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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/16/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2017 / Paper 1 / 16 / b / Solution by

Codex  0 2026-10-18

Assume that wrong-note events are independent events and that each note has the same error probability

p

, with

0 \leq p \leq 1

. Then their count

N

has a binomial distribution. Choose the

n

mistaken positions in

(n 100)

ways and multiply the probabilities of the errors and correct notes:

P (N = n) = (n 100) p^{n} (1 - p)^{100 - n}, n = 0, 1, \dots, 100.

The endpoint cases

p = 0

or

p = 1

are understood by continuity, giving deterministic counts. More than half means at least

51

mistakes, so

P (booing) = n = 51 \sum 100 (n 100) p^{n} (1 - p)^{100 - n} .

The single-note error probability alone would not determine this count distribution if mistakes were correlated; independence is the additional standard modeling assumption.

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