Probability GCSE Foundation: A Resit Guide

A probability question can look almost too simple. A bag, a spinner, perhaps two coins. Then one phrase such as “without replacement” changes everything.

For resit students, probability GCSE foundation revision is less about learning dozens of rules and more about building a reliable method. Identify every possible outcome, decide whether outcomes are equally likely, then use the right representation. That might be a list, a Venn diagram or a simple tree diagram.

This guide explains the core ideas, with worked examples at GCSE foundation level. The goal is not to guess what might happen. It is to describe uncertainty accurately enough to earn the marks.

Your probability revision checklist

You should be able to:

• place probabilities on the scale from $0$ to $1$;
• calculate probability from equally likely outcomes;
• use the fact that probabilities covering every outcome total $1$;
• list outcomes systematically;
• use experimental results to calculate relative frequency;
• complete and interpret Venn diagrams;
• multiply along branches and add separate routes on probability trees;
• recognise when probabilities change because an item is not replaced.

Begin with the MathsGenie probability revision guide if you need a short explanation before attempting questions. Students following another exam board can use the dedicated AQA probability resources or OCR probability resources.

Basic probability for GCSE foundation

Probability measures how likely an event is. It can be written as a fraction, decimal or percentage, and it lies from $0$ to $1$ inclusive.

• $0$ means impossible.
• $1$ means certain.
• A value close to $0$ means unlikely.
• A value close to $1$ means likely.

When outcomes are equally likely, use:

$$P(event)=\frac{number\; of\; favourable\; outcomes}{total\; number\; of\; possible\; outcomes}$$

Worked example: choosing a counter

A bag contains $5$ red counters, $3$ blue counters and $2$ green counters. One counter is chosen at random. Find the probability of choosing blue.

First find the total number of counters:

$$5+3+2=10$$

There are $3$ favourable outcomes, so:

$$P(blue)=\frac{3}{10}$$

The denominator represents all possible counters, not merely the counters that are not blue.

Probabilities that total one

If a set of events covers every possible outcome and the events cannot overlap, its probabilities add to $1$.

Suppose a biased spinner can land on red, blue or yellow. If

$$P(red)=0.35$$

and

$$P(blue)=0.4,$$

then:

$$P(yellow)=1-0.35-0.4=0.25$$

The same idea gives the probability of something not happening:

$$P(notA)=1-P(A)$$

This is called the complement. It is often quicker than counting many unwanted outcomes separately.

Listing outcomes without missing any

Listing outcomes is easy when there are only a few possibilities. The difficulty is being systematic.

Worked example: two coins

Two fair coins are tossed. Use $H$ for heads and $T$ for tails. The possible outcomes are:

$$HH,HT,TH,TT$$

Notice that $HT$ and $TH$ are different outcomes. The first records heads then tails; the second records tails then heads.

The probability of exactly one head is therefore:

$$P(exactly\; one\; head)=\frac{2}{4}=\frac{1}{2}$$

A common resit mistake is to list only $HH$, $HT$ and $TT$. Writing outcomes in an organised order prevents this.

For two different events, a sample space table can help. Put the outcomes of one event across the top and the outcomes of the other down the side. Complete every cell before counting the outcomes that satisfy the question.

Relative frequency and experimental probability

Sometimes probability comes from observations rather than a perfectly fair theoretical model. Relative frequency is calculated using:

$$relative\; frequency=\frac{number\; of\; times\; the\; event\; occurs}{total\; number\; of\; trials}$$

Worked example: estimating a probability

A drawing pin is dropped $80$ times and lands point-up $28$ times. Estimate the probability that it lands point-up.

$$P(point-up)\approx \frac{28}{80}=0.35$$

The approximation sign matters because this is an estimate based on an experiment. Another $80$ drops may produce a different result.

This estimate can predict an expected frequency. If the drawing pin is dropped another $200$ times, the estimated number of point-up results is:

$$200\times 0.35=70$$

An expected frequency is not a promise. It is a sensible long-run estimate. In general, results tend to become more stable as the number of trials increases, although a small experiment can still give an unusual outcome.

Using Venn diagrams in foundation probability

A Venn diagram shows how sets overlap. The rectangle represents the universal set, meaning everything under consideration. Circles represent sets, and their overlap contains items belonging to both sets.

The notation may look unfamiliar, but the underlying questions are usually concrete:

• $A\cap B$ means in both $A$ and $B$;
• $A\cup B$ means in $A$, in $B$, or in both;
• $A^{\prime }$ means not in $A$.

Worked example: completing a Venn diagram

A group contains $30$ students. Of these, $18$ study French, $12$ study Spanish and $5$ study both languages.

Start with the overlap. This is the most important habit in Venn diagram questions.

The number studying French only is:

$$18-5=13$$

The number studying Spanish only is:

$$12-5=7$$

The number studying at least one language is:

$$13+5+7=25$$

Therefore, the number studying neither language is:

$$30-25=5$$

If one student is selected at random, the probability that the student studies French or Spanish is:

$$P(F\cup S)=\frac{25}{30}=\frac{5}{6}$$

The word “or” includes the overlap unless the question says “but not both”. Counting $18+12$ would count the $5$ students in the overlap twice.

Simple probability trees

Probability trees display events that happen in stages. Each branch represents a possible result.

Two rules do most of the work:

• probabilities on branches leaving the same point add to $1$;
• multiply probabilities along a route through the tree.

When more than one route gives the required result, calculate each route and then add them.

Worked example: independent events

A fair coin is tossed twice. Find the probability of getting two heads.

The probability of heads on each toss is $\frac{1}{2}$. Multiply along the heads-then-heads route:

$$P(HH)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$

Now find the probability of exactly one head. There are two valid routes:

$$P(HT)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$ $$P(TH)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$$

Add the separate routes:

$$P(exactly\; one\; head)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Worked example: without replacement

A bag contains $3$ red counters and $2$ blue counters. Two counters are selected without replacement. Find the probability that both are red.

For the first choice:

$$P(first\; red)=\frac{3}{5}$$

After one red counter is removed, $2$ red counters remain among $4$ counters. Therefore:

$$P(second\; red\mid first\; red)=\frac{2}{4}$$

Multiply along the route:

$$P(two\; red)=\frac{3}{5}\times \frac{2}{4}=\frac{6}{20}=\frac{3}{10}$$

The phrase “without replacement” changes both the numerator and denominator on the second stage. With replacement, the second probability would remain $\frac{3}{5}$.

Use the MathsGenie probability video lessons to see these decisions modelled before trying questions independently.

How to approach probability questions in a resit

A dependable process is more valuable than a clever shortcut.

Translate the wording

Underline phrases such as “at least one”, “exactly one”, “neither”, “both” and “without replacement”. They determine which outcomes belong in your answer.

Choose a representation

Use a list for a small set of outcomes, a Venn diagram for overlapping groups and a probability tree for events in stages. The diagram is not decoration. It stores the structure of the problem.

Show the operation

Write the multiplication along each route and the addition between routes. Even if your final arithmetic goes wrong, clear working may earn method marks, depending on the mark scheme.

MathsGenie’s exam-style probability question bank lets you practise this process repeatedly, while the worked solutions and mark schemes help you see where marks are awarded. Once the topic feels secure, mix it into full papers rather than keeping it isolated.

Common mistakes in probability GCSE foundation questions

Using the wrong total

The denominator must represent the total number of possible equally likely outcomes. Recount after an item is removed.

Assuming every object has a different outcome

If a bag contains several red counters, the event “red” includes every red counter. Count objects when objects are selected; count listed outcomes when outcomes are listed.

Adding when you should multiply

Multiply along one route through probability trees. Add different routes that produce the required event.

Forgetting the overlap in a Venn diagram

Enter the intersection first. Otherwise, items in both sets are easily counted twice.

Treating relative frequency as exact

Experimental probability is an estimate. It describes what happened in the trials and can support a prediction, but it does not guarantee the next result.

Giving an impossible answer

A probability cannot be below $0$ or above $1$. Use this quick check before moving on.

Turn understanding into marks

Probability becomes calmer once each question has somewhere to go. Lists organise outcomes. Venn diagrams organise overlap. Probability trees organise stages. Relative frequency connects experiments to sensible estimates.

For a resit, the best next step is short, regular practice followed by honest checking. Use MathsGenie’s free revision lessons, practice questions, mini tests, video solutions and mark schemes to find the precise step that needs attention. Then move into past papers and predicted papers under timed conditions.

You do not need certainty before you begin. Probability itself is a reminder that uncertainty can still be handled with structure. Open the relevant MathsGenie probability lesson, complete a small set of questions, mark every answer, and return to the errors tomorrow. That is how an uncertain topic becomes dependable marks.