Free Grade 3 Sample Space Diagrams in Context Quizzes
Free Grade 3 quizzes for Sample Space Diagrams in Context: 12 real questions with a full answer key and worked solutions. Probability measures how likely something is, from 0 (impossible) to 1 (certain). P(event) = favourable outcomes ÷ total outcomes. For independent events, multiply.
Free
Foundation
35 min
24
Learning objectives
Every objective is checkable, and each one maps to an activity and an assessment.
- UnderstandExplain sample space diagrams in context
Explain sample space diagrams in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
- ApplyCalculate sample space diagrams in context
Calculate sample space diagrams in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
- AnalyseCompare sample space diagrams in context
Compare sample space diagrams in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
- EvaluateJustify sample space diagrams in context
Justify sample space diagrams in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
Prerequisites
Close these gaps first — each has its own free lesson, worksheet and quiz.
How Sample Space Diagrams in Context works
Probability measures how likely something is, from 0 (impossible) to 1 (certain). P(event) = favourable outcomes ÷ total outcomes. For independent events, multiply.
12 practice questions
Easy to hard, 24 marks in total. Try them first, then open the answers.
- 1.A bag has 2 red, 8 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 2.A bag has 5 red, 7 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 3.A bag has 5 red, 3 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 4.A bag has 7 red, 8 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 5.A bag has 2 red, 5 blue and 2 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 6.A bag has 5 red, 3 blue and 0 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 7.A bag has 6 red, 2 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 8.A bag has 3 red, 8 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 9.A bag has 4 red and 7 blue counters. Two are picked with replacement. Find P(both red).[3]
- 10.A bag has 3 red and 2 blue counters. Two are picked with replacement. Find P(both red).[3]
- 11.A bag has 8 red and 8 blue counters. Two are picked with replacement. Find P(both red).[3]
- 12.A bag has 7 red and 8 blue counters. Two are picked with replacement. Find P(both red).[3]
Show answers and working
1. P(red) = 2/13
Total counters = 13. P(red) = 2/13 = 2/13.
2. P(red) = 1/3
Total counters = 15. P(red) = 5/15 = 1/3.
3. P(red) = 5/11
Total counters = 11. P(red) = 5/11 = 5/11.
4. P(red) = 7/18
Total counters = 18. P(red) = 7/18 = 7/18.
5. P(red) = 2/9, P(not red) = 7/9
Total counters = 9. P(red) = 2/9 = 2/9. P(not red) = 1 − 2/9 = 7/9.
6. P(red) = 5/8, P(not red) = 3/8
Total counters = 8. P(red) = 5/8 = 5/8. P(not red) = 1 − 5/8 = 3/8.
7. P(red) = 6/11, P(not red) = 5/11
Total counters = 11. P(red) = 6/11 = 6/11. P(not red) = 1 − 6/11 = 5/11.
8. P(red) = 3/14, P(not red) = 11/14
Total counters = 14. P(red) = 3/14 = 3/14. P(not red) = 1 − 3/14 = 11/14.
9. 16/121
P(red) = 4/11 each time. Independent, so multiply: 4/11 × 4/11 = 16/121.
10. 9/25
P(red) = 3/5 each time. Independent, so multiply: 3/5 × 3/5 = 9/25.
11. 1/4
P(red) = 8/16 each time. Independent, so multiply: 8/16 × 8/16 = 1/4.
12. 49/225
P(red) = 7/15 each time. Independent, so multiply: 7/15 × 7/15 = 49/225.
Worked examples
A bag has 2 red, 5 blue and 5 green counters. One is picked at random. Find P(red).
- Total counters = 12.
- P(red) = 2/12 = 1/6.
- Answer: P(red) = 1/6
A bag has 7 red, 5 blue and 4 green counters. One is picked at random. Find P(red) and P(not red).
- Total counters = 16.
- P(red) = 7/16 = 7/16.
- P(not red) = 1 − 7/16 = 9/16.
- Answer: P(red) = 7/16, P(not red) = 9/16
A bag has 2 red and 7 blue counters. Two are picked with replacement. Find P(both red).
- P(red) = 2/9 each time.
- Independent, so multiply: 2/9 × 2/9 = 4/81.
- Answer: 4/81
Interactive practice
Free, no account required — answer on screen or print it.
Common mistakes
What learners typically get wrong, and how to fix it.
Writes probability as a ratio like 3:5.
Correction: Use a fraction, decimal or percentage.
Adds probabilities for 'and' events.
Correction: 'And' means multiply when events are independent.
Teacher tips
- · Run real experiments with dice and compare to theory.
Parent tips
- · Talk about the chance of rain from the forecast.
Real-life applications
- · Weather forecasts, games, insurance.
Assessment objectives
How mastery is evidenced, and which standards it maps to.
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Frequently asked
- Is this Grade 3 Sample Space Diagrams in Context quiz really free?
- Yes. Every quizzes on FreeMathWorksheet is free to use, print and share — no account, no paywall and no watermark.
- What should a learner already know before this?
- Start with Working with Sample Space Diagrams, Introducing Sample Space Diagrams, Basic Probability. Each one has its own free lesson, worksheet and quiz.
- How is the quiz sequenced?
- Recognition first, then understanding, application, reasoning and finally mastery — the same ladder used across every free resource on the platform.
- What comes next after Sample Space Diagrams in Context?
- Move on to Tree Diagrams, Venn Diagrams, Conditional Probability, which build directly on this idea.
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