Past exam of the mathematics course of the University of Cambridge / 2004 / ia / Paper 2 / 10F / Solution

For events with $\mathbb{P}(B)>0$, the conditional probability isUse the sample space $\Omega =\{1,2,3,4\}\times \{0,1{\}}^4$, where the first coordinate is the chosen coin and the other coordinates record which of its two physically distinguished faces lands upwards on each toss. There are $4\cdot 16=64$ equally likely elementary outcomes. Label the double-headed coin by one. For that coin both face indices mean heads; for each other coin only face index one means heads. In particular, strings of observed heads and tails are not equally likely outcomes.

Let $E$ be the event that the first three tosses show heads. With coin one all sixteen face-index strings are permitted. For each ordinary coin the first three indices must be one and the fourth is free, giving two strings. Thus $|E|=16+3\cdot 2=22$. Of these outcomes, sixteen from the double-headed coin and one from each ordinary coin also have a fourth head. HenceEquivalently, Bayes' theorem gives conditional probability $8/11$ of having selected the double-headed coin after $E$; the fourth-head probability is then $8/11+(3/11)(1/2)=19/22$. The tosses are independent conditional on the selected coin, while mixing over the same coin explains their unconditional dependence.