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1994 Paper 1 Q11
D: 1500.0 B: 1469.5

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The diagram shows a small railway wagon \(A\) of mass \(m\) standing at the bottom of a smooth railway track of length \(d\) inclined at an angle \(\theta\) to the horizontal. A light inextensible string, also of length \(d\), is connected to the wagon and passes over a light frictionless pulley at the top of the incline. On the other end of the string is a ball \(B\) of mass \(M\) which hangs freely. The system is initially at rest and is then released.
  1. Find the condition which \(m,M\) and \(\theta\) must satisfy to ensure that the ball will fall to the ground. Assuming that this condition is satisfied, show that the velocity \(v\) of the ball when it hits the ground satisfies \[ v^{2}=\frac{2g(M-m\sin\theta)d\sin\theta}{M+m}. \]
  2. Find the condition which \(m,M\) and \(\theta\) must satisfy if the wagon is not to collide with the pulley at the top of the incline.

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1994 Paper 1 Q12
D: 1500.0 B: 1468.0

There are 28 colleges in Cambridge, of which two (New Hall and Newnham) are for women only; the others admit both men and women. Seven women, Anya, Betty, Celia, Doreen, Emily, Fariza and Georgina, are all applying to Cambridge. Each has picked three colleges at random to enter on her application form.

  1. What is the probability that Anya's first choice college is single-sex?
  2. What is the probability that Betty has picked Newnham?
  3. What is the probability that Celia has picked at least one single-sex college?
  4. Doreen's first choice is Newnham. What is the probability that one of her other two choices is New Hall?
  5. Emily has picked Newnham. What is the probability that she has also picked New Hall?
  6. Fariza's first choice college is single-sex. What is the probability that she has also chosen the other single-sex college?
  7. One of Georgina's choices is a single-sex college. What is the probability that she has also picked the other single-sex college?

Solution:

  1. \(\frac{2}{28} = \frac{1}{14}\)
  2. \(1-\frac{27}{28}\frac{26}{27}\frac{25}{26} = \frac{3}{28}\)
  3. \(1 - \frac{26}{28}\frac{25}{27}\frac{24}{26} = \frac{13}{63}\)
  4. \(1-\frac{26}{27}\frac{25}{26} = \frac{2}{27}\)
  5. \(\frac{1}{27}\)
  6. There are \(\binom{2}{1} \binom{26}{2} + \binom{2}{2}\binom{26}{1}\) ways to choose at least one single sex college and \( \binom{2}{2}\binom{26}{1}\) ways to choose both, therefore \begin{align*} P &= \frac{ \binom{2}{2}\binom{26}{1}}{\binom21 \binom{26}2+ \binom{2}{2}\binom{26}{1}} \\ &= \frac{26}{2\cdot \frac{26\cdot25}{2}+26 }\\ &= \frac{1}{25+1} = \frac{1}{26} \end{align*}

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1994 Paper 1 Q13
D: 1500.0 B: 1512.0

I have a bag containing \(M\) tokens, \(m\) of which are red. I remove \(n\) tokens from the bag at random without replacement. Let \[ X_{i}=\begin{cases} 1 & \mbox{ if the ith token I remove is red;}\\ 0 & \mbox{ otherwise.} \end{cases} \] Let \(X\) be the total number of red tokens I remove.

  1. Explain briefly why \(X=X_{1}+X_{2}+\cdots+X_{n}.\)
  2. Find the expectation \(\mathrm{E(}X_{i}).\)
  3. Show that \(\mathrm{E}(X)=mn/M\).
  4. Find \(\mathrm{P}(X=k)\) for \(k=0,1,2,\ldots,n\).
  5. Deduce that \[ \sum_{k=1}^{n}k\binom{m}{k}\binom{M-m}{n-k}=m\binom{M-1}{n-1}. \]

Solution:

  1. The left hand side counts the number of red tokens we have taken. The right hand side counts the number of red tokens we have taken at each point, across all points. Therefore these must be the same.
  2. \(\E[X_i] = \mathbb{P}(i\text{th token is red}) = \frac{m}{M}\) (since there is nothing special about the \(i\)th token.
  3. Therefore \(\E[X] = \E[X_1 + \cdots + X_n] = n\E[X_i] = \frac{nm}{M}\)
  4. \(\mathbb{P}(X=k) = \binom{m}{k}\binom{M-m}{n-k}/\binom{M}{n}\) since this is the number of ways we can choose \(k\) of the \(m\) red objects, \(n-k\) of the \(M-m\) non-red objects divided by the total number of ways we can choose our \(n\) tokens.
  5. \(\,\) \begin{align*} && \frac{mn}{M} &= \E[X] \\ &&&= \sum_{k=1}^n k \mathbb{P}(X=k) \\ &&&= \sum_{k=1}^n k \binom{m}{k}\binom{M-m}{n-k}/\binom{M}{n} \\ \Rightarrow && \sum_{k=1}^n k \binom{m}{k}\binom{M-m}{n-k} &= m \frac{n}{M} \binom{M}{n} = m \binom{M-1}{n-1} \end{align*}
This question is a nice example of how to find the mean of the hypergeometric distribution

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1994 Paper 1 Q14
D: 1500.0 B: 1532.7

Each of my \(n\) students has to hand in an essay to me. Let \(T_{i}\) be the time at which the \(i\)th essay is handed in and suppose that \(T_{1},T_{2},\ldots,T_{n}\) are independent, each with probability density function \(\lambda\mathrm{e}^{-\lambda t}\) (\(t\geqslant0\)). Let \(T\) be the time I receive the first essay to be handed in and let \(U\) be the time I receive the last one.

  1. Find the mean and variance of \(T_{i}.\)
  2. Show that \(\mathrm{P}(U\leqslant u)=(1-\mathrm{e}^{-\lambda u})^{n}\) for \(u\geqslant0,\) and hence find the probability density function of \(U\).
  3. Obtain \(\mathrm{P}(T>t),\) and hence find the probability density function of \(T\).
  4. Write down the mean and variance of \(T\).

Solution:

  1. \(T_i \sim \textrm{Exp}(\lambda)\) so \(\E[T_i] = \lambda^{-1}, \var[T_i] = \lambda^{-2}\)
  2. \(\,\) \begin{align*} && \mathbb{P}(U \leq u) &= \mathbb{P}(T_i \leq u\quad \forall i) \\ &&&= \prod \mathbb{P}(T_i \leq u) \\ &&&= \prod \int_0^u \lambda e^{-\lambda t} \d t \\ &&&= (1-e^{-\lambda u})^n \\ \\ \Rightarrow && f_U(u) &= n\lambda e^{-\lambda u}(1-e^{-\lambda u})^{n-1} \end{align*}
  3. \(\,\) \begin{align*} && \mathbb{P}(T > t) &= \mathbb{P}(T_i > t \quad \forall i) \\ &&&= \prod \mathbb{P}(T_i > t) \\ &&&= e^{-n\lambda t} \\ \Rightarrow && f_T(t) &= n\lambda e^{-n\lambda t} \end{align*}
  4. Therefore \(\E[T] = \frac{1}{n\lambda}, \var[T] = \frac{1}{(n\lambda)^2}\)

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