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

Conditional-probability simulations

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In a simulation of a flu outbreak in Melbourne, the probability that a person who has been exposed to the virus will contract it is 0.6. Among those who contract the virus, the probability that they will transmit it to a close contact is 0.75.

If 1000 people are exposed to the virus, what is the exact expected number of people who will both contract the virus and transmit it to a close contact?

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Worked solution

  1. The probability that a person exposed to the virus will contract it is 0.6. This is given as a decimal probability.
  2. Among those who contract the virus, the probability they will transmit it to a close contact is 0.75. This is a conditional probability.
  3. To find the probability that a person both contracts the virus and transmits it, multiply the two probabilities: P(contract and transmit) = P(contract) × P(transmit | contract).
  4. Substitute the values: 0.6 × 0.75. First, convert to fractions: 0.6 = 3/5, 0.75 = 3/4.
  5. Multiply the fractions: (3/5) × (3/4) = 9/20.
  6. Convert 9/20 to decimal: 9 ÷ 20 = 0.45.
  7. There are 1000 people exposed to the virus. The expected number who both contract and transmit the virus is: 1000 × 0.45.
  8. Perform the multiplication: 1000 × 0.45 = 450.
  9. The result is 450 people. This is the expected number, based on repeated chance and conditional probability.
  10. Check: 0.6 × 0.75 = 0.45, and 1000 × 0.45 = 450. The arithmetic is consistent.
  11. The expected number of people who both contract and transmit the virus is 450.
  12. The answer is exact because all inputs are exact decimals or fractions, and the operations preserve exactness.

Reference answer: 450

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