NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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A school in Adelaide is ordering 65 whiteboard markers and 49 highlighters. Two items are selected at random without replacement to test quality.

Using a table to list all possible outcomes, what is the exact probability that the first item selected is a whiteboard marker and the second is a highlighter? Express your answer as a simplified fraction.

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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 school has 65 whiteboard markers and 49 highlighters, so the total number of items is 65 + 49 = 114.
  2. We are selecting two items at random without replacement. The first item must be a whiteboard marker, and the second must be a highlighter.
  3. The probability that the first item selected is a whiteboard marker is the number of whiteboard markers divided by the total number of items: 65 / 114.
  4. After selecting one whiteboard marker, there are now 113 items left, with 49 highlighters still available.
  5. The probability that the second item selected is a highlighter, given that the first was a whiteboard marker, is 49 / 113.
  6. To find the combined probability of both events occurring in sequence, multiply the two probabilities: (65 / 114) × (49 / 113).
  7. Multiply the numerators: 65 × 49 = 3185.
  8. Multiply the denominators: 114 × 113 = 12882.
  9. The combined probability is 3185 / 12882.
  10. Simplify the fraction by finding the greatest common divisor (GCD) of 3185 and 12882. First, factor 3185: 5 × 7 × 7 × 13 = 5 × 7² × 13. Factor 12882: 2 × 3 × 11 × 193. No common prime factors exist between 3185 and 12882.
  11. Since the GCD is 1, the fraction 3185 / 12882 is already in its simplest form.
  12. Therefore, the exact probability that the first item is a whiteboard marker and the second is a highlighter is 3185 / 12882.

Reference answer: 3185/12882

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