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2007 Paper 3 Q11
D: 1700.0 B: 1500.0

  1. A wheel consists of a thin light circular rim attached by light spokes of length \(a\) to a small hub of mass \(m\). The wheel rolls without slipping on a rough horizontal table directly towards a straight edge of the table. The plane of the wheel is vertical throughout the motion. The speed of the wheel is \(u\), where \(u^2
  2. Two particles, each of mass \(m/2\), are attached to a light circular hoop of radius \(a\), at the ends of a diameter. The hoop rolls without slipping on a rough horizontal table directly towards a straight edge of the table. The plane of the hoop is vertical throughout the motion. When the centre of the hoop is vertically above the edge of the table it has speed \(u\), where \(u^2

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2007 Paper 3 Q12
D: 1700.0 B: 1487.4

I choose a number from the integers \(1, 2, \ldots\), \((2n-1)\) and the outcome is the random variable~\(N\). Calculate \( \E(N)\) and \(\E(N^2)\). I then repeat a certain experiment \(N\) times, the outcome of the \(i\)th experiment being the random variable \(X_i\) ($1\le i \le N\(). For each \)i$, the random variable \(X_i\) has mean \(\mu\) and variance~\(\sigma^2\), and \(X_i\) is independent of \(X_j\) for \(i\ne j\) and also independent of \(N\). The random variable \(Y\) is defined by \(Y= \sum\limits_{i=1}^NX_i\). Show that \(\E(Y)=n\mu\) and that \(\mathrm{Cov}(Y,N) = \frac13n(n-1)\mu\). Find \(\var(Y) \) in terms of \(n\), \(\sigma^2\) and \(\mu\).

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2007 Paper 3 Q13
D: 1700.0 B: 1500.0

A frog jumps towards a large pond. Each jump takes the frog either \(1\,\)m or \(2\,\)m nearer to the pond. The probability of a \(1\,\)m jump is \(p\) and the probability of a \(2\,\)m jump is \(q\), where \(p+q=1\), the occurence of long and short jumps being independent.

  1. Let \(p_n(j)\) be the probability that the frog, starting at a point \((n-\frac12)\,\)m away from the edge of the pond, lands in the pond for the first time on its \(j\)th jump. Show that \(p_2(2)=p\).
  2. Let \(u_n\) be the expected number of jumps, starting at a point \((n-\frac12)\,\)m away from the edge of the pond, required to land in the pond for the first time. Write down the value of \(u_1\). By finding first the relevant values of \(p_n(m)\), calculate \(u_2\) and show that $u_3= 3-2q+q^2\(.
  3. Given that \)u_n\( can be expressed in the form \)u_n= A(-q)^{n-1} +B +Cn$, where \(A\), \(B\) and \(C\) are constants (independent of \(n\)), show that \(C= (1+q)^{-1}\) and find \(A\) and \(B\) in terms of \(q\). Hence show that, for large \(n\), \(u_n \approx \dfrac n{p+2q}\) and explain carefully why this result is to be expected.

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2007 Paper 3 Q14
D: 1700.0 B: 1500.0

  1. My favourite dartboard is a disc of unit radius and centre \(O\). I never miss the board, and the probability of my hitting any given area of the dartboard is proportional to the area. Each throw is independent of any other throw. I throw a dart \(n\) times (where \(n>1\)). Find the expected area of the smallest circle, with centre \(O\), that encloses all the \(n\) holes made by my dart. Find also the expected area of the smallest circle, with centre \(O\), that encloses all the \((n-1)\) holes nearest to \(O\).
  2. My other dartboard is a square of side 2 units, with centre \(Q\). I never miss the board, and the probability of my hitting any given area of the dartboard is proportional to the area. Each throw is independent of any other throw. I throw a dart \(n\) times (where \(n>1\)). Find the expected area of the smallest square, with centre \(Q\), that encloses all the \(n\) holes made by my dart.
  3. Determine, without detailed calculations, whether the expected area of the smallest circle, with centre \(Q\), on my square dartboard that encloses all the \(n\) holes made by my darts is larger or smaller than that for my circular dartboard.

Solution:

  1. Firstly, we consider the probability that all darts lie within a distance \(s\) from the centre, ie \begin{align*} \mathbb{P}(\text{all darts within }s) &= \prod_{k=1}^s \mathbb{P}(\text{dart within }s) \\ &= \left ( \frac{\pi s^2}{\pi} \right)^n \\ &= s^{2n} \end{align*} Therefore the pdf is \(2ns^{2n-1}\), and the expected area is \(\int_{s=0}^1 \pi s^2 \cdot 2n s^{2n-1} \d s = 2n \pi \frac{1}{2n+2} = \frac{n}{n+1} \pi\). \begin{align*} \mathbb{P}(\text{n-1 within }s) &= \underbrace{s^{2n}}_{\text{all within }s} + \underbrace{ns^{2n-2}(1-s^2)}_{\text{all but 1 within }s}\\ &= ns^{2n-2}-(n-1)s^{2n} \end{align*} Therefore the pdf is \(n(2n-2)s^{2n-3} - 2n(n-1)s^{2n-1} = 2n(n-1)(s^{2n-3}-s^{2n-1})\) and the expected area is: \begin{align*} \int \pi s^2 \cdot2n(n-1)(s^{2n-3}-s^{2n-1})\d s &= 2n(n-1) \pi \left ( \frac{1}{2n} - \frac{1}{2n+2} \right) \\ &= n(n-1)\pi \frac{2}{n(n+1)} \\ &= \frac{n-1}{n+1} \pi \end{align*}
  2. Now consider a square of side-length \(s\), we must have \(\mathbb{P}(\text{all darts within square}) = \left ( \frac{s^2}{4} \right)^n\) and therefore the pdf is \(n \frac{s^{n-1}}{4^n}\). Therefore the expected area is \(\displaystyle \int_0^2 s^2 \cdot n \frac{s^{n-1}}{4^n} \d s = \frac{n}{n+1} \frac{2^{2n+1}}{2^{2n}} = \frac{4n}{n+1}\)
  3. It is clearly larger as the square dartboard contains all of the circular dartboard, and there will be some probability that the darts land outside the circular dartboard, making the circle much larger.

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