NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Conditional probability

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In Melbourne, a study of public transport users found that 60% of people who take the train also use a bus for part of their journey, while 40% of all commuters use both the train and the bus. Given that a commuter uses the train, what is the probability that they also use the bus?

Express your answer as an exact fraction.

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Worked solution

  1. Given: 60% of train users also use a bus (P(bus | train) = 0.60). 40% of all commuters use both train and bus (P(train and bus) = 0.40).
  2. Let P(train) be the probability that a commuter uses the train. This is unknown but needed to apply conditional probability.
  3. Use the definition of conditional probability: P(bus | train) = P(train and bus) / P(train).
  4. Substitute known values: 0.60 = 0.40 / P(train).
  5. Solve for P(train): P(train) = 0.40 / 0.60 = 4/6 = 2/3.
  6. Now, the question asks: Given a commuter uses the train, what is the probability they also use the bus? This is P(bus | train).
  7. From the given data, it is directly stated that 60% of train users use a bus. So P(bus | train) = 60%.
  8. Convert 60% to a fraction: 60/100 = 3/5.
  9. The answer is 3/5, as this is the exact fraction for 60%.
  10. Check: P(train and bus) = P(bus | train) × P(train) = (3/5) × (2/3) = 6/15 = 2/5 = 0.40, which matches the given value.
  11. The calculation confirms consistency: the given 40% using both is consistent with 60% of train users using a bus.
  12. Therefore, given a commuter uses the train, the probability they also use the bus is exactly 3/5.

Reference answer: 3/5

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