Free Grade 3 Conditional Probability in Context Quizzes
Free Grade 3 quizzes for Conditional Probability 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
25 min
24
Learning objectives
Every objective is checkable, and each one maps to an activity and an assessment.
- UnderstandExplain conditional probability in context
Explain conditional probability in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
- ApplyCalculate conditional probability in context
Calculate conditional probability in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
- AnalyseCompare conditional probability in context
Compare conditional probability in context accurately in a familiar context.
Assessed by: Exit ticket of four short items
- EvaluateJustify conditional probability in context
Justify conditional probability 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 Conditional Probability 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 1 red, 4 blue and 4 green counters. One is picked at random. Find P(red).[1]
- 2.A bag has 3 red, 7 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 3.A bag has 1 red, 7 blue and 1 green counters. One is picked at random. Find P(red).[1]
- 4.A bag has 7 red, 8 blue and 2 green counters. One is picked at random. Find P(red).[1]
- 5.A bag has 5 red, 8 blue and 5 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 6.A bag has 1 red, 6 blue and 0 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 7.A bag has 5 red, 7 blue and 2 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 8.A bag has 2 red, 4 blue and 0 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 9.A bag has 6 red and 2 blue counters. Two are picked with replacement. Find P(both red).[3]
- 10.A bag has 5 red and 2 blue counters. Two are picked with replacement. Find P(both red).[3]
- 11.A bag has 4 red and 6 blue counters. Two are picked with replacement. Find P(both red).[3]
- 12.A bag has 7 red and 7 blue counters. Two are picked with replacement. Find P(both red).[3]
Show answers and working
1. P(red) = 1/9
Total counters = 9. P(red) = 1/9 = 1/9.
2. P(red) = 3/13
Total counters = 13. P(red) = 3/13 = 3/13.
3. P(red) = 1/9
Total counters = 9. P(red) = 1/9 = 1/9.
4. P(red) = 7/17
Total counters = 17. P(red) = 7/17 = 7/17.
5. P(red) = 5/18, P(not red) = 13/18
Total counters = 18. P(red) = 5/18 = 5/18. P(not red) = 1 − 5/18 = 13/18.
6. P(red) = 1/7, P(not red) = 6/7
Total counters = 7. P(red) = 1/7 = 1/7. P(not red) = 1 − 1/7 = 6/7.
7. P(red) = 5/14, P(not red) = 9/14
Total counters = 14. P(red) = 5/14 = 5/14. P(not red) = 1 − 5/14 = 9/14.
8. P(red) = 1/3, P(not red) = 2/3
Total counters = 6. P(red) = 2/6 = 1/3. P(not red) = 1 − 1/3 = 2/3.
9. 9/16
P(red) = 6/8 each time. Independent, so multiply: 6/8 × 6/8 = 9/16.
10. 25/49
P(red) = 5/7 each time. Independent, so multiply: 5/7 × 5/7 = 25/49.
11. 4/25
P(red) = 4/10 each time. Independent, so multiply: 4/10 × 4/10 = 4/25.
12. 1/4
P(red) = 7/14 each time. Independent, so multiply: 7/14 × 7/14 = 1/4.
Worked examples
A bag has 6 red, 3 blue and 3 green counters. One is picked at random. Find P(red).
- Total counters = 12.
- P(red) = 6/12 = 1/2.
- Answer: P(red) = 1/2
A bag has 4 red, 4 blue and 1 green counters. One is picked at random. Find P(red) and P(not red).
- Total counters = 9.
- P(red) = 4/9 = 4/9.
- P(not red) = 1 − 4/9 = 5/9.
- Answer: P(red) = 4/9, P(not red) = 5/9
A bag has 6 red and 7 blue counters. Two are picked with replacement. Find P(both red).
- P(red) = 6/13 each time.
- Independent, so multiply: 6/13 × 6/13 = 36/169.
- Answer: 36/169
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 Conditional Probability 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 Conditional Probability, Introducing Conditional Probability, Venn Diagrams. 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 Conditional Probability in Context?
- Move on to Sample Space Diagrams, Tree Diagrams, Venn Diagrams, which build directly on this idea.
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