Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 16 / c / i

Table of contents
- Solution i

Solution

With equally many balls of each colour, uniform sampling without replacement gives

p_{w} = \frac{N /2}{N} \frac{N /2 - 1}{N - 1} \frac{N /2 - 2}{N - 2} = \frac{N - 4}{8 ( N - 1 )} .

Here an admissible population producing positive probability for three white balls has even

N \geq 6

, so the cancellation is legitimate. Equating this to

2/19

gives

19 (N - 4) = 16 (N - 1)

, hence

N = 20.

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