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0580/41 Oct/Nov 2023, Question 4: Tree Diagrams for Independent Events

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 11 marks

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The question, in short

Complete a tree diagram for two spinners and use it to find probabilities, an expected count, a two-spin total probability, and an nth-spin expression.

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

Part (ai) · 2 marks

Fill in the missing branch probabilities on the tree diagram for Spinner A and Spinner B prime outcomes.

  1. 1Spinner A primes: 2, 3, 5 so P(not prime) = 1 − 3/5 = 2/5
  2. 2Spinner B primes: 3, 3, 3, 5, 5 so P(prime) = 5/9
  3. 3P(not prime) on B = 4/9 (4, 4, 4, 4)
  4. 4Write 5/9 and 4/9 on both pairs of Spinner B branches

AnswerSpinner A: not prime 2/5; Spinner B (on both branches): prime 5/9, not prime 4/9

Part (aiia) · 2 marks

Find the probability that both spinners land on prime numbers.

  1. 1P(both prime) = 3/5 × 5/9
  2. 2= 15/45 = 1/3

Answer1/3

Part (aiib) · 1 mark

Find the probability that the two numbers are not both prime.

  1. 1Primes on A are 2, 3, 5, so P(prime) = 3/5.
  2. 2Primes on B are 3, 3, 3, 5, 5, so P(prime) = 5/9.
  3. 3P(both prime) = 3/5 × 5/9 = 1/3
  4. 4P(not both prime) = 1 − 1/3 = 2/3

Answer2/3

Part (b) · 1 mark

Find the expected number of prime results when Spinner A is spun 120 times.

  1. 1P(prime on A) = 3/5
  2. 2Expected = 3/5 × 120 = 72

Answer72

Part (c) · 3 marks

Find the probability that two spins of Spinner B give a total of 9 or more.

  1. 1Sums of 9 or more: 4 + 5, 5 + 4, 5 + 5
  2. 2P(4) = 4/9, P(5) = 2/9
  3. 3P = 4/9 × 2/9 + 2/9 × 4/9 + 2/9 × 2/9
  4. 4= 8/81 + 8/81 + 4/81 = 20/81

Answer20/81

Part (d) · 2 marks

Give an expression in n for the probability Spinner B first lands on 4 on the nth spin.

  1. 1P(not 4) = 5/9, P(4) = 4/9
  2. 2First n − 1 spins are not 4, then the nth spin is 4
  3. 3P = (5/9)ⁿ⁻¹ × 4/9

Answer(5/9)ⁿ⁻¹ × 4/9

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Also searched as: 0580/41/O/N/23 · 0580_w23_qp_41 · 0580 w23 41 · question 4.
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