones until they are automatic. To turn a fraction into a decimal, divide the numerator by the denominator; to turn a decimal into a percentage, multiply by $100$100; and to turn a percentage into a fraction, write it over $100$100 and simplify. The words on the scale matter too: examiners expect you to use impossible, unlikely, even chance, likely and certain precisely, and to place a given probability correctly between $0$0 and $1$1 when asked to mark it on a scale.

It also helps to picture the scale as a number line. A probability close to $0$0 sits near the left-hand "impossible" end, a probability close to $1$1 sits near the right-hand "certain" end, and $\tfrac{1}{2}$21​ sits exactly in the middle. When a question gives you a worded description — "there is a small chance of snow" — you should be able to translate it into a region of the scale, and when it gives you a value such as $0.85$0.85 you should be able to describe it in words as "likely, but not certain".

Common error: Writing a probability such as "$2$2" or "$1.5$1.5". If your answer is bigger than $1$1, you have almost certainly added when you should have divided, or used a frequency instead of a probability.

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Equally Likely Outcomes and the Probability Formula

When every outcome of an experiment is equally likely, the probability of an event is

$P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$P(event)=total number of possible outcomesnumber of favourable outcomes​

This is called the theoretical probability because it is worked out from the structure of the experiment, not from carrying it out. The word "favourable" simply means "an outcome that counts as the event happening". The whole formula rests on one assumption — that the outcomes really are equally likely — so a fair dice, a fair coin and a spinner with equal sections all qualify, but a bent coin or an unevenly weighted spinner does not. Whenever you are about to use this formula, pause and confirm the equally-likely condition holds; if it does not, you must estimate the probability from data instead, which is the subject of the next lesson.

A second habit worth forming early is to count the total before you count the favourable outcomes. The total is the denominator, and almost every careless error in basic probability comes from getting it wrong — adding the wrong items, forgetting a colour, or counting categories instead of individual outcomes. Once the total is fixed, identifying the favourable outcomes is usually straightforward, and you can always sanity-check that your answer lies between $0$0 and $1$1.

Worked Example 1

A fair six-sided dice is rolled once. Work out the probability of rolling a $5$5.

Solution:

• Total equally likely outcomes $= 6$=6 (the faces $1, 2, 3, 4, 5, 6$1,2,3,4,5,6).
• Favourable outcomes (rolling a $5$5) $= 1$=1.

$P(5) = \dfrac{1}{6}$P(5)=61​

Worked Example 2

A fair six-sided dice is rolled once. Work out the probability of rolling an even number.

Solution:

• The even outcomes are $\{2, 4, 6\}${2,4,6}, so there are $3$3 favourable outcomes.

$P(\text{even}) = \dfrac{3}{6} = \dfrac{1}{2}$P(even)=63​=21​

Always simplify a probability fraction where you can — it shows clear mathematical communication and makes follow-up working easier.

Worked Example 3

A bag contains $4$4 red, $5$5 blue and $3$3 green counters. A counter is chosen at random. Work out the probability that it is blue.

Solution:

• Total counters $= 4 + 5 + 3 = 12$=4+5+3=12.
• Blue counters $= 5$=5.

$P(\text{blue}) = \dfrac{5}{12}$P(blue)=125​

Common error: Forgetting to find the total first. The denominator is always the total number of items, here $12$12, not the number of colours ($3$3).

Worked Example 4

The letters of the word MATHEMATICS are written on separate cards and one card is taken at random. Work out the probability that it shows the letter M.

Solution:

• MATHEMATICS has $11$11 letters: M, A, T, H, E, M, A, T, I, C, S.
• The letter M appears $2$2 times.

$P(M) = \dfrac{2}{11}$P(M)=112​

Worked Example 5

A spinner has $8$8 equal sections numbered $1$1 to $8$8. Find the probability of spinning a prime number.

Solution:

• The prime numbers from $1$1 to $8$8 are $2, 3, 5, 7$2,3,5,7, so there are $4$4 favourable outcomes.

$P(\text{prime}) = \dfrac{4}{8} = \dfrac{1}{2}$P(prime)=84​=21​

Common error: Counting $1$1 as prime. The number $1$1 is not prime, so it must not be included.

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Impossible and Certain Events

• An impossible event has probability $0$0: there are no favourable outcomes.
• A certain event has probability $1$1: every outcome is favourable.

Worked Example 6

A bag contains only red and blue counters. One counter is taken at random. Write down the probability that it is yellow, and give a reason for your answer.

Solution: There are no yellow counters, so the number of favourable outcomes is $0$0.

$P(\text{yellow}) = \dfrac{0}{\text{total}} = 0$P(yellow)=total0​=0

The event is impossible because no counter in the bag is yellow.

Worked Example 7

A standard dice is rolled. Write down the probability of rolling a number less than $7$7.

Solution: Every face ($1$1 to $6$6) is less than $7$7, so all $6$6 outcomes are favourable.

$P(\text{less than } 7) = \dfrac{6}{6} = 1$P(less than 7)=66​=1

The event is certain.

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The Probability of an Event Not Happening

The outcomes of an event either happen or they do not, and the two possibilities together are certain. So

$P(\text{not } A) = 1 - P(A)$P(not A)=1−P(A)

Worked Example 8

The probability that it rains tomorrow is $0.3$0.3. Work out the probability that it does not rain tomorrow.

Solution: $P(\text{not rain}) = 1 - 0.3 = 0.7$P(not rain)=1−0.3=0.7

Worked Example 9

A bag of counters has $P(\text{red}) = \dfrac{2}{5}$P(red)=52​. Work out the probability of not picking a red counter.

Solution: $P(\text{not red}) = 1 - \dfrac{2}{5} = \dfrac{5}{5} - \dfrac{2}{5} = \dfrac{3}{5}$P(not red)=1−52​=55​−52​=53​

Common error: Writing $1 - \dfrac{2}{5} = \dfrac{1}{3}$1−52​=31​. You must keep a common denominator of $5$5, not subtract the numerators and the denominators.

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Sample Spaces and Sample Space Diagrams

A sample space is a complete list of every possible outcome. Listing it carefully makes sure you do not miss any outcomes or count any twice. For a single experiment you can list the outcomes in curly brackets; for two experiments together a sample space diagram (a grid) is the clearest method.

Worked Example 10

Two fair coins are tossed. Write down the sample space.

Solution: Using H for heads and T for tails, list every combination of first and second coin:

$S = \{HH, HT, TH, TT\}$S={HH,HT,TH,TT}

There are $4$4 equally likely outcomes.

Common error: Listing only $\{HH, HT, TT\}${HH,HT,TT}. The outcomes HT (first head, second tail) and TH (first tail, second head) are different, so there are $4$4 outcomes, not $3$3.

Worked Example 11

Two fair six-sided dice are rolled and their scores are added. Complete a sample space diagram and find the probability that the total is $7$7.

Solution: Each cell shows the total of the row score and the column score.

$+$+	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

There are $6 \times 6 = 36$6×6=36 equally likely outcomes. The total $7$7 appears in the cells $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$(1,6),(2,5),(3,4),(4,3),(5,2),(6,1) — that is $6$6 outcomes.

$P(\text{total} = 7) = \dfrac{6}{36} = \dfrac{1}{6}$P(total=7)=366​=61​

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Extended Worked Examples

Worked Example 12

A fair coin is tossed and a fair four-sided dice (numbered $1$1 to $4$4) is rolled. Draw a sample space diagram and find the probability of getting a head and an even number.

Solution:

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

There are $2 \times 4 = 8$2×4=8 equally likely outcomes. "Head and even" is satisfied by H2 and H4 — $2$2 outcomes.

$P(\text{head and even}) = \dfrac{2}{8} = \dfrac{1}{4}$P(head and even)=82​=41​

Worked Example 13

A spinner has $5$5 equal sections numbered $1, 2, 3, 4, 5$1,2,3,4,5. The spinner is spun and a fair coin is tossed. (a) How many outcomes are in the sample space? (b) Work out the probability of getting a head and an odd number.

Solution: (a) Each of the $5$5 spinner outcomes pairs with each of the $2$2 coin outcomes, so there are $5 \times 2 = 10$5×2=10 outcomes. (b) Odd numbers are $\{1, 3, 5\}${1,3,5}, so the favourable outcomes are H1, H3, H5 — that is $3$3 outcomes.

$P(\text{head and odd}) = \dfrac{3}{10}$P(head and odd)=103​

Worked Example 14

A fair eight-sided dice is numbered $1$1 to $8$8. Work out: (a) $P(\text{factor of } 8)$P(factor of 8) (b) $P(\text{multiple of } 3)$P(multiple of 3) (c) $P(\text{square number})$P(square number)

Solution: (a) Factors of $8$8 are $\{1, 2, 4, 8\}${1,2,4,8}, so $P = \dfrac{4}{8} = \dfrac{1}{2}$P=84​=21​. (b) Multiples of $3$3 from $1$1 to $8$8 are $\{3, 6\}${3,6}, so $P = \dfrac{2}{8} = \dfrac{1}{4}$P=82​=41​. (c) Square numbers from $1$1 to $8$8 are $\{1, 4\}${1,4}, so $P = \dfrac{2}{8} = \dfrac{1}{4}$P=82​=41​.

Common error: Missing $1$1 as a factor of $8$8 and as a square number. The number $1$1 is a factor of every number and is itself a square ($1 = 1^2$1=12).

Worked Example 15

A biased spinner has four colours. The probabilities of three of them are $P(\text{red}) = 0.2$P(red)=0.2, $P(\text{blue}) = 0.3$P(blue)=0.3 and $P(\text{green}) = 0.15$P(green)=0.15. (a) Work out $P(\text{yellow})$P(yellow). (b) The spinner is spun once. Write down which colour is most likely.

Solution: (a) The four probabilities sum to $1$1, so $P(\text{yellow}) = 1 - (0.2 + 0.3 + 0.15) = 1 - 0.65 = 0.35$P(yellow)=1−(0.2+0.3+0.15)=1−0.65=0.35 (b) The largest probability is $0.35$0.35, so yellow is most likely.

Common error: Forgetting that the missing colour can be the most likely one. Always work out the unknown probability before deciding.

Answering at different grade levels

Specimen question modelled on the OCR J560 paper format: A bag contains $6$6 red, $3$3 blue and $1$1 green counter. A counter is taken at random. (a) Write down the probability that the counter is green. (b) Work out the probability that the counter is not red. (c) Give a reason why the event "the counter is yellow" has probability $0$0.

Grades 3–4 response: (a) Total $= 6 + 3 + 1 = 10$=6+3+1=10, so $P(\text{green}) = \dfrac{1}{10}$P(green)=101​. (b) Not red $= 3 + 1 = 4$=3+1=4, so $P(\text{not red}) = \dfrac{4}{10} = \dfrac{2}{5}$P(not red)=104​=52​. (c) There are no yellow counters.

Grades 5–6 response: (a) The sample space has $10$10 equally likely outcomes, so $P(\text{green}) = \dfrac{1}{10}$P(green)=101​. (b) Using the complement, $P(\text{not red}) = 1 - \dfrac{6}{10} = \dfrac{4}{10} = \dfrac{2}{5}$P(not red)=1−106​=104​=52​. (c) "Yellow" is not in the sample space $\{\text{red, blue, green}\}${red, blue, green}, so it has $0$0 favourable outcomes and probability $0$0.

Grades 7–9 response: (a) $P(\text{green}) = \dfrac{1}{10}$P(green)=101​. (b) $P(\text{not red}) = 1 - P(\text{red}) = 1 - \dfrac{6}{10} = \dfrac{2}{5}$P(not red)=1−P(red)=1−106​=52​; the complement is faster and less error-prone than re-counting. (c) The event "yellow" is the empty event within this sample space; it contains no outcomes, so $P(\text{yellow}) = 0$P(yellow)=0, the lower bound of the probability scale.

Examiner-style commentary: Marks are gained for showing the total, using the complement in (b) rather than re-counting, and giving a precise reason in (c) — "there are none" earns the mark, "it is unlikely" does not.

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Common Mistakes and Misconceptions

• Probability outside $0$0 to $1$1: any value below $0$0 or above $1$1 is impossible; re-check whether you divided by the total.
• Wrong denominator: the denominator is the total number of outcomes, not the number of categories.
• Treating HT and TH as the same: order matters when two separate items are involved, so two coins give $4$4 outcomes.
• Counting $1$1 as prime: $1$1 is not prime; the smallest prime is $2$2.
• Subtracting denominators for the complement: keep a common denominator when computing $1 - P(A)$1−P(A) with fractions.
• Confusing probability with frequency: a probability is a number between $0$0 and $1$1; a frequency is a whole-number count.

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Going Further

The idea that the probabilities of all outcomes add to $1$1 is the engine behind almost everything that follows. In later lessons you will use it to find a missing probability in a distribution, to derive $P(\text{not } A)$P(not A), and to check tree diagrams (the branches from any point must total $1$1). Sample-space listing also scales up: when listing by hand becomes unwieldy, you will switch to multiplication and tree diagrams, but the principle of counting equally likely outcomes never changes. A useful habit is to ask, before any probability question, "Are these outcomes really equally likely?" — if they are not, you must use experimental probability instead, which is the subject of the next lesson.

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Exam Tips

• For AO1, state the formula $P(\text{event}) = \dfrac{\text{favourable}}{\text{total}}$P(event)=totalfavourable​ and show the total clearly; method marks follow the working, not just the answer.
• When a question says "Write down", no working is expected — but a single clear value is; when it says "Work out" or "Calculate", show your method.
• For AO2 "Give a reason for your answer" parts, use precise language: impossible, certain, equally likely. Vague answers lose the mark.
• Draw the full sample space diagram for two-event problems; OCR awards a mark for the complete listing even before the probability is found — this is classic AO3 structured reasoning.
• Leave probabilities as exact fractions unless told otherwise; simplify to show control of the arithmetic.

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Summary

• Probability runs on a scale from $0$0 (impossible) to $1$1 (certain) and can be a fraction, decimal or percentage.
• For equally likely outcomes, $P(\text{event}) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$P(event)=total outcomesfavourable outcomes​.
• $P(\text{not } A) = 1 - P(A)$P(not A)=1−P(A), and the probabilities of all outcomes sum to $1$1.
• A sample space lists every possible outcome; a sample space diagram organises the outcomes of two experiments into a grid.
• Always check outcomes really are equally likely before using the theoretical formula.

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This content is aligned with the OCR GCSE Mathematics (J560) specification.