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 Melbourne, a gym has 54 basketballs and 17 footballs. A ball is selected at random, its type recorded, and then returned to the gym before another ball is selected.

What is the exact probability that the second ball is a football, given that the first ball 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 gym has 54 basketballs and 17 footballs, so the total number of balls is 54 + 17 = 71.
  2. A ball is selected at random, its type is recorded, and it is returned to the gym before the next selection. This means the selections are independent and the composition of the gym remains unchanged.
  3. The first ball selected is a basketball. Since the ball is returned, the total number of balls remains 71, and the number of footballs remains 17.
  4. We are asked to find the probability that the second ball is a football, given that the first ball was a basketball.
  5. Because the first ball is returned, the second draw is independent of the first. The probability of drawing a football on the second draw is unaffected by the first outcome.
  6. The probability of drawing a football is the number of footballs divided by the total number of balls: 17 / 71.
  7. The fraction 17/71 is already in simplest form because 17 is a prime number and does not divide 71.
  8. Therefore, the conditional probability that the second ball is a football given that the first was a basketball is 17/71.
  9. The result is a simplified fraction and matches the required format.
  10. Check: since the draw is with replacement, the events are independent, so P(second = football | first = basketball) = P(second = football) = 17/71.
  11. The final answer is 17/71.
  12. The answer is confirmed as correct and complete.

Reference answer: 17/71

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