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2006 Paper 1 Q11
D: 1500.0 B: 1470.9

Particles \(A_1\), \(A_2\), \(A_3\), \(\ldots\), \(A_n\) (where \(n\ge 2\)) lie at rest in that order in a smooth straight horizontal trough. The mass of \(A_{n-1}\) is \(m\) and the mass of \(A_n\) is \(\lambda m\), where \(\lambda>1\). Another particle, \(A_0\), of mass \(m\), slides along the trough with speed \(u\) towards the particles and collides with \(A_1\). Momentum and energy are conserved in all collisions.

  1. Show that it is not possible for there to be exactly one particle moving after all collisions have taken place.
  2. Show that it is not possible for \(A_{n-1}\) and \(A_n\) to be the only particles moving after all collisions have taken place.
  3. Show that it is not possible for \(A_{n-2}\), \(A_{n-1}\) and \(A_n\) to be the only particles moving after all collisions have taken place.
  4. Given that there are exactly two particles moving after all collisions have taken place, find the speeds of these particles in terms of \(u\) and \(\lambda\).

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2006 Paper 1 Q12
D: 1500.0 B: 1499.3

Oxtown and Camville are connected by three roads, which are at risk of being blocked by flooding. On two of the three roads there are two sections which may be blocked. On the third road there is only one section which may be blocked. The probability that each section is blocked is \(p\). Each section is blocked independently of the other four sections. Show that the probability that Oxtown is cut off from Camville is \(p^3 \l 2-p \r^2\). I want to travel from Oxtown to Camville. I choose one of the three roads at random and find that my road is not blocked. Find the probability that I would not have reached Camville if I had chosen either of the other two roads. You should factorise your answer as fully as possible. Comment briefly on the value of this probability in the limit \(p\to1\).

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2006 Paper 1 Q13
D: 1484.0 B: 1468.0

A very generous shop-owner is hiding small diamonds in chocolate bars. Each diamond is hidden independently of any other diamond, and on average there is one diamond per kilogram of chocolate.

  1. I go to the shop and roll a fair six-sided die once. I decide that if I roll a score of \(N\), I will buy \(100N\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6} \l \frac{1 - \e^{-0.6} }{ 1 - \e^{-0.1}} \r \] Show also that the expected number of diamonds I find is 0.35.
  2. Instead, I decide to roll a fair six-sided die repeatedly until I score a 6. If I roll my first 6 on my \(T\)th throw, I will buy \(100T\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6 - 5\e^{-0.1}} \] Calculate also the expected number of diamonds that I find. (You may find it useful to consider the the binomial expansion of \(\l 1 - x \r^{-2}\).)

Solution: Not that the number of diamonds per kilogram is \(1\) so we are assuming it is \(Po(M)\) where \(M\) is the mass in kg. In particular \(\E[X] = M\) and \(\mathbb{P}(X = 0) = e^{-M}\)

  1. \(\,\) \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^6\mathbb{P}(\text{no diamonds and roll }n) \\ &&&= \sum_{n=1}^6 \tfrac16 e^{-\frac{n}{10}} \\ &&&= \frac{e^{-0.1}}6 \left ( \frac{1-e^{-0.6}}{1-e^{-0.1}}\right) \\ && \E[\text{diamonds}] &= \sum_{n=1}^6 \E(\text{diamonds}|N = n)\mathbb{P}(N = n) \\ &&&= \sum_{n=1}^6 0.1n \cdot \frac16 \\ &&&= 0.1 \cdot \frac{7}{2} = 0.35 \end{align*}
  2. \(\mathbb{P}(T = k) = \left ( \frac56 \right)^{t-1} \frac16\), so \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^\infty\mathbb{P}(\text{no diamonds and }T=n) \\ &&&= \sum_{n=1}^\infty e^{-0.1n} \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{e^{-0.1}}{6} \frac1{1- \frac56 e^{-0.1}} \\ &&&= \frac{e^{-0.1}}{6 - 5e^{-0.1}} \\ \\ && \E[\text{diamonds}] &= \sum_{n=1}^\infty \E(\text{diamonds}|T = n)\mathbb{P}(T = n) \\ &&&= \sum_{n=1}^\infty 0.1n \cdot \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{0.1}{6} \sum_{n=1}^\infty n \cdot \left ( \frac56 \right)^{n-1} \\ &&&= \frac{1}{60} \frac{1}{(1- \tfrac56)^2} \\ &&&= \frac{6}{10} = \frac35 \end{align*}

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2006 Paper 1 Q14
D: 1500.0 B: 1502.6

  1. A bag of sweets contains one red sweet and \(n\) blue sweets. I take a sweet from the bag, note its colour, return it to the bag, then shake the bag. I repeat this until the sweet I take is the red one. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?
  2. Instead, I take sweets from the bag, without replacing them in the bag, until I take the red sweet. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?

Solution:

  1. This is the probability of having the sequence \(\underbrace{BB\cdots B}_{r-1 \text{ times}}R\) which has probability \(\displaystyle \left ( \frac{n}{n+1} \right)^{r-1}\frac{1}{n+1}\). Maximising this, is equivalent to maximising \(\log\) of it, ie \begin{align*} && y &= (r-1) \ln n - r \ln (n+1) \\ \Rightarrow && \frac{\d y}{\d n} &= \frac{r-1}{n} - \frac{r}{n+1} \\ &&&= \frac{(r-1)(n+1)-rn}{n(n+1)} \\ &&&= \frac{r-n-1}{n(n+1)} \end{align*} Therefore this is maximised when \(n = r-1\)

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