Probability: revision notes and practice

What you will learn

• Calculate probabilities using a stated, equally likely sample space.
• Use complements and distinguish independent from mutually exclusive events.
• Follow successive draws with or without replacement.
• Higher extension: interpret probability when extra information restricts the possibilities.

Key rules

Equally likely outcomes

Probability = number of favourable outcomes ÷ total outcomes, provided each outcome is equally likely.

Find the complement

P(not A) = 1 − P(A). The probabilities of all possible outcomes add to 1.

Combine events

For independent events, P(A and B) = P(A) × P(B). For mutually exclusive events, P(A or B) = P(A) + P(B).

Worked examples

Example 1: A bag contains 3 red, 2 blue and 5 green counters. One is chosen at random. Find P(not red).

1. There are 10 counters and each counter is equally likely to be chosen.
2. There are 2 + 5 = 7 counters that are not red.
3. P(not red) = 7/10.

7/10

Example 2: A fair coin is tossed and a fair six-sided die is rolled independently. Find P(heads and a 6).

1. P(heads) = 1/2 and P(6) = 1/6.
2. The events are independent, so multiply: 1/2 × 1/6.

1/12

A practice question

A bag contains 7 red, 8 blue and 7 green counters. One counter is chosen at random. What is the probability that it is red or blue? Give a fraction in its simplest form.

1. Use the probability scale correctly

A probability is between 0 and 1 inclusive. Zero means impossible under the stated model, and 1 means certain. A probability of 1/2 means an event has the same likelihood as its complement, not that it must occur once in every pair of trials.

Give numerical probabilities as fractions, decimals or percentages. For example, 1/4 = 0.25 = 25%. Probabilities outside the allowed range signal an error. Start by stating what experiment is being performed and what event is being asked for.

2. Count equally likely outcomes

When all elementary outcomes are equally likely, divide the number that satisfy the event by the total number of outcomes. A fair six-sided die has two results greater than 4, namely 5 and 6, so the probability is 2/6 = 1/3.

Count outcomes rather than just categories. A bag with three red counters and one blue counter has four equally likely counter selections, but the two colours are not equally likely. The chance of red is 3/4, not 1/2. If a spinner's sections have unequal angles, count labels alone only if the model justifies it.

P(event) = favourable outcomes ÷ total equally likely outcomes

3. Find complements and check totals

The complement of an event is everything in the sample space that is not that event. Exactly one of the event and its complement happens, so their probabilities sum to 1. If P(red) = 0.35, then P(not red) = 0.65.

Use a complement when it is simpler than listing all wanted outcomes. For a fair die, not rolling a 6 has probability 1 − 1/6 = 5/6. In a complete probability table, missing entries can be found by subtracting the known probabilities from 1.

P(not A) = 1 − P(A)

4. Add alternatives without counting twice

Mutually exclusive events cannot occur together in the same trial. On one die roll, rolling 1 and rolling 2 are mutually exclusive, so P(1 or 2) = 1/6 + 1/6 = 1/3. Here 'or' collects alternative acceptable outcomes.

If events overlap, simple addition counts their overlap twice. On a die, 'even' and 'greater than 4' overlap at 6. The union is {2, 4, 5, 6}, so its probability is 4/6 = 2/3. Listing a sample space or drawing a Venn diagram can prevent double counting.

5. Multiply along independent stages

Independent events do not change one another's probabilities. A fair coin toss and a fair die roll can be modelled independently, so P(heads and 6) = 1/2 × 1/6 = 1/12. In a probability tree, multiply the branch probabilities along one complete path.

At each branching point, the outgoing probabilities must add to 1. Add complete path probabilities when several disjoint paths meet the question's condition. Independence is different from mutual exclusivity: two independent events can happen together, while mutually exclusive events cannot.

For independent events, P(A and B) = P(A) × P(B).

6. Update draws without replacement

Without replacement, the selected counter stays out, so the total and sometimes the relevant colour count change. With three red and two blue counters, P(two reds) = 3/5 × 2/4 = 3/10. After a red first draw, only two reds remain among four counters.

For one red and one blue in either order, there are two disjoint paths: red then blue and blue then red. Their probabilities are 3/5 × 2/4 and 2/5 × 3/4. Adding gives 3/10 + 3/10 = 3/5. With replacement, the original counts are restored before the second draw.

7. List outcomes and use 'at least one'

An organised table or list can make combined outcomes clear. Two fair coin tosses have four equally likely ordered outcomes: HH, HT, TH and TT. Exactly one head has probability 2/4, because HT and TH are different paths.

At least one head includes HH as well as HT and TH. Its probability is 3/4, or 1 minus P(no heads). For larger independent experiments, a complement often saves work: at least one success means every possibility except no successes. Read 'exactly', 'at least' and 'in this order' carefully.

8. Interpret experimental frequency

Relative frequency is the observed event count divided by the number of trials. If a spinner lands on red 18 times in 60 trials, the estimate is 18/60 = 0.3. It estimates the underlying probability; it does not prove that the spinner's exact chance is 0.3.

Larger unbiased samples tend to give more stable estimates, but short runs still vary. If the probability is 0.3, the expected count in 200 future trials is 200 × 0.3 = 60. Expected does not mean guaranteed. Previous independent results do not make a future result overdue.

9. Higher extension: conditional probability

Extra information can restrict the sample space. If a fair die result is known to be even, the remaining possibilities are {2, 4, 6}. The probability it is greater than 4 is then 1/3, because only 6 qualifies within those three equally likely possibilities.

In a two-way table, 'given the person is in group B' means use group B's total as the denominator. Do not divide by the total of everyone unless the question concerns everyone. Tree diagrams already use stage-specific probabilities; conditional questions ask you to interpret the restriction explicitly.

Check your understanding

How to earn the marks

• State fairness, independence or replacement assumptions when relevant.
• Check every probability lies between 0 and 1.
• Multiply along a tree path; add disjoint acceptable paths.
• Update both numerator and denominator after each unreplaced draw.

A common mistake

Without replacement, the number of counters changes after a draw. The next probability may change too.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Mathematics — P4, P7, P8

  Simple events and combined independent or dependent events are available at Foundation. Explicit conditional probability is Higher content and is not in this bank.

• Edexcel GCSE Mathematics — P4, P7, P8

  Simple events and combined independent or dependent events are available at Foundation. Explicit conditional probability is Higher content and is not in this bank.

• OCR GCSE Mathematics — 11.01, 11.02

  Simple events and combined independent or dependent events are available at Foundation. Explicit conditional probability is Higher content and is not in this bank.

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