Free Grade 3 Conditional Probability in Context Techniques Flashcards
Free Grade 3 flashcards for Conditional Probability in Context 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
Core
35 min
24
Learning objectives
Every objective is checkable, and each one maps to an activity and an assessment.
- RememberIdentify conditional probability in context techniques
Identify conditional probability in context techniques accurately in a familiar context.
Assessed by: Exit ticket of four short items
- UnderstandExplain conditional probability in context techniques
Explain conditional probability in context techniques accurately in a familiar context.
Assessed by: Exit ticket of four short items
- ApplyCalculate conditional probability in context techniques
Calculate conditional probability in context techniques accurately in a familiar context.
Assessed by: Exit ticket of four short items
- AnalyseCompare conditional probability in context techniques
Compare conditional probability in context 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 Conditional Probability in Context 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, 1 blue and 4 green counters. One is picked at random. Find P(red).[1]
- 2.A bag has 2 red, 2 blue and 0 green counters. One is picked at random. Find P(red).[1]
- 3.A bag has 2 red, 7 blue and 4 green counters. One is picked at random. Find P(red).[1]
- 4.A bag has 3 red, 1 blue and 0 green counters. One is picked at random. Find P(red).[1]
- 5.A bag has 8 red, 3 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 6.A bag has 8 red, 4 blue and 3 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 7.A bag has 2 red, 6 blue and 4 green counters. One is picked at random. Find P(red) and P(not red).[2]
- 8.A bag has 2 red, 8 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 2 blue counters. Two are picked with replacement. Find P(both red).[3]
- 10.A bag has 6 red and 7 blue counters. Two are picked with replacement. Find P(both red).[3]
- 11.A bag has 4 red and 7 blue counters. Two are picked with replacement. Find P(both red).[3]
- 12.A bag has 7 red and 3 blue counters. Two are picked with replacement. Find P(both red).[3]
Show answers and working
1. P(red) = 1/2
Total counters = 10. P(red) = 5/10 = 1/2.
2. P(red) = 1/2
Total counters = 4. P(red) = 2/4 = 1/2.
3. P(red) = 2/13
Total counters = 13. P(red) = 2/13 = 2/13.
4. P(red) = 3/4
Total counters = 4. P(red) = 3/4 = 3/4.
5. P(red) = 4/7, P(not red) = 3/7
Total counters = 14. P(red) = 8/14 = 4/7. P(not red) = 1 − 4/7 = 3/7.
6. P(red) = 8/15, P(not red) = 7/15
Total counters = 15. P(red) = 8/15 = 8/15. P(not red) = 1 − 8/15 = 7/15.
7. P(red) = 1/6, P(not red) = 5/6
Total counters = 12. P(red) = 2/12 = 1/6. P(not red) = 1 − 1/6 = 5/6.
8. P(red) = 1/6, P(not red) = 5/6
Total counters = 12. P(red) = 2/12 = 1/6. P(not red) = 1 − 1/6 = 5/6.
9. 4/9
P(red) = 4/6 each time. Independent, so multiply: 4/6 × 4/6 = 4/9.
10. 36/169
P(red) = 6/13 each time. Independent, so multiply: 6/13 × 6/13 = 36/169.
11. 16/121
P(red) = 4/11 each time. Independent, so multiply: 4/11 × 4/11 = 16/121.
12. 49/100
P(red) = 7/10 each time. Independent, so multiply: 7/10 × 7/10 = 49/100.
Worked examples
A bag has 5 red, 2 blue and 3 green counters. One is picked at random. Find P(red).
- Total counters = 10.
- P(red) = 5/10 = 1/2.
- Answer: P(red) = 1/2
A bag has 1 red, 3 blue and 2 green counters. One is picked at random. Find P(red) and P(not red).
- Total counters = 6.
- P(red) = 1/6 = 1/6.
- P(not red) = 1 − 1/6 = 5/6.
- Answer: P(red) = 1/6, P(not red) = 5/6
A bag has 2 red and 3 blue counters. Two are picked with replacement. Find P(both red).
- P(red) = 2/5 each time.
- Independent, so multiply: 2/5 × 2/5 = 4/25.
- Answer: 4/25
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
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- Start with Conditional Probability in Context Essentials, Working with Conditional Probability. Each one has its own free lesson, worksheet and quiz.
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- Move on to Conditional Probability in Context Word Problems, Conditional Probability in Context Common Errors, Introducing Conditional Probability, Working with Conditional Probability, which build directly on this idea.
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