Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 7 / b

To obtain the $(k+1)$st head on toss $n+1$, the first $n$ tosses must contain exactly $k$ heads, and the last toss must be a head. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ choices of the first $k$ head positions. By independence, each such arrangement followed by a head has probability $r^k(1-r{)}^{n-k}r$. The arrangements are disjoint, soThis is the negative binomial stopping argument, with $n+1$ counting all tosses up to and including the $(k+1)$st success. It is not merely the probability of $k+1$ heads somewhere among $n+1$ tosses: the last toss is prescribed to be a head.

Put $D=1-(1-p)(1-q)=p+q-pq$. By the law of total probability,For $0\le t<1$, termwise differentiation of the geometric series, valid inside its radius of convergence, givesTaking $t=(1-p)(1-q)$ therefore yieldsSince $D-p=(1-p)q$, this is the zero-based geometric distribution with parameterThis illustrates Bernoulli thinning of a geometric count. Here $0<p_{\ast }<1$, as $0<p,q<1$.

For the reverse conditional distribution, Bayes' theorem gives, for $n\ge k$,ThusFor $n<k$ the conditional probability is zero. The conditional distribution of $X-k$ is the failure-count form of a negative binomial distribution, with $k+1$ successes and success parameter $r$: the geometric-binomial posterior is shifted negative binomial.