NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Conditional probability

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In a survey of 240 residents in Adelaide, 140 people have a gym membership and 90 people have a healthy eating plan. Of those who have a gym membership, 60 also follow a healthy eating plan.

If a resident is selected at random and is known to have a healthy eating plan, what is the probability that they also have a gym membership? Express your answer as an exact fraction.

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Worked solution

  1. Given: 240 residents in Adelaide. 140 have a gym membership (G). 90 have a healthy eating plan (H). 60 have both gym membership and healthy eating plan (G ∩ H).
  2. We are asked to find the probability that a randomly selected resident who has a healthy eating plan also has a gym membership. This is a conditional probability: P(G | H).
  3. The formula for conditional probability is P(G | H) = P(G ∩ H) / P(H).
  4. P(G ∩ H) is the proportion of people who have both a gym membership and a healthy eating plan. This is 60 / 240.
  5. P(H) is the proportion of people who have a healthy eating plan. This is 90 / 240.
  6. Substitute into the formula: P(G | H) = (60 / 240) / (90 / 240).
  7. The 240 in the denominator cancels out: (60 / 240) ÷ (90 / 240) = 60 / 90.
  8. Simplify 60 / 90 by dividing numerator and denominator by 30: 60 ÷ 30 = 2, 90 ÷ 30 = 3.
  9. The simplified fraction is 2 / 3.
  10. Check: 60 people have both, out of 90 who have a healthy eating plan. 60 / 90 = 2 / 3. This matches.
  11. The probability that a resident with a healthy eating plan also has a gym membership is 2 / 3.
  12. Final answer is 2 / 3.

Reference answer: 2/3

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