【A-Level】CIE Mathematics (9709) S1 Probability in Summary and Past Paper

📘 1. Probability

1.1 Experiments, events and outcomes

• Experiment: A process that leads to well-defined results.
• Outcome: A possible result of an experiment.
• Sample Space (S): The set of all possible outcomes.
• Event: A subset of the sample space.

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1.2 Mutually exclusive events and the addition law

• Mutually Exclusive Events: Two events that cannot occur at the same time.
• Addition Law:
  • For any two events A and B: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
  • If A and B are mutually exclusive: $$P(A \cup B) = P(A) + P(B)$$

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1.3 Independent events and the multiplication law

• Independent Events: The occurrence of one does not affect the probability of the other.
• Multiplication Law: $$P(A \cap B) = P(A) \times P(B) \quad \text{(if independent)}$$

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1.4 Conditional probability

• Conditional Probability: The probability of A given B has occurred: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$

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1.5 Dependent events and conditional probability

• If events are dependent, use: $$P(A \cap B) = P(A) \times P(B|A) = P(B) \times P(A|B)$$

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📚2.Past Paper Questions & Solutions (Last 5 Years)

2.1. Question from 9709/w19/62 (November 2019 Paper 62)

Question:
Events A and B are such that $ P(A) = 0.6 $, $ P(B) = 0.7 $, and $ P(A|B) = 0.8 $. Find $ P(A \cup B) $.

Solution:

$$P(A|B) = \frac{P(A \cap B)}{P(B)} \Rightarrow 0.8 = \frac{P(A \cap B)}{0.7} \Rightarrow P(A \cap B) = 0.56$$ $$P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.7 - 0.56 = 0.74$$

坑点:

• Confusing $ P(A|B) $ with $ P(B|A) $.
• Forgetting to subtract $ P(A \cap B) $ in the union formula.

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2.2. Question from 9709/s20/63 (June 2020 Paper 63)

Question:
A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Find the probability that both are red.

Solution:

$$P(\text{both red}) = \frac{5}{8} \times \frac{4}{7} = \frac{20}{56} = \frac{5}{14}$$

坑点:

• Without replacement → events are dependent.
• Not adjusting the denominator after the first draw.

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2.3. Question from 9709/w20/62 (November 2020 Paper 62)

Question:
Events X and Y are independent. $ P(X) = 0.4 $, $ P(Y) = 0.5 $. Find $ P(X \cup Y) $.

Solution:

$$P(X \cap Y) = P(X) \times P(Y) = 0.4 \times 0.5 = 0.2$$ $$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = 0.4 + 0.5 - 0.2 = 0.7$$

坑点:

• Assuming independence means mutually exclusive (they are not!).

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2.4. Question from 9709/s21/62 (June 2021 Paper 62)

Question:
The probability that a student passes Math is 0.8, and Physics is 0.6. The probability that he passes at least one subject is 0.9. Are the events independent?

Solution:
Let M = Math, P = Physics.

$$P(M \cup P) = P(M) + P(P) - P(M \cap P)$$ $$0.9 = 0.8 + 0.6 - P(M \cap P) \Rightarrow P(M \cap P) = 0.5$$

If independent, $ P(M \cap P) = 0.8 \times 0.6 = 0.48 \neq 0.5 $ → Not independent.

坑点:

• Misapplying the independence test.

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2.5. Question from 9709/w21/62 (November 2021 Paper 62)

Question:
A fair die is rolled twice. Find the probability that the sum is 10 given that the first roll is 1.

Solution:

$$P(\text{sum} = 10 | \text{first} = 4) = \frac{P(\text{first}=4 \cap \text{sum}=10)}{P(\text{first}=4)} = \frac{P(4,6)}{1/6} = \frac{1/36}{1/6} = \frac{1}{6}$$

坑点:

• Not correctly identifying the conditional event.

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⚠️ Common Pitfalls & Tips

• Mutually Exclusive vs Independent:

  • Mutually exclusive: $ P(A \cap B) = 0 $
  • Independent: $ P(A \cap B) = P(A)P(B) $

• Conditional Probability:

  • Always use: $ P(A|B) = \frac{P(A \cap B)}{P(B)} $

• Tree Diagrams: Useful for sequential events.

• Venn Diagrams: Help visualize unions and intersections.

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📌 Revision Advice

1. Practice Venn and Tree Diagrams for visual reasoning.
2. Memorize the formulas for union, intersection, and conditional probability.
3. Identify dependency before applying multiplication rule.
4. Solve past papers under timed conditions.
5. Review mark schemes to understand common mistakes.