NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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In a community garden in Perth, participants can choose one type of vegetable to grow and one type of compost to use. There are 33 different vegetable varieties available, and 14 different compost types.

Each participant selects exactly one vegetable and one compost type. Using a two-way table to represent all possible combinations, what is the exact probability that a randomly selected participant chose a specific vegetable and a specific compost type?

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

Worked solution and answer (click to reveal)

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Worked solution

  1. The total number of possible combinations of one vegetable and one compost type is found by multiplying the number of vegetable varieties by the number of compost types.
  2. There are 33 vegetable varieties and 14 compost types, so total combinations = 33 × 14.
  3. Calculate 33 × 14: 33 × 10 = 330, and 33 × 4 = 132. Add: 330 + 132 = 462.
  4. The total number of possible outcomes (combinations) is 462.
  5. There is only one specific combination that matches the chosen vegetable and chosen compost type.
  6. The probability of selecting that one specific combination is 1 divided by the total number of combinations.
  7. So, probability = 1 / 462.
  8. The fraction 1/462 is already in simplest form because 1 and 462 share no common factors other than 1.
  9. Therefore, the exact probability is 1/462.
  10. Check: 1/462 ≈ 0.002164, which is a small but valid probability for a single specific outcome.
  11. The two-way table would have 33 rows (vegetables) and 14 columns (composts), giving 33 × 14 = 462 cells, each representing one unique combination.
  12. Thus, the probability that a randomly selected participant chose a specific vegetable and a specific compost type is 1/462.

Reference answer: 1/462

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