Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 2 / 11

Let $C$ mean control-group membership and $E$ the adverse event. Equal allocation gives $P(C)=P(C^c)=1/2$, so the law of total probability gives $P(E)=(p_0+p_1)/2$. Bayes' theorem then givesThis is defined when $p_0+p_1>0$. If both are zero, the conditioning event has zero probability and the stated conditional probability is undefined.

Normalize the Rayleigh distribution by substituting $s=x^2/(2{\sigma }^2)$:For its sketch, the density starts at zero, is positive for $x>0$, and tends to zero at infinity. Differentiation givesIt therefore has a single maximum at $x=\sigma$, with height $e^{-1/2}/\sigma$.

The expected value isIntegrating by parts with $xdx{e}^{-x^2/(2{\sigma }^2)}=-{\sigma }^{2}d(e^{-x^2/(2{\sigma }^2)})$ makes the boundary term vanish and leaves ${\int }_0^{\infty }{e}^{-x^2/(2{\sigma }^2)}dx$. The Gaussian integral gives

Use the usual model of independent games, each with Sheila-win probability $\theta$. After $m+n$ games, an outcome with $m$ wins and $n$ losses has $\left(\genfrac{}{}{0ex}{}{m+n}{m}\right)$ possible orders. Since both totals are below eleven, none of those sequences has already ended the match. ThusFor a completed Sheila win with $n$ losses, the last game must be her eleventh win. The preceding $10+n$ games contain ten wins and $n$ losses, so their probability is $\left(\genfrac{}{}{0ex}{}{10+n}{10}\right){\theta }^{10}(1-\theta {)}^n$, followed by one further factor $\theta$. The allowed loss counts are $0\le n\le 10$, and these terminal events are disjoint. HenceThis is the match win probability with a fixed winning score, obtained by the negative binomial stopping argument. Constant marginal game-win probabilities alone would not justify the product without the usual independence assumption.