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4MA1/2HR Jan 2020, Question 16: Tree Diagrams for Independent Events

Pearson Edexcel International GCSE Mathematics A 4MA1 · Paper 2HR (Higher) · Calculator · 8 marks

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The question, in short

Cody draws one counter from bag A (10 counters, 7 odd) and one from bag B (12 counters, 7 odd); complete the tree, find the probability the two numbers total odd, then use a combined even probability of 3/100 to find Harriet's least counter count.

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

Part (a) · 2 marks

Fill in the missing branch probabilities on the tree diagram for drawing from bag A then bag B.

  1. 1Bag A has 10 counters and 7 are odd, so P(odd from A) = 7/10
  2. 2P(even from A) = 1 − 7/10 = 3/10
  3. 3Bag B has 12 counters and 7 are odd, so P(odd from B) = 7/12
  4. 4P(even from B) = 1 − 7/12 = 5/12
  5. 5The counter taken from bag A does not change bag B, so both pairs of bag B branches are the same
  6. 6Missing values: 3/10, then 7/12 and 5/12, then 7/12 and 5/12

AnswerBag A even 3/10; after an odd counter: odd 7/12, even 5/12; after an even counter: odd 7/12, even 5/12

Part (b) · 3 marks

Work out the probability that the numbers on the two drawn counters add to an odd total.

  1. 1P(odd from A and even from B) = (7/10) × (5/12) = 35/120
  2. 2P(even from A and odd from B) = (3/10) × (7/12) = 21/120
  3. 3Total probability = 35/120 + 21/120 = 56/120

Answer56/120

Part (c) · 3 marks

Given the probability all three counters are even is 3/100, find the smallest possible number of counters in Harriet's bag.

  1. 1P(all three even) = P(even from A) × P(even from B) × P(even from Harriet's bag)
  2. 2(3/10) × (5/12) × x = 3/100
  3. 3(15/120) × x = 3/100
  4. 4x = (3/100) ÷ (15/120)
  5. 5x = (3/100) × (120/15) = (3 × 120) / (100 × 15) = 360 / 1500 = 6/25
  6. 6So probability of even from Harriet’s bag is 6/25
  7. 7Least number of counters = denominator when fraction is in simplest form = 25

Answer25

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