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2011 Paper 3 Q11
D: 1700.0 B: 1484.0

A thin uniform circular disc of radius \(a\) and mass \(m\) is held in equilibrium in a horizontal plane a distance \(b\) below a horizontal ceiling, where \(b>2a\). It is held in this way by \(n\) light inextensible vertical strings, each of length \(b\); one end of each string is attached to the edge of the disc and the other end is attached to a point on the ceiling. The strings are equally spaced around the edge of the disc. One of the strings is attached to the point \(P\) on the disc which has coordinates \((a,0,-b)\) with respect to cartesian axes with origin on the ceiling directly above the centre of the disc. The disc is then rotated through an angle \(\theta\) (where \(\theta<\pi\)) about its vertical axis of symmetry and held at rest by a couple acting in the plane of the disc. Show that the string attached to~\(P\) now makes an angle \(\phi\) with the vertical, where \[ b\sin\phi = 2a \sin\tfrac12 \theta\,. \] Show further that the magnitude of the couple is \[ \frac {mga^2\sin\theta}{\sqrt{b^2-4a^2\sin^2 \frac12\theta \ } \ }\,. \] The disc is now released from rest. Show that its angular speed, \(\omega\), when the strings are vertical is given by \[ \frac{a^2\omega^2}{4g} = b-\sqrt{b^2 - 4a^2\sin^2 \tfrac12\theta \;}\,. \]

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2011 Paper 3 Q12
D: 1700.0 B: 1516.0

The random variable \(N\) takes positive integer values and has pgf (probability generating function) \(\G(t)\). The random variables \(X_i\), where \(i=1\), \(2\), \(3\), \(\ldots,\) are independently and identically distributed, each with pgf \({H}(t)\). The random variables \(X_i\) are also independent of \(N\). The random variable \(Y\) is defined by \[ Y= \sum_{i=1}^N X_i \;. \] Given that the pgf of \(Y\) is \(\G(H(t))\), show that \[ \E(Y) = \E(N)\E(X_i) \text{ and } \var(Y) = \var(N)\big(\E(X_i)\big)^2 + \E(N) \var(X_i) \,.\] A fair coin is tossed until a head occurs. The total number of tosses is \(N\). The coin is then tossed a further \(N\) times and the total number of heads in these \(N\) tosses is \(Y\). Find in this particular case the pgf of \(Y\), \(\E(Y)\), \(\var(Y)\) and \(\P(Y=r)\).

Solution: Recall that for a random variable \(Z\) with pgf \(F(t)\) we have \(F(1) = 1\), \(\E[Z] = F'(1)\) and \(\E[Z^2] = F''(1) +F'(1)\) so \begin{align*} && \E[Y] &= G'(H(1))H'(1) \\ &&&= G'(1)H'(1) \\ &&&= \E[N]\E[X_i] \\ \\ && \E[Y^2] &= G''(H(1))(H'(1))^2+G'(H(1))H''(1) + G'(H(1))H'(1) \\ &&&= G''(1)(H'(1))^2+G'(1)H''(1) + G'(1)H'(1) \\ &&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N](\E[X_i^2]-\E[X_i]) + \E[N]\E[X_i] \\ &&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] \\ && \var[Y] &= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] - (\E[N])^2(\E[X_i])^2\\ &&&= (\var[N]+(\E[N])^2-\E[N])(\E[X_i])^2 + \E[N](\var[X_i]+\E[X_i]^2) - (\E[N])^2(\E[X_i])^2\\ &&&= \var[N](\E[X_i])^2 + \E[N]\var[X_i] \end{align*} Notice that \(N \sim Geo(\tfrac12)\) and \(Y = \sum_{i=1}^N X_i\) where \(X_i\) are Bernoulli. We have that \(G(t) = \frac{\frac12}{1-\frac12z}\) and \(H(t) = \frac12+\frac12p\) so the pgf of \(Y\) is \(G(H(t) = \frac{\frac12}{1 - \frac14-\frac14p} = \frac{2}{3-p}\). \begin{align*} && \E[X_i] &= \frac12\\ && \var[X_i] &= \frac14 \\ && \E[N] &= 2 \\ && \var[N] &= 2 \\ \\ && \E[Y] &= 2 \cdot \frac12 = 1 \\ && \var[Y] &= 2 \cdot \frac14 + 2 \frac14 = 1 \\ && \mathbb{P}(Y=r) &= \tfrac23 \left ( \tfrac13 \right)^r \end{align*}

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2011 Paper 3 Q13
D: 1700.0 B: 1500.0

In this question, the notation \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to \(x\), so for example \(\lfloor \pi\rfloor = 3\) and \(\lfloor 3 \rfloor =3\).

  1. A bag contains \(n\) balls, of which \(b\) are black. A sample of \(k\) balls is drawn, one after another, at random with replacement. The random variable \(X\) denotes the number of black balls in the sample. By considering \[ \frac{\P(X=r+1)}{\P(X=r)}\,, \] show that, in the case that it is unique, the most probable number of black balls in the sample is \[ \left\lfloor \frac{(k+1)b}{n}\right\rfloor. \] Under what circumstances is the answer not unique?
  2. A bag contains \(n\) balls, of which \(b\) are black. A sample of \(k\) balls (where \(k\le b\)) is drawn, one after another, at random without replacement. Find, in the case that it is unique, the most probable number of black balls in the sample. Under what circumstances is the answer not unique?

Solution:

  1. \(\mathbb{P}(X = r) = \binom{k}{r}p^rq^{k-r}\) where \(p = \frac{b}{n}, q = 1-p\). Therefore \begin{align*} && \frac{\mathbb{P}(X=r+1)}{\mathbb{P}(X=r)} &= \frac{\binom{k}{r+1}p^{r+1}q^{k-r-1}}{\binom{k}{r}p^rq^{k-r}} \\ &&&= \frac{(k-r)p}{(r+1)q} \\ &&&= \frac{(k-r)b}{(r+1)(n-b)} \end{align*} Comparing this to \(1\) we find: \begin{align*} && 1 & < \frac{\mathbb{P}(X=r+1)}{\mathbb{P}(X=r)} \\ \Leftrightarrow && 1 &< \frac{(k-r)b}{(r+1)(n-b)} \\ \Leftrightarrow && (r+1)(n-b) &<(k-r)b \\ \Leftrightarrow && rn& < (k+1)b - n \\ \Leftrightarrow && r &< \frac{(k+1)b}{n} - 1\\ \end{align*} If this equation is true, then \(\mathbb{P}(X=r+1)\) is larger, so \(r_{max} = \left \lfloor \frac{(k+1)b}{n} \right \rfloor\)
  2. Let \(Y\) be the number of black balls in our sample, ie \(\mathbb{P}(Y = r) = \binom{b}{r}\binom{n-b}{k-r}/\binom{n}{k}\), so \begin{align*} && \frac{\mathbb{P}(Y = r+1)}{\mathbb{P}(Y=r)} &= \frac{\binom{b}{r+1}\binom{n-b}{k-(r+1)}/\binom{n}{k}}{\binom{b}{r}\binom{n-b}{k-r}/\binom{n}{k}} \\ &&&= \frac{b-r}{r+1} \frac{k-r}{n-b-k+r+1} \\ && 1 &< \frac{\mathbb{P}(Y = r+1)}{\mathbb{P}(Y=r)} \\ \Leftrightarrow && (r+1)(n-b-k+r+1) &< (b-r)(k-r) \\ \Leftrightarrow &&r(n-b-k+1)+(n-b-k+1) &< -r(b+k)+bk \\ \Leftrightarrow &&r(n+1) &< bk+b+k+1-n \\ \Leftrightarrow && r &< \frac{(b+1)(k+1)}{n+1} - \frac{n}{n+1} \end{align*} Therefore \(r = \left \lfloor \frac{ (b+1)(k+1)}{n+1}\right \rfloor\), it is not unique if \(n+1\) divides \((b+1)(k+1)\)

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