The correct probability with no working
Mathematics compulsory written paper. Question 7 is worth 7 marks: 'A bag contains 4 red and 6 blue counters. Two counters are drawn at random without replacement. (a) Find the probability that at least one is red. (b) Given that the first counter is blue, find the probability that the second is red.' The instructions say: 'Show all working. Answers without justification may not receive full credit.' The candidate writes for (a) only '2/3'. For (b), the candidate writes '4/9' with a small tree diagram showing first draw blue (6/10) and second draw red (4/9). For (c), which asks 'Hence find the probability that exactly one is red', the candidate writes '0.67 - 0.133 = 0.537'. The mark scheme allocates 3 marks for method, 2 for accurate calculation, and 2 for interpretation in (a); (b) is 2 marks; (c) is 3 marks. The candidate's final answers for (a) and (b) are correct, but (c) is slightly off because of rounding. The examiner must apply method and follow-through rules.
1. Award marks for part (a). Does the correct answer '2/3' with no working earn full credit? Explain.
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Under the instruction 'Show all working', the answer-only response cannot earn full credit. The correct '2/3' demonstrates the accuracy mark, but the 3 method marks are unsupported: there is no complementary event, no multiplication of probabilities, and no tree diagram. The 2 interpretation marks are also unsupported because the candidate has not explained 'at least one'. A defensible award is 2 out of 7, not 7 out of 7. If the rubric allowed full marks for a correct answer, the instruction would be redundant; it does not. The examiner should apply the same standard to all candidates, so a candidate who shows the complement method but makes an arithmetic slip can still earn method marks, while an answer-only response cannot.
2. Apply follow-through and rounding rules to part (c). Is 0.537 a new error or a carried-forward rounding issue?
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Part (c) shows a method: P(at least one red) minus P(both red). The candidate substitutes 0.67 instead of 2/3, giving 0.537 instead of 8/15, which is 0.5333. Follow-through applies because the earlier answer is correct and the discrepancy comes from a rounding choice, not a new conceptual error. Award the method mark and the interpretation mark; withhold the final accuracy mark if the scheme demands an exact value or three significant figures from exact values. But check the tolerance: if 0.537 is within the accepted range, award full marks. Do not penalise the same rounding twice. The examiner should record that the method is visible and that the error is numerical, not structural.
3. What exam technique advice would you give this candidate to avoid losing method and accuracy marks?
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Keep exact fractions until the final step. Write the exact expression before substituting decimals: '1 - (6/10)(5/9) = 2/3', then 'P(exactly one) = 2/3 - (4/10)(3/9) = 8/15'. State the complement explicitly and show a one-line reason for each probability. In part (a), a tree diagram or the complement calculation would have earned method marks. In part (c), use 2/3, not 0.67, because 'hence' signals exact dependence. Practise under timed conditions with the instruction 'show all working' visible. The candidate understands the mathematics; the lost marks are from communication and premature rounding, which are fixable with a written checklist.