Past exam of the year 1 of the computer science course of the University of Oxford / 2013 / Discrete mathematics, probability and continuous mathematics / 7 / c

Let $H$ mean the added ball was red, and $D$ mean the first draw was red. Their relevant probabilities are $\mathbb{P}(H)=\mathbb{P}(H^c)=1/2$, $\mathbb{P}(D\mid H)=1$, and $\mathbb{P}(D\mid H^c)=1/2$. The law of total probability gives $\mathbb{P}(D)=3/4$.

Given the observed red draw, the remaining ball is red precisely when $H$ occurred. Bayes' theorem therefore givesThe observation increases the weight of the two-red-ball box, because that box is more likely to yield a red first draw.

Let $B$ count the red balls among the $n$ added balls. By independence of random variables, $B$ has a binomial distribution with parameters $n,1/2$. The total red count is $R=1+B$, and the box has $n+1$ balls. Conditional on $R=r$, sampling without replacement giveswhere $D_1,D_2$ denote red first and second draws. Averaging these conditional probabilities and taking their ratio givesThe binomial distribution moments give $\mathbb{E}[B]=n/2$ and $\mathbb{E}[B^2]=n/4+n^2/4$. ConsequentlyTherefore the red-ball prediction with one certain red ball isIt reduces to $2/3$ at $n=1$ and approaches $1/2$ as the number of randomly coloured extra balls grows. Taking an unweighted average of the remaining red fractions would miss the change in weights caused by taking a conditional probability given the first red draw.