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0580/12 Feb/Mar 2025, Question 23: Tree Diagrams for Independent Events

Cambridge IGCSE Mathematics 0580 · Paper 12 (Core) · Non-calculator · 5 marks

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The question, in short

Complete a two-battery probability tree with faulty probability 1/10, then use it to find the chance both are faulty and the expected faulty count among 3000 batteries.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 2 marks

Fill in the missing probabilities on the tree diagram for the first and second batteries.

  1. 1P(not faulty) = 1 − 1/10 = 9/10
  2. 2Each second battery: Faulty 1/10, Not faulty 9/10

AnswerFirst battery: Not faulty 9/10; Second battery after Faulty: Faulty 1/10, Not faulty 9/10; after Not faulty: Faulty 1/10, Not faulty 9/10

Part (b) · 2 marks

Calculate the probability that both batteries Mai buys are faulty.

  1. 1P(F and F) = 1/10 × 1/10
  2. 2= 1/100

Answer1/100

Part (c) · 1 mark

Find the expected number of faulty batteries sold when 3000 are sold in a month.

  1. 1Expected number = 3000 × 1/10
  2. 2= 300

Answer300

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