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

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.

ii

There are ten balls of each colour. Under uniform sampling without replacement, all

(12 20)

subsets of twelve labelled balls are equally likely. A favourable subset chooses four of the ten white balls and eight of the ten black balls, giving

(4 10) (8 10)

P (four white) = \frac{( 4 10 ) ( 8 10 )}{( 12 20 )} .

Thus the requested integers are

N_{1} = N_{2} = 10, m_{1} = 4, m_{2} = 8

, with

N = 20

and

k = 12

There are no discussions about this article yet.

There are currently no matching articles.