Free Grade 3 Introducing Conditional Probability Techniques Quizzes
Free Grade 3 quizzes for Introducing Conditional Probability Techniques: 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
Stretch
20 min
24
Learning objectives
Every objective is checkable, and each one maps to an activity and an assessment.
- RememberIdentify introducing conditional probability techniques
Identify introducing conditional probability techniques accurately in a familiar context.
Assessed by: Exit ticket of four short items
- UnderstandExplain introducing conditional probability techniques
Explain introducing conditional probability techniques accurately in a familiar context.
Assessed by: Exit ticket of four short items
- ApplyCalculate introducing conditional probability techniques
Calculate introducing conditional probability techniques accurately in a familiar context.
Assessed by: Exit ticket of four short items
- AnalyseCompare introducing conditional probability techniques
Compare introducing conditional probability techniques 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 Introducing Conditional Probability Techniques 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 5 red, 5 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 2.A bag has 6 red, 8 blue and 3 green counters. One is picked at random. Find P(red).[1]
- 3.A bag has 2 red, 8 blue and 5 green counters. One is picked at random. Find P(red).[1]
- 4.A bag has 2 red, 5 blue and 2 green counters. One is picked at random. Find P(red).[1]
- 5.A bag has 6 red, 6 blue and 2 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 6.A bag has 7 red, 8 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 7.A bag has 7 red, 3 blue and 0 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 8.A bag has 6 red, 1 blue and 2 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 7 red and 5 blue counters. Two are picked with replacement. Find P(both red).[3]
- 11.A bag has 8 red and 5 blue counters. Two are picked with replacement. Find P(both red).[3]
- 12.A bag has 4 red and 5 blue counters. Two are picked with replacement. Find P(both red).[3]
Show answers and working
1. P(red) = 5/13
Total counters = 13. P(red) = 5/13 = 5/13.
2. P(red) = 6/17
Total counters = 17. P(red) = 6/17 = 6/17.
3. P(red) = 2/15
Total counters = 15. P(red) = 2/15 = 2/15.
4. P(red) = 2/9
Total counters = 9. P(red) = 2/9 = 2/9.
5. P(red) = 3/7, P(not red) = 4/7
Total counters = 14. P(red) = 6/14 = 3/7. P(not red) = 1 − 3/7 = 4/7.
6. P(red) = 7/18, P(not red) = 11/18
Total counters = 18. P(red) = 7/18 = 7/18. P(not red) = 1 − 7/18 = 11/18.
7. P(red) = 7/10, P(not red) = 3/10
Total counters = 10. P(red) = 7/10 = 7/10. P(not red) = 1 − 7/10 = 3/10.
8. P(red) = 2/3, P(not red) = 1/3
Total counters = 9. P(red) = 6/9 = 2/3. P(not red) = 1 − 2/3 = 1/3.
9. 16/121
P(red) = 4/11 each time. Independent, so multiply: 4/11 × 4/11 = 16/121.
10. 49/144
P(red) = 7/12 each time. Independent, so multiply: 7/12 × 7/12 = 49/144.
11. 64/169
P(red) = 8/13 each time. Independent, so multiply: 8/13 × 8/13 = 64/169.
12. 16/81
P(red) = 4/9 each time. Independent, so multiply: 4/9 × 4/9 = 16/81.
Worked examples
A bag has 2 red, 1 blue and 4 green counters. One is picked at random. Find P(red).
- Total counters = 7.
- P(red) = 2/7 = 2/7.
- Answer: P(red) = 2/7
A bag has 8 red, 5 blue and 0 green counters. One is picked at random. Find P(red) and P(not red).
- Total counters = 13.
- P(red) = 8/13 = 8/13.
- P(not red) = 1 − 8/13 = 5/13.
- Answer: P(red) = 8/13, P(not red) = 5/13
A bag has 6 red and 1 blue counters. Two are picked with replacement. Find P(both red).
- P(red) = 6/7 each time.
- Independent, so multiply: 6/7 × 6/7 = 36/49.
- Answer: 36/49
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
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- What should a learner already know before this?
- Start with Introducing Conditional Probability Essentials, 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 Introducing Conditional Probability Techniques?
- Move on to Introducing Conditional Probability Word Problems, Introducing Conditional Probability Common Errors, Working with Conditional Probability, Conditional Probability in Context, which build directly on this idea.
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