A-LevelMathematicsProbability and statistics9709/63 · May/June 2017 · Q58 Marks

Hebe attempts a crossword puzzle every day. The number of puzzles she completes in a week (7 days) is denoted by X.

(i)State two conditions that are required for X to have a binomial distribution.[2]
(ii)On average, Hebe completes 7 out of 10 of these puzzles. Use a binomial distribution to find the probability that Hebe completes at least 5 puzzles in a week.[3]
(iii)Use a binomial distribution to find the probability that, over the next 10 weeks, Hebe completes 4 or fewer puzzles in exactly 3 of the 10 weeks.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

[B1] constant probability (of completing) Any one condition of these two
[B1] independent trials/events The other condition
[M1] P(5, 6, 7) = 7C5(0.7)5(0.3)² + 7C6(0.7)6(0.3)¹ + (0.7)7 Bin term 7Cₓ(0.7)ˣ(0.3)⁷⁻ˣ, x ≠ 0, 7
[A1] Correct unsimplified answer (sum) OE
[A1] = 0.647
[M1] P(0, 1, 2, 3, 4) = 1 – their ‘0.6471' = 0.3529 Find P( ≤4) either by subtracting their (ii) from 1 or from adding Probs of 0,1,2,3,4 with n=7 (or 10) and p = 0.7
[M1] P(3) = 10C3(0.3529)3(0.6471)7 10C3 (their 0.353)³(1 – their 0.353)⁷ on its own
[A1] = 0.251

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