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 Sydney, a school has 36 basketballs and 63 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 1000 times to model conditional probability. What is the exact probability that the second ball selected is a basketball, given that the first ball selected was a football?

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 36 basketballs + 63 footballs = 99 balls.
  2. The first ball selected is a football, and it is returned to the locker room before the second selection. Since the ball is replaced, the total number of balls remains 99 for the second draw.
  3. The probability of selecting a basketball 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 basketballs is 36, and the total number of balls is 99. The probability of selecting a basketball is 36/99.
  5. Simplify the fraction 36/99 by dividing both numerator and denominator by their greatest common divisor, which is 9.
  6. 36 ÷ 9 = 4, and 99 ÷ 9 = 11. Therefore, 36/99 = 4/11.
  7. The conditional probability that the second ball is a basketball given that the first was a football is 4/11, because the first draw is replaced and does not affect the second draw.
  8. The result is a simplified fraction, and no further calculation is needed.

Reference answer: 4/11

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