Probability Revision Notes
Subject: Mathematics | Level: A-Level | Exam Board: AQA
This guide provides a comprehensive overview of A-Level Probability, focusing on the key concepts, formulas, and exam techniques required by AQA. It breaks down complex ideas into manageable chunks, using visual aids and real-world examples to build a solid foundation for tackling exam questions with confidence.
Revision Notes & Key Concepts
Key Terms & Definitions
- Sample Space
- The set of all possible outcomes of an experiment.
- Event
- A subset of the sample space; a set of one or more outcomes.
- Mutually Exclusive Events
- Events that have no outcomes in common. They cannot occur at the same time. P(A ∩ B) = 0.
- Independent Events
- The occurrence of one event has no effect on the probability of the other event occurring. P(A ∩ B) = P(A) x P(B).
- Conditional Probability
- The probability of an event occurring, given that another event has already occurred.
- Complement
- The event that A does not occur, denoted A'. P(A') = 1 - P(A).
Worked Examples
Worked Example
Question: In a college, 60% of students have a laptop (L), 55% have a tablet (T), and 30% have both. A student is chosen at random. Find the probability that the student has a laptop or a tablet, but not both.
Solution: Step 1: Identify the given probabilities. P(L) = 0.60, P(T) = 0.55, P(L ∩ T) = 0.30. Step 2: Use a Venn diagram or the addition rule to find P(L ∪ T). P(L ∪ T) = P(L) + P(T) - P(L ∩ T) = 0.60 + 0.55 - 0.30 = 0.85. Step 3: The question asks for 'laptop or tablet, but not both'. This is the union minus the intersection. P((L ∪ T) and not (L ∩ T)) = P(L ∪ T) - P(L ∩ T). Step 4: Calculate the final probability. 0.85 - 0.30 = 0.55. Final answer: 0.55.
Worked Example
Question: The events A and B are such that P(A) = 0.4 and P(B) = 0.5. Given that A and B are independent, find P(A ∪ B).
Solution: Step 1: State the condition for independence. Since A and B are independent, P(A ∩ B) = P(A) x P(B). Step 2: Calculate the probability of the intersection. P(A ∩ B) = 0.4 x 0.5 = 0.20. Step 3: Use the addition rule to find P(A ∪ B). P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 0.4 + 0.5 - 0.20 = 0.70. Final answer: 0.70.
Worked Example
Question: A bag contains 6 red and 4 blue balls. A ball is taken out, its colour is noted, and it is not replaced. A second ball is then taken out. Find the probability that the two balls are different colours.
Solution: Step 1: Draw a tree diagram. First branch: P(Red) = 6/10, P(Blue) = 4/10. Second branches (after taking a red): P(Red) = 5/9, P(Blue) = 4/9. Second branches (after taking a blue): P(Red) = 6/9, P(Blue) = 3/9. Step 2: Identify the outcomes that satisfy 'different colours'. These are 'Red then Blue' (RB) and 'Blue then Red' (BR). Step 3: Calculate the probability of each outcome by multiplying along the branches. P(RB) = (6/10) * (4/9) = 24/90. Step 4: P(BR) = (4/10) * (6/9) = 24/90. Step 5: Add the probabilities of these two mutually exclusive outcomes. P(different colours) = P(RB) + P(BR) = 24/90 + 24/90 = 48/90. This simplifies to 8/15. Final answer: 8/15.
Practice Questions
Question: Given that P(A) = 0.5, P(B) = 0.6 and P(A ∪ B) = 0.8. Find P(A|B).
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Question: A survey in a town showed that 70% of people own a car (C) and 45% own a bicycle (B). 25% own neither. Determine if the events 'owning a car' and 'owning a bicycle' are independent.
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Question: State the two conditions that must be met for a series of trials to be modelled by a binomial distribution.
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Question: A factory produces items, and 5% are defective. A quality check involves randomly selecting two items. What is the probability that at least one of the selected items is defective?
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Question: Explain the difference between mutually exclusive and independent events, using a real-world example for each.
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