NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Compound-event simulations

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A craft designer in Melbourne is creating a pattern using two custom six-sided dice, each with faces numbered from 1 to 6. The designer uses a digital simulation to model rolling both dice 360 times.

The compound event of interest is rolling a difference of 3 (absolute difference between the two numbers). What is the exact theoretical probability of this event occurring in a single trial?

Express your answer as a simplified fraction.

Concept illustrations

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Two independent fair coin tosses. HH, HT, TH and TT are equally likely. Two outcomes have exactly one head: probability = 2/4 = 1/2.HHHT✓TH✓TTExactly one head: HT or TH
HH, HT, TH and TT are equally likely. Two outcomes have exactly one head: probability = 2/4 = 1/2.
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

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Worked solution

  1. The problem involves rolling two standard six-sided dice, each with faces numbered 1 to 6. The compound event of interest is the absolute difference between the two dice being exactly 3.
  2. The total number of possible outcomes when rolling two dice is 6 × 6 = 36. This is because each die has 6 independent outcomes.
  3. We now list all outcomes where the absolute difference between the two dice is 3. This means |die1 - die2| = 3.
  4. List all pairs (die1, die2) such that |die1 - die2| = 3. Start with die1 = 1: |1 - 4| = 3 → (1,4). die1 = 2: |2 - 5| = 3 → (2,5). die1 = 3: |3 - 6| = 3 → (3,6). die1 = 4: |4 - 1| = 3 → (4,1). die1 = 5: |5 - 2| = 3 → (5,2). die1 = 6: |6 - 3| = 3 → (6,3).
  5. The valid outcomes are: (1,4), (2,5), (3,6), (4,1), (5,2), (6,3). This gives 6 favourable outcomes.
  6. The theoretical probability is the number of favourable outcomes divided by the total number of outcomes: 6 / 36.
  7. Simplify the fraction 6/36 by dividing both numerator and denominator by their greatest common divisor, which is 6.
  8. 6 ÷ 6 = 1, 36 ÷ 6 = 6. So, 6/36 = 1/6.
  9. The simplified fraction is 1/6. This is the exact theoretical probability of rolling a difference of 3 in a single trial.
  10. Check: 1/6 ≈ 0.1667. Since 6 × 0.1667 = 1.0002, which is close to 1, and 6 × 1/6 = 1, the fraction is correct.
  11. The result matches the expected probability for a difference of 3 across two fair dice.
  12. Therefore, the exact theoretical probability is 1/6.

Reference answer: 1/6

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