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

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In Canberra, a study of 1500 residents found that 900 have a university degree and 600 have a professional certification. Of those with a university degree, 360 also have a professional certification.

If a resident is selected at random and is known to have a professional certification, what is the probability that they also have a university degree? Express your answer as an exact fraction.

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Worked solution

  1. Given: 1500 residents in Canberra. 900 have a university degree (U), 600 have a professional certification (P). 360 have both U and P. We are to find the probability that a randomly selected resident with a professional certification also has a university degree.
  2. This is a conditional probability: P(U | P) = P(U and P) / P(P). The language 'given that they have a professional certification' indicates conditional probability.
  3. P(U and P) is the proportion of residents who have both qualifications. This is 360 out of 1500. So P(U and P) = 360 / 1500.
  4. P(P) is the proportion of residents with a professional certification. This is 600 out of 1500. So P(P) = 600 / 1500.
  5. Substitute into the conditional probability formula: P(U | P) = (360 / 1500) / (600 / 1500). The 1500 in denominator and numerator cancels out.
  6. Simplify: (360 / 1500) ÷ (600 / 1500) = 360 / 600.
  7. Simplify 360 / 600 by dividing numerator and denominator by 120: 360 ÷ 120 = 3, 600 ÷ 120 = 5. So 360 / 600 = 3 / 5.
  8. Check: 3 / 5 = 0.6. 360 / 600 = 0.6, so the fraction is correct.
  9. The probability that a resident with a professional certification also has a university degree is 3/5.

Reference answer: 3/5

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