 Solution

ID: 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/ii/solution

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

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