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Maths 5 min read October 27, 2026
Maths

11+ Probability: Equally Likely Outcomes and Worked Answers

Count outcomes fairly, use complements and distinguish replacement with eight original probability questions and explained answers.

A bag containing red and blue counters does not automatically give each colour a half chance. If there are more red counters than blue, a random draw is more likely to select red. Counting the colour names misses the individual objects that can actually be selected.

These original explanations and questions support preparation for 11-plus exams taken in 2027. They are not official papers or a complete school syllabus. Check your target school’s guidance and try each question before opening the explanation. Written English models show a supported response, not the only acceptable wording.

Define the experiment before calculating

For a finite set of equally likely outcomes, probability is the number of favourable outcomes divided by the total number of outcomes. Say what is random and why the outcomes can be treated equally. A fair six-sided die has six equally likely numbered faces; a spinner needs equal sectors and a fair mechanism before equal-sector counting is justified.

ProbabilityInterpretation
0Impossible in the stated experiment
1/2An even chance
1Certain in the stated experiment

Values between zero and one-half are less likely than not; values between one-half and one are more likely than not. These descriptions do not promise what will happen on a particular attempt. Even a high-probability outcome can fail to occur.

Worked example: count objects, not categories

A bag has five red counters and three blue counters, all equally likely to be drawn. The chance of red is 5/8, not 1/2. The chance of not red is 3/8. Together those complementary events cover every counter once, so their probabilities total one. This complement method works when “not” means every outcome outside the named event.

Worked example: why replacement matters

Start with three green counters and two yellow counters. Draw one uniformly at random, replace it, mix, then draw again. The probability of two greens is 3/5 × 3/5 = 9/25. Without replacement, after a green has been removed there are two greens among four counters: 3/5 × 2/4 = 3/10. Updating the second fraction is essential; the first draw changes the bag.

For two fair independent coin tosses, write the ordered outcomes HH, HT, TH and TT. “One head and one tail” covers HT and TH, not just one outcome. A list is often clearer than a shortcut, particularly when order matters or several routes lead to the same event.

Eight original questions

A bag contains 4 red and 6 blue counters. Each counter is equally likely to be drawn. What is the probability of red?

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2/5

Four of ten counters are red: 4/10 = 2/5.

Roll a fair die numbered 1 to 6. What is the probability of a number greater than 4?

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1/3

The favourable faces are 5 and 6: 2/6 = 1/3.

A fair spinner has 8 equal sectors, 3 green and 5 orange. What is the probability of not green?

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5/8

Five sectors are outside the green event. Equivalently, 1 − 3/8 = 5/8.

Draw uniformly from cards numbered 1 to 12. What is the probability of a multiple of 3?

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1/3

The favourable cards are 3, 6, 9 and 12: 4/12 = 1/3.

Toss a fair coin twice independently. What is the probability of exactly one head?

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1/2

HT and TH are two of the four equally likely ordered outcomes.

A bag has 2 red and 3 blue counters. Draw uniformly twice, replacing and mixing after the first draw. Find the probability of two reds.

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4/25

The chance is 2/5 on each independent draw: 2/5 × 2/5 = 4/25.

Use the same 2-red, 3-blue bag but do not replace the first counter. Find the probability of two reds.

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1/10

First red: 2/5. Then one red remains among four counters: 1/4. Multiply to get 2/20 = 1/10.

A fair die numbered 1 to 6 is rolled. What is the probability of rolling 7?

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0

No face is numbered 7, so the event has no favourable outcomes.

Check the model as well as the arithmetic

An answer greater than one or below zero cannot be a probability. A denominator based on colour names rather than counters needs correcting. A repeated-draw calculation needs an explicit replacement rule. None of these theoretical models predicts an actual school result or pass rate.

The Stanford probability teaching notes explain equally likely sample spaces, complements and replacement. This guide is an introduction to selected reasoning problems, not a claim that probability appears in every 11-plus specification.

Choose a useful next step

Use fraction and decimal conversions to express the same probability differently, and data interpretation to distinguish observed counts from a model of chance. Return to the maths learning hub or explore original practice and worked answers.

Sources and scope

The DfE programme of study provides a curriculum reference, not a promise about every school’s test. Examples and questions on this page are original; no school affiliation or educator endorsement is implied.

Ready to practise?

Choose original practice and use the answer explanations to decide what to work on next. These are not official school papers.