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1999 Paper 3 Q11
D: 1700.0 B: 1500.0

Calculate the moment of inertia of a uniform thin circular hoop of mass \(m\) and radius \(a\) about an axis perpendicular to the plane of the hoop through a point on its circumference. The hoop, which is rough, rolls with speed \(v\) on a rough horizontal table straight towards the edge and rolls over the edge without initially losing contact with the edge. Show that the hoop will lose contact with the edge when it has rotated about the edge of the table through an angle \(\theta\), where \[ \cos\theta = \frac 12 +\frac {v^2}{2ag}. \] %Give the corresponding result for a smooth hoop and table.

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1999 Paper 3 Q12
D: 1700.0 B: 1500.0

In the game of endless cricket the scores \(X\) and \(Y\) of the two sides are such that \[ \P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\] for some positive constant \(\lambda\), where \(j,k = 0\), \(1\), \(2\), \(\ldots\).

  1. Find \(\P(X+Y=n)\) for each \(n>0\).
  2. Show that \(2\lambda \e^{2\lambda-1}=1\).
  3. Show that \(2x \e^{2x-1}\) is an increasing function of \(x\) for \(x>0\) and deduce that the equation in (ii) has at most one solution and hence determine \(\lambda\).
  4. Calculate the expectation \(\E(2^{X+Y})\).

Solution:

  1. \begin{align*} && \mathbb{P}(X+Y = n) &= \sum_{i = 0}^n \mathbb{P}(X = i, Y = n-i) \\ &&&= \sum_{i = 0}^n e^{-1} \frac{n \lambda^n}{i! (n-i)!} \\ &&&=e^{-1} n \lambda^n \sum_{i = 0}^n\frac{1}{i! (n-i)!} \\ &&&=\frac{e^{-1} n}{n!} \lambda^n \sum_{i = 0}^n\frac{n!}{i! (n-i)!} \\ &&&= \frac{n\lambda^n}{e n!} 2^n \\ &&&= \frac{n (2 \lambda)^n}{e \cdot n!} \end{align*}
  2. \begin{align*} && 1 &= \sum_{n = 0}^{\infty} \mathbb{P}(X+Y =n ) \\ &&&= \sum_{n = 0}^{\infty}\frac{n (2 \lambda)^n}{e \cdot n!} \\ &&&= \sum_{n = 1}^\infty \frac{ (2 \lambda)^n}{e \cdot (n-1)!} \\ &&&= \frac{2 \lambda}{e}\sum_{n = 0}^\infty \frac{ (2 \lambda)^n}{n!} \\ &&&= \frac{2 \lambda}{e} e^{2\lambda} \\ &&&= 2 \lambda e^{2\lambda - 1} \end{align*} \\
  3. Consider \(f(x) = 2xe^{2x-1}\), then \begin{align*} && f'(x) &= 2e^{2x-1} + 2xe^{2x-1} \cdot 2 \\ &&&= e^{2x-1} (2 + 4x) > 0 \end{align*} Therefore \(f(x)\) is an increasing function of \(x\), which means \(f(x) = 1\) has at most one solution for \(\lambda\). Therefore \(\lambda = \frac12\)
  4. \begin{align*} \mathbb{E}(2^{X+Y}) &= \sum_{n = 0}^\infty \mathbb{P}(X+Y = n) 2^n \\ &= \sum_{n = 1}^\infty \frac{1}{e(n-1)!} 2^{n} \\ &= \frac{2}{e} \sum_{n=0}^\infty \frac{2^n}{n!} \\ &= \frac{2}{e} e^2 \\ &= 2e \end{align*}

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1999 Paper 3 Q13
D: 1700.0 B: 1484.0

The cakes in our canteen each contain exactly four currants, each currant being randomly placed in the cake. I take a proportion \(X\) of a cake where \(X\) is a random variable with density function \[{\mathrm f}(x)=Ax\] for \(0\leqslant x\leqslant 1\) where \(A\) is a constant.

  1. What is the expected number of currants in my portion?
  2. If I find all four currants in my portion, what is the probability that I took more than half the cake?

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1999 Paper 3 Q14
D: 1700.0 B: 1487.9

In the basic version of Horizons (H1) the player has a maximum of \(n\) turns, where \(n \ge 1\). At each turn, she has a probability \(p\) of success, where \(0 < p < 1\). If her first success is at the \(r\)th turn, where \(1 \le r \le n\), she collects \(r\) pounds and then withdraws from the game. Otherwise, her winnings are nil. Show that in H1, her expected winnings are $$ p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds}, $$ where \(q=1-p\). The rules of H2 are the same as those of H1, except that \(n\) is randomly selected from a Poisson distribution with parameter \(\lambda\). If \(n=0\) her winnings are nil. Otherwise she plays H1 with the selected \(n\). Show that in H2, her expected winnings are $$ {1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)} -{{\lambda}q}{\e^{-{\lambda}p}} \quad\hbox{pounds}. $$

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