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

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Required skills (click to reveal)

A student in Canberra uses a digital simulation to model a walking route that involves two independent decisions. At the first intersection, the student turns left with a probability of 83/100 or goes straight with a probability of 17/100.

At the second intersection, the student turns left with a probability of 91/100 or goes straight with a probability of 9/100. 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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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 student makes two independent decisions at separate intersections. The first decision has a probability of turning left of 83/100. The second decision has a probability of turning left of 91/100.
  2. The compound event of interest is turning left at both intersections. Since the two decisions are independent, the probability of both events occurring is the product of their individual probabilities.
  3. The formula for the probability of two independent events A and B both occurring is P(A and B) = P(A) × P(B).
  4. Substitute the given probabilities: P(both left) = (83/100) × (91/100).
  5. Multiply the numerators: 83 × 91. Calculate 83 × 90 = 7470, then 83 × 1 = 83, so 7470 + 83 = 7553.
  6. Multiply the denominators: 100 × 100 = 10,000.
  7. The combined probability is 7553/10,000.
  8. Check if the fraction 7553/10,000 can be simplified. Find the greatest common divisor (GCD) of 7553 and 10,000.
  9. Test for common factors. 7553 is odd, so not divisible by 2. Sum of digits: 7+5+5+3 = 20, not divisible by 3. Ends in 3, not 0 or 5, so not divisible by 5. Check divisibility by 7: 7553 ÷ 7 ≈ 1079, 7×1079 = 7553? 7×1000=7000, 7×79=553, 7000+553=7553. Yes, 7×1079 = 7553.
  10. Now check if 10,000 is divisible by 7. 10,000 ÷ 7 ≈ 1428.57, not an integer. So 7 is not a common factor of both numerator and denominator.
  11. Check for other common factors. 10,000 = 2^4 × 5^4. 7553 is not divisible by 2 or 5. Since 7553 is not divisible by 2, 4, 5, or 10, and no other common prime factors exist, the fraction 7553/10,000 is already in simplest form.
  12. Therefore, the exact probability of turning left at both intersections is 7553/10,000.

Reference answer: 7553/10000

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