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

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In a simulation of a conditional probability model for a library's book return system in Melbourne, 49% of books returned on time are checked out again within a week, while 90% of books returned late are checked out again within a week. If a book is selected at random from those returned in the past month, and it was checked out again within a week, what is the exact probability that it was returned on time?

Express your answer as a simplified fraction.

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Restrict the sample space before counting. All 12 dots are cyclists. Three filled dots also represent red clothing. Within the cyclist group, P(red given cycles) = 3/12 = 1/4.12 cyclists · 3 wearing red
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Choose 0 or 1 with equal probability: 0 means tails and 1 means heads. Repeat independent trials and count outcomes. No experiment results are asserted here.

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Worked solution

  1. Let T be the event that a book was returned on time, and L be the event that it was returned late. Let Q be the event that a book was checked out again within a week.
  2. Given: P(Q|T) = 49% = 0.49, P(Q|L) = 90% = 0.90. We are to find P(T|Q), the probability that a book was returned on time given that it was checked out again within a week.
  3. We apply Bayes' Theorem: P(T|Q) = [P(Q|T) × P(T)] / [P(Q|T) × P(T) + P(Q|L) × P(L)].
  4. We are not given P(T) or P(L) directly. However, since every book is either returned on time or late, P(T) + P(L) = 1. We assume no prior information, so we use a uniform prior: P(T) = P(L) = 0.5.
  5. Substitute values into the formula: P(T|Q) = [0.49 × 0.5] / [0.49 × 0.5 + 0.90 × 0.5].
  6. Calculate numerator: 0.49 × 0.5 = 0.245.
  7. Calculate denominator: 0.49 × 0.5 = 0.245, 0.90 × 0.5 = 0.45. Sum = 0.245 + 0.45 = 0.695.
  8. Now compute P(T|Q) = 0.245 / 0.695.
  9. Simplify 0.245 / 0.695. Multiply numerator and denominator by 1000 to eliminate decimals: 245 / 695.
  10. Simplify 245/695 by dividing numerator and denominator by 5: 49/139.
  11. Check that 49 and 139 have no common factors. 49 = 7², 139 is prime. So 49/139 is in simplest form.
  12. Therefore, the probability that a book returned on time given it was checked out again within a week is 49/139.

Reference answer: 49/139

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