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2011 Paper 1 Q11
D: 1516.0 B: 1516.0

A thin non-uniform bar \(AB\) of length \(7d\) has centre of mass at a point \(G\), where \(AG=3d\). A light inextensible string has one end attached to \(A\) and the other end attached to \(B\). The string is hung over a smooth peg \(P\) and the bar hangs freely in equilibrium with \(B\) lower than~\(A\). Show that \[ 3\sin\alpha = 4\sin\beta\,, \] where \(\alpha\) and \(\beta\) are the angles \(PAB\) and \(PBA\), respectively. Given that \(\cos\beta=\frac45\) and that \(\alpha\) is acute, find in terms of \(d\) the length of the string and show that the angle of inclination of the bar to the horizontal is \(\arctan \frac17\,\).

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2011 Paper 1 Q12
D: 1500.0 B: 1470.2

I am selling raffle tickets for \(\pounds1\) per ticket. In the queue for tickets, there are \(m\) people each with a single \(\pounds1\) coin and \(n\) people each with a single \(\pounds2\) coin. Each person in the queue wants to buy a single raffle ticket and each arrangement of people in the queue is equally likely to occur. Initially, I have no coins and a large supply of tickets. I stop selling tickets if I cannot give the required change.

  1. In the case \(n=1\) and \(m\ge1\), find the probability that I am able to sell one ticket to each person in the queue.
  2. By considering the first three people in the queue, show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=2\) and \(m\ge2\) is \(\dfrac{m-1}{m+1}\,\).
  3. Show that the probability that I am able to sell one ticket to each person in the queue in the case \(n=3\) and \(m\ge3\) is \(\dfrac{m-2}{m+1}\,\).

Solution:

  1. The only way you wont be able to sell to them is if they are first, ie \(\frac1{m+1}\)
  2. If \(n=2\), the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie \(2\) or \(122\). These have probabilities \(\frac{2}{m+2}\) and \(\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}\). Therefore the total probability you don't sell all the tickets is \(\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}\). Therefore the probability you do sell all the tickets is \(1 - \frac{2}{m+1} = \frac{m-1}{m+1}\)
  3. The only ways to fail when \(n=3\) are: \(2\), \(122\), or if all three \(2\)s appear before three \(1\)s. this can happen in \(11222\), \(12122\) These happen with probability: \begin{align*} 2: && \frac{3}{m+3} \\ 122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\ 11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ 12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ \end{align*} Therefore the total probability is: \begin{align*} P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\ &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\ &= \frac{3}{m+1} \end{align*} and the result follows

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2011 Paper 1 Q13
D: 1484.0 B: 1471.5

In this question, you may use without proof the following result: \[ \int \sqrt{4-x^2}\, \d x = 2 \arcsin (\tfrac12 x ) + \tfrac 12 x \sqrt{4-x^2} +c\,. \] A random variable \(X\) has probability density function \(\f\) given by \[ \f(x) = \begin{cases} 2k & -a\le x <0 \\[3mm] k\sqrt{4-x^2} & \phantom{-} 0\le x \le 2 \\[3mm] 0 & \phantom{-}\text{otherwise}, \end{cases} \] where \(k\) and \(a\) are positive constants.

  1. Find, in terms of \(a\), the mean of \(X\).
  2. Let \(d\) be the value of \(X\) such that \(\P(X> d)=\frac1 {10}\,\). Show that \(d < 0\) if \(2a> 9\pi\) and find an expression for \(d\) in terms of \(a\) in this case.
  3. Given that \(d=\sqrt 2\), find \(a\).

Solution: First notice that \begin{align*} && 1 &= \int_{-a}^2 f(x) \d x \\ &&&= 2ka + k\pi \\ \Rightarrow && k &= (\pi + 2a)^{-1} \end{align*}

  1. \(\,\) \begin{align*} && \E[X] &= \int_{-a}^2 x f(x) \d x\\ &&&= \int_{-a}^0 2kx \d x + k\int_0^{2} x\sqrt{4-x^2} \d x\\ &&&= \left [kx^2 \right]_{-a}^0 +k \left [-\frac13(4-x^2)^{\frac32} \right]_0^2 \\ &&&= -ka^2 + \frac83k \\ &&&= \frac{\frac83-a^2}{\pi + 2a} \end{align*}
  2. Consider \(\mathbb{P}(X < 0)\) then \(d < 0 \Leftrightarrow \mathbb{P}(X < 0) > \frac{9}{10}\) \begin{align*} && \frac{9}{10} &< \mathbb{P}(X < 0) \\ &&&= \int_{-a}^0 2k \d x \\ &&&= \frac{2a}{\pi+2a} \\ \Leftrightarrow && 9\pi &< 2a \\ \\ && \frac{9}{10} &= \int_{-a}^d 2k \d x \\ &&&= \frac{2(d+a)}{\pi + 2a} \\ \Rightarrow && 9\pi &= 2a + 20d \\ \Rightarrow && d &= \frac{2a-9\pi}{20} \end{align*}
  3. Suppose \(d=\sqrt 2\) then \begin{align*} && \frac1{10} &= \int_{\sqrt{2}}^2 f(x) \d x \\ &&&= \int_{\sqrt{2}}^2 k\sqrt{4-x^2} \d x \\ &&&= k\left [ 2 \sin^{-1} \tfrac12 x + \tfrac12 x \sqrt{4-x^2}\right]_{\sqrt{2}}^2 \\ &&&= k\left (\pi -\frac{\pi}{2} - 1 \right) \\ \Rightarrow && \pi + 2a &= 5\pi - 10 \\ \Rightarrow && a &= 2\pi-5 \end{align*}

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