 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-2/14/b/iii/solution

There are

99

possible stopping points, so the equal defective-ball probabilities are

d_{k} = 1/99

. With

g_{0} > 0

\frac{p g _{0}}{p g _{0} + ( 1 - p ) /99} = \frac{99}{100}, 100 p g_{0} = 99 p g_{0} + 1 - p .

Thus, for a fixed good-ball stopping law,

p = \frac{1}{1 + g _{0}} .

Since

g_{0} \leq 1

, the smallest possible production probability over admissible stopping laws is

p_{m i n} = \frac{1}{2}, attained when g_{0} = 1.

If strict positivity of every

g_{k}

is additionally imposed in this general case, then

g_{0} < 1

and one only has the infimum

1/2

, approached as

g_{0} \to 1

. The paper states that strict-positivity assumption explicitly for part (a), not for the general case in part (b). Without choosing or bounding

g_{0}

, the exact fixed-law answer remains

1/ (1 + g_{0})

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