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 sports equipment allocation in Adelaide, a school has 76 basketballs and 29 footballs stored in a locker room. A student randomly selects one ball without looking, records its type, and then returns it to the locker room before selecting another ball.

This process is repeated 100 times to simulate conditional probability. What is the exact probability that the second ball selected is a football, given that the first ball selected was a basketball?

Express your answer as a simplified fraction.

Concept illustrations

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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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Explanation

Worked solution and answer (click to reveal)

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Worked solution

  1. The total number of balls in the locker room is 76 basketballs + 29 footballs = 105 balls.
  2. The first ball selected is a basketball, and it is returned to the locker room before the second selection. Since the ball is replaced, the total number of balls remains 105 for the second draw.
  3. The probability of selecting a football on the second draw is independent of the first draw because the ball is replaced. Therefore, the probability is based on the original composition of the locker room.
  4. The number of footballs is 29, and the total number of balls is 105. The probability of selecting a football is 29 / 105.
  5. Simplify the fraction 29/105. Check for common factors between 29 and 105. 29 is a prime number, and 105 = 3 × 5 × 7. Since 29 does not divide 105, the fraction is already in simplest form.
  6. The conditional probability that the second ball is a football given that the first was a basketball is equal to the probability of selecting a football on any single draw, because the draws are independent due to replacement.
  7. Therefore, P(second = football | first = basketball) = 29 / 105.
  8. The fraction 29/105 cannot be simplified further, as 29 is prime and does not divide 105.
  9. The result is an exact probability, expressed as a simplified fraction.
  10. Final answer: 29/105.
  11. Check: 29 ÷ 105 ≈ 0.2762, and 29/105 is less than 1/3 (≈0.333), which is reasonable given 29 out of 105 is about 27.6%.
  12. The second ball selected is a football with probability 29/105, given the first was a basketball.

Reference answer: 29/105

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