Free Grade 3 Introducing Conditional Probability Common Errors Quizzes
Free Grade 3 quizzes for Introducing Conditional Probability Common Errors: 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
Core
15 min
24
Learning objectives
Every objective is checkable, and each one maps to an activity and an assessment.
- RememberIdentify introducing conditional probability common errors
Identify introducing conditional probability common errors accurately in a familiar context.
Assessed by: Exit ticket of four short items
- UnderstandExplain introducing conditional probability common errors
Explain introducing conditional probability common errors accurately in a familiar context.
Assessed by: Exit ticket of four short items
- ApplyCalculate introducing conditional probability common errors
Calculate introducing conditional probability common errors accurately in a familiar context.
Assessed by: Exit ticket of four short items
- AnalyseCompare introducing conditional probability common errors
Compare introducing conditional probability common errors 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 Common Errors 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 4 red, 6 blue and 0 green counters. One is picked at random. Find P(red).[1]
- 2.A bag has 2 red, 6 blue and 4 green counters. One is picked at random. Find P(red).[1]
- 3.A bag has 3 red, 8 blue and 4 green counters. One is picked at random. Find P(red).[1]
- 4.A bag has 4 red, 3 blue and 1 green counters. One is picked at random. Find P(red).[1]
- 5.A bag has 7 red, 2 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 6.A bag has 3 red, 5 blue and 2 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 7.A bag has 3 red, 4 blue and 1 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 8.A bag has 3 red, 4 blue and 5 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 9.A bag has 1 red and 5 blue counters. Two are picked with replacement. Find P(both red).[3]
- 10.A bag has 1 red and 7 blue counters. Two are picked with replacement. Find P(both red).[3]
- 11.A bag has 7 red and 1 blue counters. Two are picked with replacement. Find P(both red).[3]
- 12.A bag has 6 red and 8 blue counters. Two are picked with replacement. Find P(both red).[3]
Show answers and working
1. P(red) = 2/5
Total counters = 10. P(red) = 4/10 = 2/5.
2. P(red) = 1/6
Total counters = 12. P(red) = 2/12 = 1/6.
3. P(red) = 1/5
Total counters = 15. P(red) = 3/15 = 1/5.
4. P(red) = 1/2
Total counters = 8. P(red) = 4/8 = 1/2.
5. P(red) = 7/12, P(not red) = 5/12
Total counters = 12. P(red) = 7/12 = 7/12. P(not red) = 1 − 7/12 = 5/12.
6. P(red) = 3/10, P(not red) = 7/10
Total counters = 10. P(red) = 3/10 = 3/10. P(not red) = 1 − 3/10 = 7/10.
7. P(red) = 3/8, P(not red) = 5/8
Total counters = 8. P(red) = 3/8 = 3/8. P(not red) = 1 − 3/8 = 5/8.
8. P(red) = 1/4, P(not red) = 3/4
Total counters = 12. P(red) = 3/12 = 1/4. P(not red) = 1 − 1/4 = 3/4.
9. 1/36
P(red) = 1/6 each time. Independent, so multiply: 1/6 × 1/6 = 1/36.
10. 1/64
P(red) = 1/8 each time. Independent, so multiply: 1/8 × 1/8 = 1/64.
11. 49/64
P(red) = 7/8 each time. Independent, so multiply: 7/8 × 7/8 = 49/64.
12. 9/49
P(red) = 6/14 each time. Independent, so multiply: 6/14 × 6/14 = 9/49.
Worked examples
A bag has 6 red, 5 blue and 1 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 2 red, 7 blue and 4 green counters. One is picked at random. Find P(red) and P(not red).
- Total counters = 13.
- P(red) = 2/13 = 2/13.
- P(not red) = 1 − 2/13 = 11/13.
- Answer: P(red) = 2/13, P(not red) = 11/13
A bag has 5 red and 7 blue counters. Two are picked with replacement. Find P(both red).
- P(red) = 5/12 each time.
- Independent, so multiply: 5/12 × 5/12 = 25/144.
- Answer: 25/144
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 Word Problems, Introducing Conditional Probability Techniques, 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 Common Errors?
- Move on to Working with Conditional Probability, Conditional Probability in Context, which build directly on this idea.
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