NavGanitMateby NavGreen Compute, AustraliaBased on Victorian Curriculum Mathematics v2.0
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Compound-event simulations

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Required skills (click to reveal)

A student in Melbourne uses a digital simulation to model a walking route that involves two independent choices at different intersections. At the first intersection, the student turns left with a probability of 69/100 or goes straight with a probability of 31/100.

At the second intersection, the student turns left with a probability of 43/100 or goes straight with a probability of 57/100. The compound event of interest is the student turning left at both intersections.

Using theoretical probability based on all possible outcomes, what is the exact probability of this compound event occurring in a single trial? Express your answer as a simplified fraction.

Concept illustrations

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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.
Six equally likely die outcomes. Outcomes 5 and 6 are greater than 4: 2 favourable outcomes out of 6, so probability = 1/3.12345✓6✓Each outcome has probability 1/6
Outcomes 5 and 6 are greater than 4: 2 favourable outcomes out of 6, so probability = 1/3.

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Explanation

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Worked solution

  1. The student makes two independent choices at different intersections. The first choice has a probability of turning left of 69/100, and going straight of 31/100. The second choice has a probability of turning left of 43/100, and going straight of 57/100.
  2. The compound event of interest is turning left at both intersections. Since the two choices are independent, the probability of both events occurring is the product of their individual probabilities.
  3. The probability of turning left at the first intersection is 69/100. This is a fraction representing a probability, so it is already in decimal form as 0.69, but we keep it as a fraction for exactness.
  4. The probability of turning left at the second intersection is 43/100. This is also a fraction, and we keep it as such for exact calculation.
  5. To find the probability of both events occurring, multiply the two probabilities: (69/100) × (43/100).
  6. Multiply the numerators: 69 × 43. First, compute 70 × 43 = 3010, then subtract 1 × 43 = 43, so 3010 - 43 = 2967.
  7. Multiply the denominators: 100 × 100 = 10,000.
  8. The product is 2967/10,000.
  9. Check if the fraction 2967/10,000 can be simplified. Find the greatest common divisor (GCD) of 2967 and 10,000.
  10. Factor 2967: divide by 3: 2+9+6+7=24, divisible by 3. 2967 ÷ 3 = 989. 989 ÷ 23 = 43 (since 23×43 = 989). So 2967 = 3 × 23 × 43. 10,000 = 2^4 × 5^4. No common prime factors.
  11. Since 2967 and 10,000 share no common factors, the fraction 2967/10,000 is already in simplest form.
  12. Therefore, the exact probability of turning left at both intersections is 2967/10,000.

Reference answer: 2967/10000

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