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

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A sports equipment manager in Perth is simulating the probability that a randomly selected basketball is used in a game, given that it was previously selected for training. There are 31 basketballs in total, and 56% of them are used in training on any given day.

If a basketball is selected at random and found to have been used in training, what is the exact probability that it is also selected for a game, given that 40% of basketballs used in training are subsequently selected for a game? 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
All 12 dots are cyclists. Three filled dots also represent red clothing. Within the cyclist group, P(red given cycles) = 3/12 = 1/4.
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. Given: 31 basketballs in total. 56% are used in training. Convert percentage to decimal: 56% = 56/100 = 0.56.
  2. Calculate number of basketballs used in training: 31 × 0.56 = 17.36. Since number of basketballs must be an integer, and 17.36 is not a whole number, this suggests the 56% is a proportion, not a count. Therefore, we work with proportions, not counts.
  3. Given: 40% of basketballs used in training are subsequently selected for a game. Convert 40% to decimal: 40% = 40/100 = 0.40.
  4. We are to find the conditional probability that a basketball is selected for a game, given that it was used in training. This is P(Game | Training).
  5. By definition of conditional probability: P(Game | Training) = P(Game and Training) / P(Training).
  6. P(Training) = 0.56 (from 56% of basketballs used in training).
  7. P(Game and Training) = P(Game | Training) × P(Training). But we are given that 40% of training-used basketballs are selected for a game, so P(Game | Training) = 0.40.
  8. Therefore, P(Game | Training) = 0.40. This is directly given as 40%.
  9. Convert 0.40 to a simplified fraction: 0.40 = 40/100 = 4/10 = 2/5.
  10. Check: 2/5 = 0.40, which matches 40%. The fraction 2/5 is in simplest form (2 and 5 are coprime).
  11. The question asks for the exact probability that a basketball selected at random and found to have been used in training is also selected for a game. This is P(Game | Training) = 2/5.
  12. Final answer: 2/5.

Reference answer: 2/5

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