Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 14

Distinguish the production type as well as the stopping position. The sample space has the six outcomesHere $x_k$ in an event means $X=x_k$. Their joint probabilities are $p{g}_k$ and $(1-p)d_k$, respectively. The conditional stopping laws obey ${\sum }_{k=0}^2{g}_k={\sum }_{k=0}^2{d}_k=1$; with the stated strict positivity all six outcomes are possible.

Multiply the production probability by the conditional stopping probability:

The good and defective cases are disjoint and exhaustive. The law of total probability therefore gives

Divide the relevant joint probability by the marginal stopping probability. By Bayes' theorem,The denominator is positive under the stated assumptions.

The condition $X>x_0$ selects positions $x_1$ and $x_2$. Their combined good-ball joint probability is $p(g_1+g_2)$, and their total probability is $p(g_1+g_2)+(1-p)(d_1+d_2)$. Thus

For $0\le k\le n$, sum the marginal stopping probabilities over the selected tail:Equivalently it is $p(1-{\sum }_{j=0}^k{g}_j)+(1-p)(1-{\sum }_{j=0}^k{d}_j)$. At $k=n$, both empty tail sums are zero, as required.

The event includes stopping positions zero through $k$. Applying Bayes' theorem to their joint and marginal probabilities givesThis conditional probability is defined whenever its conditioning event has positive probability. At $k=n$, it reduces to $p$.

There are $99$ possible stopping points, so the equal defective-ball probabilities are $d_k=1/99$. With $g_0>0$, the desired conditional probability givesThus, for a fixed good-ball stopping law,Since $g_0\le 1$, the smallest possible production probability over admissible stopping laws isIf 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 $1/(1+g_0)$.