Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 7

Let $G$ denote carrying the gene and $+$ a positive result. The law of total probability givesThe last value is to three significant figures. The much larger non-carrier population contributes most of the positive tests in this mathematical model.

By Bayes' theorem, the required conditional probability isThis is about $1.94\%$; it is a different conditioning direction from the test's $99\%$ positive probability among carriers.

The negative-test probability is $P(-)=1-0.05094=0.94906$. Applying Bayes' theorem again,

For $B$ to make a first throw, $A$ must first pass, with probability $1/2$. For $B$ then to win on that throw, $B$ must roll one of two winning faces, with probability $1/3$. The throws are independent, givingA first throw of $3$ by $A$ would make $B$ win without throwing, and is excluded by the requested event.

Before $A$'s second throw, $A$ can win either immediately, with probability $1/3$, or by passing to $B$ and having $B$ throw $3$, with probability $(1/2)(1/6)$. These are disjoint events, so

This is a two-turn renewal game. A full pair of throws has probability $(1/2{)}^2=1/4$ of passing twice and returning to the original state. The probability that $A$ wins before that restart is $5/12$. Thus first-step analysis gives $p=5/12+p/4$, orEquivalently sum the geometric series $(5/12){\sum }_{j\ge 0}(1/4{)}^j$. The probability of surviving infinitely many pairs is zero, so the eventual winner is well defined with probability one.