NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Probability experiments and simulations

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In a sports equipment allocation simulation conducted in Melbourne, a coach is distributing 51 basketballs and 78 footballs to two teams using a chance experiment. The coach conducts a two-step experiment: first, a ball is randomly selected from the combined pool of 129 balls (51 basketballs and 78 footballs) and removed without replacement; second, another ball is selected from the remaining 128.

The coach wants to know the exact probability that both selected balls are footballs. What is the exact probability of this outcome?

Express your answer as a simplified fraction.

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A fair-coin simulation model. 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.Choose 0 or 1Equal chance→0=T 1=HMap outcome→RepeatCount heads
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. The total number of balls is 51 basketballs + 78 footballs = 129 balls.
  2. The first ball is selected at random from 129 balls. The probability that it is a football is 78/129.
  3. After removing one football, 77 footballs remain out of 128 total balls.
  4. The probability that the second ball is also a football, given the first was a football, is 77/128.
  5. The combined probability of both events (first and second ball being footballs) is found by multiplying the two probabilities: (78/129) × (77/128).
  6. Multiply numerators: 78 × 77 = 6006.
  7. Multiply denominators: 129 × 128 = 16512.
  8. The fraction is now 6006/16512.
  9. Simplify the fraction by finding the greatest common divisor (GCD) of 6006 and 16512. First, factor both numbers.
  10. GCD(6006, 16512) = 6. Divide both numerator and denominator by 6: 6006 ÷ 6 = 1001, 16512 ÷ 6 = 2752.
  11. The simplified fraction is 1001/2752. This is the exact probability that both selected balls are footballs.
  12. Check: 1001/2752 ≈ 0.3637, and (78/129) × (77/128) ≈ 0.6047 × 0.6016 ≈ 0.3637. The result is consistent.

Reference answer: 1001/2752

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