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

For a discrete random variable taking nonnegative integer values, its probability generating function isIt is defined at least for $|s|\le 1$. The usual probability generating function is intended for nonnegative integer-valued random variables; a general discrete real-valued random variable need not have this form. The coefficient of $s^k$ recovers $\mathbb{P}(X=k)$.

For the binomial distribution, the binomial theorem givesDifferentiation of the probability generating function recovers the first expected value and second factorial moment:Since $X^2=X(X-1)+X$, the variance isThusThe formulas also hold for $n=0,1$ and the degenerate choices $p=0,1$, with derivatives read directly from the resulting polynomial.

The conditional probability rescales the probability of the part of $A$ lying in $B$ by the probability of $B$:The denominator makes $B$ itself have conditional probability one.

The events $A\cap B$ and $A\cap B^c$ are disjoint and their union is $A$. Additivity of probability givesBoth conditioning events have positive probability because $0<\mathbb{P}(B)<1$. Apply the definition of conditional probability to each term to obtain the law of total probability:

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.