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1997 Paper 3 Q12
D: 1700.0 B: 1500.0

  1. I toss a biased coin which has a probability \(p\) of landing heads and a probability \(q=1-p\) of landing tails. Let \(K\) be the number of tosses required to obtain the first head and let \[ \mathrm{G}(s)=\sum_{k=1}^{\infty}\mathrm{P}(K=k)s^{k}. \] Show that \[ \mathrm{G}(s)=\frac{ps}{1-qs} \] and hence find the expectation and variance of \(K\).
  2. I sample cards at random with replacement from a normal pack of \(52\). Let \(N\) be the total number of draws I make in order to sample every card at least once. By expressing \(N\) as a sum \(N=N_{1}+N_{2}+\cdots+N_{52}\) of random variables, or otherwise, find the expectation of \(N\). Estimate the numerical value of this expectation, using the approximations \(\mathrm{e}\approx2.7\) and \(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\approx0.5+\ln n\) if \(n\) is large.


Solution:

  1. Let \(N_i\) be the number of draws between the \((i-1)\)th new card and the \(i\)th new card. (Where \(N_1 = 1\)0 then \(N_i \sim K\) with \(p = \frac{53-i}{52}\)). Therefore \begin{align*} \E[N] &= \E[N_1 + \cdots + N_{52}] \\ &= \E[N_1] + \cdots + \E[N_i] + \cdots + \E[N_{52}] \\ &= 1 + \frac{52}{51} + \cdots + \frac{52}{53-k} + \cdots + \frac{52}{1} \\ &= 52 \left (1 + \frac{1}{2} + \cdots + \frac{1}{52} \right) \\ &= 52 \cdot \left ( 1 + \ln 52 \right) \end{align*} Notice that \(2.7 \times 2.7 = 7.29\) and \(7.3 \times 7.3 \approx 53.3\) so \(\ln 52 \approx 4\) and so our number is \(\approx 52 \cdot 4.5 =234\). [The correct answer actual number is 235.9782]

1994 Paper 2 Q13
D: 1600.0 B: 1629.1

The makers of Cruncho (`The Cereal Which Cares') are giving away a series of cards depicting \(n\) great mathematicians. Each packet of Cruncho contains one picture chosen at random. Show that when I have collected \(r\) different cards the expected number of packets I must open to find a new card is \(n/(n-r)\) where \(0\leqslant r\leqslant n-1.\) Show by means of a diagram, or otherwise, that \[ \frac{1}{r+1}\leqslant\int_{r}^{r+1}\frac{1}{x}\,\mathrm{d}x\leqslant\frac{1}{r} \] and deduce that \[ \sum_{r=2}^{n}\frac{1}{r}\leqslant\ln n\leqslant\sum_{r=1}^{n-1}\frac{1}{r} \] for all \(n\geqslant2.\) My children will give me no peace until we have the complete set of cards, but I am the only person in our household prepared to eat Cruncho and my spouse will only buy the stuff if I eat it. If \(n\) is large, roughly how many packets must I expect to consume before we have the set?