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1992 Paper 3 Q14
D: 1700.0 B: 1500.0
circular motion involute parametric acceleration components friction differential equation string particle on table unwinding

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A horizontal circular disc of radius \(a\) and centre \(O\) lies on a horizontal table and is fixed to it so that it cannot rotate. A light inextensible string of negligible thickness is wrapped round the disc and attached at its free end to a particle \(P\) of mass \(m\). When the string is all in contact with the disc, \(P\) is at \(A\). The string is unwound so that the part not in contact with the disc is taut and parallel to \(OA\). \(P\) is then at \(B\). The particle is projected along the table from \(B\) with speed \(V\) perpendicular to and away from \(OA\). In the general position, the string is tangential to the disc at \(Q\) and \(\angle AOQ=\theta.\) Show that, in the general position, the \(x\)-coordinate of \(P\) with respect to the axes shown in the figure is \(a\cos\theta+a\theta\sin\theta,\) and find \(y\)-coordinate of \(P\). Hence, or otherwise, show that the acceleration of \(P\) has components \(a\theta\dot{\theta}^{2}\) and \(a\dot{\theta}^{2}+a\theta\ddot{\theta}\) along and perpendicular to \(PQ,\) respectively. The friction force between \(P\) and the table is \(2\lambda mv^{2}/a,\) where \(v\) is the speed of \(P\) and \(\lambda\) is a constant. Show that \[ \frac{\ddot{\theta}}{\dot{\theta}}=-\left(\frac{1}{\theta}+2\lambda\theta\right)\dot{\theta} \] and find \(\dot{\theta}\) in terms of \(\theta,\lambda\) and \(a\). Find also the tension in the string when \(\theta=\pi.\)

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1992 Paper 3 Q15
D: 1700.0 B: 1500.0
cumulative distribution function probability density function geometric probability circular region median conditional probability area calculation uniform distribution

A goat \(G\) lies in a square field \(OABC\) of side \(a\). It wanders randomly round its field, so that at any time the probability of its being in any given region is proportional to the area of this region. Write down the probability that its distance, \(R\), from \(O\) is less than \(r\) if \(0 < r\leqslant a,\) and show that if \(r\geqslant a\) the probability is \[ \left(\frac{r^{2}}{a^{2}}-1\right)^{\frac{1}{2}}+\frac{\pi r^{2}}{4a^{2}}-\frac{r^{2}}{a^{2}}\cos^{-1}\left(\frac{a}{r}\right). \] Find the median of \(R\) and probability density function of \(R\). The goat is then tethered to the corner \(O\) by a chain of length \(a\). Find the conditional probability that its distance from the fence \(OC\) is more than \(a/2\).

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1992 Paper 3 Q16
D: 1700.0 B: 1484.0
probability Poisson distribution binomial distribution conditional probability Bayes' theorem compound distribution thinning property

The probability that there are exactly \(n\) misprints in an issue of a newspaper is \(\mathrm{e}^{-\lambda}\lambda^{n}/n!\) where \(\lambda\) is a positive constant. The probability that I spot a particular misprint is \(p\), independent of what happens for other misprints, and \(0 < p < 1.\)

  1. If there are exactly \(m+n\) misprints, what is the probability that I spot exactly \(m\) of them?
  2. Show that, if I spot exactly \(m\) misprints, the probability that I have failed to spot exactly \(n\) misprints is \[ \frac{(1-p)^{n}\lambda^{n}}{n!}\mathrm{e}^{-(1-p)\lambda}. \]

Solution:

  1. \(\binom{m+n}{m} p^m (1-p)^n\)
  2. \(\,\) \begin{align*} \mathbb{P}(\text{failed to spot }n\text{ misprints}|\text{spotted }m\text{ misprints}) &= \frac{\mathbb{P}(\text{failed to spot }n\text{ misprints and spotted }m\text{ misprints}) }{\mathbb{P}(\text{spotted }m\text{ misprints})} \\ &= \frac{\binom{m+n}{n}p^m(1-p)^n e^{-\lambda} \lambda^{m+n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}p^m(1-p)^k e^{-\lambda} \lambda^{m+k}/(n+k)!} \\ &= \frac{\binom{m+n}{n}(1-p)^n \lambda^{n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}(1-p)^k \lambda^{k}/(n+k)!} \\ &= \frac{(1-p)^n \lambda^{n}/n!}{\sum_{k=0}^{\infty} (1-p)^k \lambda^{k}/k!} \\ &= \frac{(1-p)^n\lambda^n}{n!} e^{-(1-p)\lambda} \end{align*} Alternatively, given the missed misprints and spotted misprints are independent, we can view them as both following \(Po(p\lambda)\) and \(Po((1-p)\lambda)\) and so we obtain exactly this result, without calculation.

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