Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Discrete mathematics, continuous mathematics and probability / 7

Let $(B_i)$ be a finite or countable partition of a sample space into events of positive probability. The events $A\cap B_i$ are disjoint and their union is $A$. Countable additivity and the definition of conditional probability therefore give the law of total probabilityCells of zero probability contribute zero through $\mathbb{P}(A\cap B_i)$ and can be omitted rather than assigning an undefined conditional probability.

The two definitions of conditional probability giveSubstitution proves Bayes' theorem:The positive probabilities assumed in the question make both conditional expressions defined.

Mutual independence means that for every nonempty $I\subseteq \{1,2,3\}$,Beyond the trivial single-event identities, this requires the three pairwise equalities and the triple equality $\mathbb{P}(A_1\cap A_2\cap A_3)=\mathbb{P}(A_1)\mathbb{P}(A_2)\mathbb{P}(A_3)$. Pairwise independence alone omits that last requirement.

Each position is equally likely to receive any coin, so the law of total probability gives $\mathbb{P}(A_i)=(1/3+1/2+2/3)/3=1/2$. For two specified positions, the ordered pair of distinct coin biases is uniform over six possibilities. Assuming independent tosses conditional on the order,Thus the events are not independent, even pairwise. The covariance of draws without replacement explains the difference: the head indicator random variables have covariance $13/54-1/4=-1/108$. Independent physical tosses given the assigned coins do not imply independence after that hidden assignment is averaged out.

Let $C_j$ identify the selected coin and let $H$ be the event of two heads. The prior coin probabilities are equal, and the conditional probabilities of $H$ are $1/9,1/4,4/9$. By Bayes' theorem,The full posterior distribution over the three biases is $(4,9,16)/29$.

Given the coin, the next toss is independent of its earlier tosses and has that coin's head probability. Averaging using the posterior distribution gives the posterior predictive probabilityObserving two heads raises the probability of having chosen a high-bias coin.

If the selected coin has bias $p$, the other two biases sum to $3/2-p$. A uniform choice between those two coins therefore has conditional head probability $3/4-p/2$. Average over the posterior distribution, using the previous posterior predictive probability:The observation makes a high-bias coin more likely to have been removed from the remaining pair.