Handling Data 7
4 marks · HardA rare disease affects 2 out of every 1000 people. A test for the disease gives a positive result in 99% of cases when the person actually has the disease, but also gives a false positive result in 2% of cases when the person does not have the disease. If a person is selected at random and tests positive, how likely is it that they actually have the disease? Explain your reasoning.
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About 9%
Must show: Calculation using correct probabilities for true positive and false positive; explains denominator (total positives) and numerator (true positives only) for conditional probability.
Also accept: ~9%; 0.09
Common mistakes: 0.99; 0.02; overstates probability
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Marking points
M1 · 1 mark · Clear statement or calculation of probabilities for having the disease (prevalence), true positive, and false positive.
M2 · 1 mark · Finds total probability of testing positive by adding probabilities for true and false positives.
A1 · 1 mark · Correct calculation of conditional probability using formula: P(have disease | test positive) = P(have & positive)/P(positive). (follow-through)
B1 · 1 mark · Valid conclusion, stating the probability (~9%) with reasoning.
Worked solution
Let D = have disease, T+ = test positive. P(D) = 0.002, P(T+|D) = 0.99, P(T+|not D) = 0.02 P(D and T+) = 0.002 × 0.99 = 0.00198 P(not D and T+) = 0.998 × 0.02 = 0.01996 P(T+) = 0.00198 + 0.01996 = 0.02194 P(D|T+) = P(D and T+)/P(T+) = 0.00198 / 0.02194 ≈ 0.0903 So the probability is approximately 9%.