Free Grade 3 Introducing Conditional Probability Lesson Plans
Free Grade 3 lesson plans for Introducing Conditional Probability: 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
30 min
24
Learning objectives
Every objective is checkable, and each one maps to an activity and an assessment.
- RememberIdentify introducing conditional probability
Identify introducing conditional probability accurately in a familiar context.
Assessed by: Exit ticket of four short items
- UnderstandExplain introducing conditional probability
Explain introducing conditional probability accurately in a familiar context.
Assessed by: Exit ticket of four short items
- ApplyCalculate introducing conditional probability
Calculate introducing conditional probability accurately in a familiar context.
Assessed by: Exit ticket of four short items
- AnalyseCompare introducing conditional probability
Compare introducing conditional probability 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 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 8 red, 2 blue and 0 green counters. One is picked at random. Find P(red).[1]
- 2.A bag has 8 red, 1 blue and 5 green counters. One is picked at random. Find P(red).[1]
- 3.A bag has 2 red, 6 blue and 2 green counters. One is picked at random. Find P(red).[1]
- 4.A bag has 5 red, 8 blue and 2 green counters. One is picked at random. Find P(red).[1]
- 5.A bag has 7 red, 6 blue and 2 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 6.A bag has 5 red, 4 blue and 4 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 7.A bag has 6 red, 7 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 8.A bag has 4 red, 1 blue and 5 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 9.A bag has 8 red and 8 blue counters. Two are picked with replacement. Find P(both red).[3]
- 10.A bag has 5 red and 8 blue counters. Two are picked with replacement. Find P(both red).[3]
- 11.A bag has 5 red and 5 blue counters. Two are picked with replacement. Find P(both red).[3]
- 12.A bag has 7 red and 1 blue counters. Two are picked with replacement. Find P(both red).[3]
Show answers and working
1. P(red) = 4/5
Total counters = 10. P(red) = 8/10 = 4/5.
2. P(red) = 4/7
Total counters = 14. P(red) = 8/14 = 4/7.
3. P(red) = 1/5
Total counters = 10. P(red) = 2/10 = 1/5.
4. P(red) = 1/3
Total counters = 15. P(red) = 5/15 = 1/3.
5. P(red) = 7/15, P(not red) = 8/15
Total counters = 15. P(red) = 7/15 = 7/15. P(not red) = 1 − 7/15 = 8/15.
6. P(red) = 5/13, P(not red) = 8/13
Total counters = 13. P(red) = 5/13 = 5/13. P(not red) = 1 − 5/13 = 8/13.
7. P(red) = 3/8, P(not red) = 5/8
Total counters = 16. P(red) = 6/16 = 3/8. P(not red) = 1 − 3/8 = 5/8.
8. P(red) = 2/5, P(not red) = 3/5
Total counters = 10. P(red) = 4/10 = 2/5. P(not red) = 1 − 2/5 = 3/5.
9. 1/4
P(red) = 8/16 each time. Independent, so multiply: 8/16 × 8/16 = 1/4.
10. 25/169
P(red) = 5/13 each time. Independent, so multiply: 5/13 × 5/13 = 25/169.
11. 1/4
P(red) = 5/10 each time. Independent, so multiply: 5/10 × 5/10 = 1/4.
12. 49/64
P(red) = 7/8 each time. Independent, so multiply: 7/8 × 7/8 = 49/64.
Worked examples
A bag has 8 red, 8 blue and 4 green counters. One is picked at random. Find P(red).
- Total counters = 20.
- P(red) = 8/20 = 2/5.
- Answer: P(red) = 2/5
A bag has 2 red, 5 blue and 0 green counters. One is picked at random. Find P(red) and P(not red).
- Total counters = 7.
- P(red) = 2/7 = 2/7.
- P(not red) = 1 − 2/7 = 5/7.
- Answer: P(red) = 2/7, P(not red) = 5/7
A bag has 4 red and 2 blue counters. Two are picked with replacement. Find P(both red).
- P(red) = 4/6 each time.
- Independent, so multiply: 4/6 × 4/6 = 4/9.
- Answer: 4/9
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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- What comes next after Introducing Conditional Probability?
- Move on to Working with Conditional Probability, Conditional Probability in Context, Sample Space Diagrams, Tree Diagrams, which build directly on this idea.
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