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2007 Paper 3 Q14
D: 1700.0 B: 1500.0
geometric probability order statistics expected value uniform distribution CDF integration dartboard smallest enclosing region

  1. My favourite dartboard is a disc of unit radius and centre \(O\). I never miss the board, and the probability of my hitting any given area of the dartboard is proportional to the area. Each throw is independent of any other throw. I throw a dart \(n\) times (where \(n>1\)). Find the expected area of the smallest circle, with centre \(O\), that encloses all the \(n\) holes made by my dart. Find also the expected area of the smallest circle, with centre \(O\), that encloses all the \((n-1)\) holes nearest to \(O\).
  2. My other dartboard is a square of side 2 units, with centre \(Q\). I never miss the board, and the probability of my hitting any given area of the dartboard is proportional to the area. Each throw is independent of any other throw. I throw a dart \(n\) times (where \(n>1\)). Find the expected area of the smallest square, with centre \(Q\), that encloses all the \(n\) holes made by my dart.
  3. Determine, without detailed calculations, whether the expected area of the smallest circle, with centre \(Q\), on my square dartboard that encloses all the \(n\) holes made by my darts is larger or smaller than that for my circular dartboard.

Solution:

  1. Firstly, we consider the probability that all darts lie within a distance \(s\) from the centre, ie \begin{align*} \mathbb{P}(\text{all darts within }s) &= \prod_{k=1}^s \mathbb{P}(\text{dart within }s) \\ &= \left ( \frac{\pi s^2}{\pi} \right)^n \\ &= s^{2n} \end{align*} Therefore the pdf is \(2ns^{2n-1}\), and the expected area is \(\int_{s=0}^1 \pi s^2 \cdot 2n s^{2n-1} \d s = 2n \pi \frac{1}{2n+2} = \frac{n}{n+1} \pi\). \begin{align*} \mathbb{P}(\text{n-1 within }s) &= \underbrace{s^{2n}}_{\text{all within }s} + \underbrace{ns^{2n-2}(1-s^2)}_{\text{all but 1 within }s}\\ &= ns^{2n-2}-(n-1)s^{2n} \end{align*} Therefore the pdf is \(n(2n-2)s^{2n-3} - 2n(n-1)s^{2n-1} = 2n(n-1)(s^{2n-3}-s^{2n-1})\) and the expected area is: \begin{align*} \int \pi s^2 \cdot2n(n-1)(s^{2n-3}-s^{2n-1})\d s &= 2n(n-1) \pi \left ( \frac{1}{2n} - \frac{1}{2n+2} \right) \\ &= n(n-1)\pi \frac{2}{n(n+1)} \\ &= \frac{n-1}{n+1} \pi \end{align*}
  2. Now consider a square of side-length \(s\), we must have \(\mathbb{P}(\text{all darts within square}) = \left ( \frac{s^2}{4} \right)^n\) and therefore the pdf is \(n \frac{s^{n-1}}{4^n}\). Therefore the expected area is \(\displaystyle \int_0^2 s^2 \cdot n \frac{s^{n-1}}{4^n} \d s = \frac{n}{n+1} \frac{2^{2n+1}}{2^{2n}} = \frac{4n}{n+1}\)
  3. It is clearly larger as the square dartboard contains all of the circular dartboard, and there will be some probability that the darts land outside the circular dartboard, making the circle much larger.

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2002 Paper 3 Q13
D: 1700.0 B: 1516.0
exponential distribution probability minimum of random variables CDF independence median memoryless property continuous random variable

A continuous random variable is said to have an exponential distribution with parameter \(\lambda\) if its density function is \(\f(t) = \lambda \e ^{- \lambda t} \; \l 0 \le t < \infty \r\,\). If \(X_1\) and \(X_2\), which are independent random variables, have exponential distributions with parameters \(\lambda_1\) and \(\lambda_2\) respectively, find an expression for the probability that either \(X_1\) or \(X_2\) (or both) is less than \(x\). Prove that if \(X\) is the random variable whose value is the lesser of the values of \(X_1\) and \(X_2\), then \(X\) also has an exponential distribution. Route A and Route B buses run from my house to my college. The time between buses on each route has an exponential distribution and the mean time between buses is 15 minutes for Route A and 30 minutes for Route B. The timings of the buses on the two routes are independent. If I emerge from my house one day to see a Route A bus and a Route B bus just leaving the stop, show that the median wait for the next bus to my college will be approximately 7 minutes.

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1994 Paper 1 Q14
D: 1500.0 B: 1532.7
probability continuous probability distributions exponential distribution order statistics CDF PDF mean and variance independence

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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