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2008 Paper 2 Q11
D: 1600.0 B: 1500.0

A wedge of mass \(km\) has the shape (in cross-section) of a right-angled triangle. It stands on a smooth horizontal surface with one face vertical. The inclined face makes an angle \(\theta\) with the horizontal surface. A particle \(P\), of mass \(m\), is placed on the inclined face and released from rest. The horizontal face of the wedge is smooth, but the inclined face is rough and the coefficient of friction between \(P\) and this face is \(\mu\).

  1. When \(P\) is released, it slides down the inclined plane at an acceleration \(a\) relative to the wedge. Show that the acceleration of the wedge is \[ \frac {a \cos\theta}{k+1}\,. \] To a stationary observer, \(P\) appears to descend along a straight line inclined at an angle~\(45^\circ\) to the horizontal. Show that \[ \tan\theta = \frac k {k+1}\,. \] In the case \(k=3\), find an expression for \(a\) in terms of \(g\) and \(\mu\).
  2. What happens when \(P\) is released if \(\tan\theta \le \mu\)?

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2008 Paper 2 Q12
D: 1600.0 B: 1500.0

In the High Court of Farnia, the outcome of each case is determined by three judges: the ass, the beaver and the centaur. Each judge decides its verdict independently. Being simple creatures, they make their decisions entirely at random. Past verdicts show that the ass gives a guilty verdict with probability \(p\), the beaver gives a guilty verdict with probability \(p/3\) and the centaur gives a guilty verdict with probability \(p^2\). Let \(X\) be the number of guilty verdicts given by the three judges in a case. Given that \(\E(X)= 4/3\), find the value of \(p\). The probability that a defendant brought to trial is guilty is \(t\). The King pronounces that the defendant is guilty if at least two of the judges give a guilty verdict; otherwise, he pronounces the defendant not guilty. Find the value of \(t\) such that the probability that the King pronounces correctly is \(1/2\).

Solution: \begin{align*} && \mathbb{E}(X) &= p + \frac{p}{3} + p^2 = \frac43p+p^2 \\ \Rightarrow && \frac43 &= \frac43p+p^2 \\ \Rightarrow && 0 &= 3p^2+4p-4 \\ &&&= (3p-2)(p+2) \\ \Rightarrow && p &= \frac23, -2 \end{align*} Since \(p \in [0,1]\) we must have \(p = \frac23\). \begin{align*} && \mathbb{P}(\text{correct verdict}) &= t p+ (1-t) (1-p) \\ &&&= t(2p-1)+(1-p)\\ \Rightarrow && \frac12 &= t(2p-1)+(1-p) \\ \Rightarrow && t &= \frac{\frac12-(1-p)}{2p-1} \\ &&&= \frac{2p-1}{2(2p-1)} = \frac12 \end{align*} (so it doesn't depend at all on what the judges are doing, the only way to be fair is if the trials happen at random!)

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2008 Paper 2 Q13
D: 1600.0 B: 1516.0

Bag \(P\) and bag \(Q\) each contain \(n\) counters, where \(n\ge2\). The counters are identical in shape and size, but coloured either black or white. First, \(k\) counters (\(0\le k\le n\)) are drawn at random from bag \(P\) and placed in bag \(Q\). Then, \(k\) counters are drawn at random from bag \(Q\) and placed in bag \(P\).

  1. If initially \(n-1\) counters in bag \(P\) are white and one is black, and all \(n\) counters in bag \(Q\) are white, find the probability in terms of \(n\) and \(k\) that the black counter ends up in bag \(P\). Find the value or values of \(k\) for which this probability is maximised.
  2. If initially \(n-1\) counters in bag \(P\) are white and one is black, and \(n-1\) counters in bag \(Q\) are white and one is black, find the probability in terms of \(n\) and \(k\) that the black counters end up in the same bag. Find the value or values of \(k\) for which this probability is maximised.

Solution:

  1. \(\,\) \begin{align*} \mathbb{P}(\text{black counter in }P) &= \mathbb{P}(\text{black counter moves twice})+\mathbb{P}(\text{black counter doesn't move}) \\ &= \mathbb{P}(\text{black counter moves out})\mathbb{P}(\text{black counter moves back}) + (1-\mathbb{P}(\text{black counter moves out})) \\ &= \frac{k}n\cdot \frac{k}{n+k}+\frac{n-k}{n} \\ &= \frac{k^2+n^2-k^2}{n(n+k)} \\ &= \frac{n^2}{n(n+k)} = \frac{n}{n+k} \end{align*} This is maximised if \(k\) is as small as possibe, ie \(k = 0\) (ie it doesn't leave it's bag)
  2. \(\,\) \begin{align*} && \mathbb{P}(\text{both counters in same bag}) &= \mathbb{P}(\text{both in }P)+ \mathbb{P}(\text{both in }Q) \\ &&&= \mathbb{P}(B_P \to Q \to P, B_Q \to P)+\mathbb{P}(B_P \text{ stays}, B_Q \to P)+\mathbb{P}(B_P \to Q, \text{both stay}) \\ &&&= \frac{k}{n} \cdot \frac{k(k-1)}{(n+k)(n+k-1)} + \frac{n-k}{n} \frac{k}{n+k} + \frac{k}{n} \frac{n(n-1)}{(n+k)(n+k-1)} \\ &&&= \frac{(k^3-k^2)+(n-k)k(n+k-1)+kn(n-1)}{n(n+k)(n+k-1)}\\ &&&= \frac{2kn(n-1)}{n(n+k)(n+k-1)}\\ &&&= \frac{2k(n-1)}{(n+k)(n+k-1)} \end{align*} \begin{align*} && \frac{P_{k+1}}{P_k} &= \frac{2(k+1)(n-1)}{(n+k+1)(n+k)} \frac{(n+k)(n+k-1)}{2k(n-1)} \\ &&&= \frac{(k+1)(n+k-1)}{k(n+k+1)} \\ &&& \geq 1 \\ \Leftrightarrow && (k+1)(n+k-1) &\geq k(n+k+1) \\ \Leftrightarrow && n-1 &\geq k \\ \end{align*} Therefore this probability is increasing while \(k \leq n-1\), ie it's maximised \(k = n-1\) or \(k=n\)

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