1MA1 · P1–P4, P7ProbabilityFoundation & HigherGrades 1–4

Probability Basics

Probability measures how likely something is, on a scale from 0 (impossible) to 1 (certain). For fair dice, coins and bags of counters, you can calculate it exactly by counting outcomes. For two events at once — two dice, a coin and a spinner — a $\textbf{sample space diagram}$ lists every combination so nothing is missed.

40 min Probability

The big picture

When outcomes are equally likely, probability is simply a count: favourable outcomes divided by total outcomes. The probabilities of all the different possible outcomes always add up to 1, which gives the most useful shortcut in the topic: P(not A) $= 1 -$ P(A).

Edexcel regularly gives a probability table with an unknown, such as $x$ and $2x$ , and asks you to find it using the fact that the probabilities sum to 1. Answers must be fractions, decimals or percentages — never ratios or words like "1 in 4".

What you'll learn

Use the probability scale and write probabilities as fractions, decimals or percentages
Calculate probabilities of equally likely outcomes
Use the fact that probabilities of mutually exclusive, exhaustive outcomes sum to 1
List outcomes systematically and use sample space diagrams for two events

The probability scale

Every probability lies between 0 and 1: 0 means impossible, 1 means certain, and 0.5 means an even chance.

Write probabilities as fractions, decimals or percentages: $\tfrac14 = 0.25 = 25\%$ .

Never write a probability greater than 1, negative, or as a ratio such as $1 : 4$ .

Tip — "1 in 4" or "1 out of 4" may be how people talk, but Edexcel only credits $\tfrac14$ , $0.25$ or $25\%$ .

Equally likely outcomes

If all outcomes are equally likely, P(event) $= \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ .

A bag with 4 red, 5 blue and 3 green counters has 12 counters, so P(blue) $= \tfrac{5}{12}$ .

P(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}

Equally likely outcomes.

Example 1A letter is chosen at random from the word PROBABILITY. Find the probability that it is a B, and that it is a vowel.

111 letters in total.

2B appears twice:

\tfrac{2}{11}

3Vowels O, A, I, I:

\tfrac{4}{11}

AnswerP(B)

= \tfrac{2}{11}

; P(vowel)

= \tfrac{4}{11}

Your turnSame method, new numbers — try it before moving on.

A letter is chosen at random from the word MATHEMATICS. What is the probability it is an M?

Your answer

Show worked solution

Solution

1MATHEMATICS has 11 letters, including 2 Ms.
2 $P(M) = \frac{2}{11}$ .

Answer: $\frac{2}{11}$

Outcomes that sum to 1

Events are $\textbf{mutually exclusive}$ if they cannot happen at the same time, like getting a 2 and getting a 5 on one roll.

If a list of mutually exclusive outcomes covers every possibility, their probabilities add to 1.

So P(not A) $= 1 -$ P(A).

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

Complement.

Example 1A spinner lands on red, blue, green or yellow. P(red)

= 0.2

, P(blue)

= 0.35

, P(green)

= x

and P(yellow)

= 2x

. Find P(yellow).

0.2 + 0.35 + x + 2x = 1

3x = 0.45

, so

x = 0.15

3P(yellow)

= 2x

Answer

0.3

Your turnSame method, new numbers — try it before moving on.

Four outcomes have probabilities $0.15$ , $0.4$ , $x$ and $x$ . Find $x$ .

Your answer

Show worked solution

Solution

1 $0.55 + 2x = 1$ .
2 $2x = 0.45$ , so $x = 0.225$ .

Answer: $x = 0.225$

Why this works

The "sum to 1" rule is the reason the complement rule works: an event either happens or it does not, and those two outcomes cover everything.

Sample spaces

For two events, list outcomes systematically, or draw a $\textbf{sample space diagram}$ — a grid with one event along each side.

Two fair dice give $6 \times 6 = 36$ equally likely outcomes.

The product rule for counting: if one event has $m$ outcomes and another has $n$ , together they have $m \times n$ combinations.

Example 1Two fair six-sided dice are rolled and the scores added. Find the probability that the total is 7.

136 outcomes in the grid.

2Totals of 7:

(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

— six outcomes.

Answer

\tfrac{6}{36} = \tfrac16

Your turnSame method, new numbers — try it before moving on.

Two fair six-sided dice are rolled and the scores added. What is the probability of a total of 9?

Your answer

Show worked solution

Solution

1Totals of 9: (3,6), (4,5), (5,4), (6,3).
2 $\frac{4}{36} = \frac{1}{9}$ .

Answer: $\frac{1}{9}$

Tip — List pairs in a fixed order — first die 1 to 6, then second die — so repeated or missing combinations are easy to spot.

Think like an examiner

Give probabilities as fractions, decimals or percentages only.
Show the equation when a table of probabilities contains an unknown.
Show the sample space or list, not just the final count.

Probability

0 \le P(A) \le 1

Probability scale.

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

Complement.

m \times n

Combinations of two events.

Watch out for these

Writing a probability as "3 out of 8" or

3 : 8

Write

\tfrac38

Counting

(2, 5)

and

(5, 2)

as the same outcome with two dice.

They are different outcomes; both count.

Giving a probability greater than 1.

Recheck: probabilities cannot exceed 1.

Forgetting that probabilities in a complete table sum to 1.

Use this to find unknowns.

Stretch yourself

A bag contains red, white and blue beads. The probability of red is $\tfrac14$ and the probability of white is $\tfrac25$ . There are 21 blue beads. How many beads are in the bag altogether?

Hint — Find the probability of blue first, then use it as a fraction of the total.

Reveal solution

SolutionP(blue)

= 1 - \tfrac14 - \tfrac25 = 1 - \tfrac{5}{20} - \tfrac{8}{20} = \tfrac{7}{20}

. So

\tfrac{7}{20}

of the beads are blue: 21 beads, meaning

\tfrac{1}{20}

is 3 beads. Total

= 60

beads (15 red, 24 white, 21 blue).

\blacksquare

Questions students ask

What does "at random" mean?

Every item has the same chance of being chosen.

Are all outcomes always equally likely?

No. A biased dice or unequal spinner sections give unequal probabilities; then you need experiments or given values.

How many outcomes are there for three coins?

2 \times 2 \times 2 = 8

Key takeaways

Probabilities lie between 0 and 1, written as fractions, decimals or percentages.
Equally likely outcomes: favourable ÷ total.
Mutually exclusive, exhaustive outcomes sum to 1; P(not A) $= 1 -$ P(A).
Use sample space diagrams for two events.
Combinations of two events: $m \times n$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 22 marks

A bag contains 4 red and 6 green balls only. One ball is taken at random.

(a)Write down the probability that it is green.[1]
Show mark scheme (a)
- B1 $\frac{6}{10}$ or equivalent
Answer: $\frac{3}{5}$
(b)Write down the probability that it is blue.[1]
Show mark scheme (b)
- B1 $0$

Question 2Grade 33 marks

A spinner lands on 1, 2, 3 or 4 with probabilities $0.1$ , $0.3$ , $2x$ and $x$ .

Work out the probability that it lands on 3.[3]
Show mark scheme
- M1 $0.4 + 3x = 1$
- M1 $x = 0.2$
- A1 $0.4$
Answer: $0.4$

Question 3Grade 42 marks

Two fair coins are flipped and a fair six-sided dice is rolled.

Work out the probability of getting two heads and a six.[2]
Show mark scheme
- M1 $2 \times 2 \times 6 = 24$ outcomes, or $\frac{1}{2} \times \frac{1}{2} \times \frac{1}{6}$
- A1 $\frac{1}{24}$
Answer: $\frac{1}{24}$

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

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Answer a few questions, then review anything you missed.

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