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A-LevelMathematicsProbability and statistics9709/63 · May/June 2019 · Q69 Marks

A fair five-sided spinner has sides numbered 1, 1, 1, 2, 3. A fair three-sided spinner has sides numbered 1, 2, 3. Both spinners are spun once and the score is the product of the numbers on the sides the spinners land on.

  1. (i)Draw up the probability distribution table for the score.[4]
  2. (ii)Find the mean and the variance of the score.[3]
  3. (iii)Find the probability that the score is greater than the mean score.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

  1. [B1] Probability distribution table with correct scores, allow extra score values if probability of zero stated
  2. [B1] 2 probabilities (with correct score) correct
  3. [B1] 3 or more correct probabilities with correct scores
  4. [B1] FT Σp = 1, at least 4 probabilities
  5. [B1] mean = (3+8+12+4+12+9)/15 = 48/15 = (3.2)
  6. [M1] FT Substitute their attempts at scores in correct var formula, must have “– mean²” (condone probabilities not summing to 1)
  7. [A1] 224/15 - 3.2² = 4.69 (352/75)
  8. [M1] Identifying relevant scores from their mean and their table
  9. [A1] Correct answer
  10. [SC B1] for 4/15 with no working

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About this question

Topic
Probability and statistics, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source
A-Level Mathematics, May/June 2019 Paper 6 Variant 3, question 6.

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