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

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

With

N /2

white balls, uniform sampling without replacement gives the probability

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

Rearranging,

4 p_{w} (N - 1) = N - 2

, so

N = \frac{2 - 4 p _{w}}{1 - 4 p _{w}} = \frac{2 ( 1 - 2 p _{w} )}{1 - 4 p _{w}} .

A finite admissible population has even

N \geq 2

and

0 \leq p_{w} < 1/4

, so the denominator is nonzero.

ii

Solution

Interchanging the two colours preserves the probability distribution when their population counts are equal. Thus the probability of two black balls is also

p_{w}

. The three mutually exclusive possibilities are two white, two black, or different colours, so

p_{d} = 1 - 2 p_{w} .

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