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

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In a probability simulation conducted in Adelaide, a game involves drawing two balls consecutively without replacement from a bag containing 5 white balls and 3 black balls. The game is designed so that a player wins if the two balls drawn are of different colours.

Using a digital simulation tool, 1000 trials are conducted to estimate the probability of winning. What is the exact probability of winning the game in a single trial?

Express your answer as a simplified fraction.

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Explanation

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Worked solution

  1. The bag contains 5 white balls and 3 black balls, so the total number of balls is 5 + 3 = 8.
  2. The player draws two balls consecutively without replacement. The game is won if the two balls are of different colours.
  3. The winning outcomes are: (white then black) or (black then white). These are the only two ways to draw two balls of different colours.
  4. Calculate the probability of drawing a white ball first: P(white first) = 5/8.
  5. After drawing one white ball, 7 balls remain: 4 white and 3 black. The probability of drawing a black ball next is 3/7.
  6. Multiply the probabilities for the sequence (white then black): P(white then black) = (5/8) × (3/7) = 15/56.
  7. Calculate the probability of drawing a black ball first: P(black first) = 3/8.
  8. After drawing one black ball, 7 balls remain: 5 white and 2 black. The probability of drawing a white ball next is 5/7.
  9. Multiply the probabilities for the sequence (black then white): P(black then white) = (3/8) × (5/7) = 15/56.
  10. Add the two mutually exclusive winning probabilities: P(win) = 15/56 + 15/56 = 30/56.
  11. Simplify the fraction 30/56 by dividing numerator and denominator by their greatest common divisor, which is 2: 30 ÷ 2 = 15, 56 ÷ 2 = 28. So, 30/56 = 15/28.
  12. The exact probability of winning in a single trial is 15/28.

Reference answer: 15/28

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