NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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A school in Sydney is selecting two different students at random without replacement from a group of 52 Year 9 students and 58 Year 10 students to represent the school at a national event. Using a table to list all possible outcomes, what is the exact probability that the first student selected is from Year 9 and the second is from Year 10?

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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Worked solution

  1. The total number of students is 52 (Year 9) + 58 (Year 10) = 110 students.
  2. The first student must be from Year 9. The probability of selecting a Year 9 student first is 52/110.
  3. After selecting one Year 9 student, there are now 109 students left, with 58 still in Year 10.
  4. The probability that the second student is from Year 10, given the first was from Year 9, is 58/109.
  5. The combined probability for both events (first from Year 9, second from Year 10) is found by multiplying the two probabilities: (52/110) × (58/109).
  6. Calculate the numerator: 52 × 58 = 3016.
  7. Calculate the denominator: 110 × 109 = 11990.
  8. The fraction is 3016/11990.
  9. Simplify the fraction by finding the greatest common divisor (GCD) of 3016 and 11990. First, factor both numbers: 3016 = 2^3 × 13 × 29, 11990 = 2 × 5 × 11 × 109. The common factor is 2.
  10. Divide both numerator and denominator by 2: 3016 ÷ 2 = 1508, 11990 ÷ 2 = 5995. The simplified fraction is 1508/5995.
  11. Check if 1508 and 5995 have any further common factors. 1508 = 2^2 × 13 × 29, 5995 = 5 × 11 × 109. No common prime factors remain.
  12. The exact probability is 1508/5995. This is the final answer.

Reference answer: 1508/5995

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