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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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1989 Paper 1 Q15
D: 1500.0 B: 1516.0
Poisson distribution uniform distribution binomial distribution probability conditional Bayes' theorem expectation optimisation route choice

I can choose one of three routes to cycle to school. Via Angle Avenue the distance is 5\(\,\)km, and I am held up at a level crossing for \(A\) minutes, where \(A\) is a continuous random variable uniformly distributed between \(0\) and 10. Via Bend Boulevard the distance is 4\(\,\)km, and I am delayed, by talking to each of \(B\) friends for 3\(\,\)minutes, for a total of \(3B\) minutes, where \(B\) is a random variable whose distribution is Poisson with mean 4. Via Detour Drive the distance should be only 2\(\,\)km, but in addition, due to never-ending road works, there are five places at each of which, with probability \(\frac{4}{5},\) I have to make a detour that increases the distance by 1\(\,\)km. Except when delayed by talking to friends or at the level crossing, I cycle at a steady 12\(\,\)km\(\,\)h\(^{-1}\). For each of the three routs, calculate the probability that a journey lasts at least 27 minutes. Each day I choose one of the three routes at random, and I am equally likely to choose any of the three alternatives. One day I arrive at school after a journey of at least 27 minutes. What is the probability that I came via Bend Boulevard? Which route should I use all the time: \begin{questionparts} \item if I wish my average journey time to be as small as possible; \item if I wish my journey time to be less than 32 minutes as often as possible? \end{questionpart} Justify your answers.

Solution: \(A \sim 5\cdot 5 + U[0,10]\) \(B \sim 4 \cdot 5 + 3 \textrm{Po}(4)\) \(C \sim 2 \cdot 5 + B(5, \frac{4}{5}) \cdot 5\) \begin{align*} && \mathbb{P}(A \leq 27) &= \mathbb{P}(U \leq 2) = 0.2 \\ && \mathbb{P}(B \leq 27) &= \mathbb{P}(3 \textrm{Po}(4) \leq 7) \\\ &&&= \mathbb{P}(Po(4) \leq 2) \\ &&&= e^{-4}(1 + 4 + \frac{4^2}{2}) \\ &&&= 0.23810\ldots \\ && \mathbb{P}(C \leq 27) &= \mathbb{P}(5 \cdot B(5,\tfrac45) \leq 17) \\ &&&= \mathbb{P}(B(5,\tfrac45) \leq 3) \\ &&&= \binom{5}{0} (\tfrac15)^5 + \binom{5}{1} (\tfrac45)(\tfrac 15)^4+ \binom{5}{2} (\tfrac45)^2(\tfrac 15)^3 + \binom{5}3 (\tfrac45)^3(\tfrac 15)^2+\\ &&&= 0.26272 \end{align*} \begin{align*} \mathbb{P}(\text{came via B} | \text{at least 27 minutes}) &= \frac{\mathbb{P}(\text{came via B and at least 27 minutes})}{\mathbb{P}(\text{at least 27 minutes})} \\ &= \frac{\frac13 \cdot 0.23810\ldots }{\frac13 \cdot 0.2 + \frac13 \cdot 0.23810\ldots + \frac13 \cdot 0.26272} \\ &= 0.3397\ldots \\ &= 0.340 \, \, (3\text{ s.f.}) \end{align*}

  1. \begin{align*} \mathbb{E}(A) &= 25 + 5 &= 30 \\ \mathbb{E}(B) &= 20 + 3\cdot 4 &= 32 \\ \mathbb{E}(C) &= 10 + 5 \cdot 4 &= 30 \end{align*} \(A\) and \(C\) are equally good.
  2. \begin{align*} \mathbb{P}(A \leq 32) &= \mathbb{P}(U \leq 7) &= 0.7 \\ \mathbb{P}(B \leq 32) &= \mathbb{P}(Po(4) \leq 4) \\ &= e^{-4}(1 + 4 + 8 + \frac{4^3}{6}) &= 0.4334\ldots \\ \mathbb{P}(C \leq 32) &= \mathbb{P}(B(5,\tfrac45) \leq 4) \\ &= 1 - \mathbb{P}(B(5,\tfrac45) = 5) \\ &= 1 - \frac{4^5}{5^5} &=0.67232 \end{align*} So you should choose route \(A\).

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1987 Paper 3 Q16
D: 1500.0 B: 1500.0
probability uniform distribution order statistics inclusion-exclusion Poisson distribution optimisation binomial distribution limiting probability

  1. \(X_{1},X_{2},\ldots,X_{n}\) are independent identically distributed random variables drawn from a uniform distribution on \([0,1].\) The random variables \(A\) and \(B\) are defined by \[ A=\min(X_{1},\ldots,X_{n}),\qquad B=\max(X_{1},\ldots,X_{n}). \] For any fixed \(k\), such that \(0< k< \frac{1}{2},\) let \[ p_{n}=\mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k). \] What happens to \(p_{n}\) as \(n\rightarrow\infty\)? Comment briefly on this result.
  2. Lord Copper, the celebrated and imperious newspaper proprietor, has decided to run a lottery in which each of the \(4,000,000\) readers of his newspaper will have an equal probability \(p\) of winning \(\pounds 1,000,000\) and their changes of winning will be independent. He has fixed all the details leaving to you, his subordinate, only the task of choosing \(p\). If nobody wins \(\pounds 1,000,000\), you will be sacked, and if more than two readers win \(\pounds 1,000,000,\) you will also be sacked. Explaining your reasoning, show that however you choose \(p,\) you will have less than a 60\% change of keeping your job.

Solution:

  1. \begin{align*} && p_n &= \mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k) \\ &&&= \mathrm{P}(A\leqslant k) +\P(B\geqslant1-k) - \mathrm{P}(A\leqslant k\mbox{ or }B\geqslant1-k)\\ &&&= 1-\mathrm{P}(A\geq k) +1-\P(B \leq 1-k) - \l 1- \mathrm{P}(A\geq k\mbox{ and }B\leq 1-k)\r\\ &&&= 1 - \P(X_i \geq k) - \P(X_i \leq 1-k) + \P(k \leq X_i \leq 1-k) \\ &&&= 1 - k^n - (1-k)^n + (1-2k)^n \end{align*} Therefore as \(n \to \infty\) \(p_n \to 1\), since \(k, (1-k), (1-2k)\) are all between \(0\) and \(1\) and so their powers will tend to \(0\).
  2. Let \(N = 4\,000\,000\). The probability exactly one person wins is \(Np(1-p)^{N-1}\). The probability exactly two people win is \(\binom{N}{2} p^2 (1-p)^{N-2}\). We wish to maximise the sum of these probabilities. To find this maximum, differentiate wrt \(p\). \begin{align*} \frac{\d}{\d p} : && \small N(1-p)^{N-1}-N(N-1)p(1-p)^{N-2} + N(N-1)p(1-p)^{N-2} - \frac12 N(N-1)(N-2)p^2(1-p)^{N-3} \\ &&= N(1-p)^{N-3} \l (1-p)^2 - \frac12(N-1)(N-2)p^2\r \\ \Rightarrow && \frac{(1-p)}{p} = \sqrt{\frac{(N-1)(N-2)}{2}} \\ \Rightarrow && p = \frac{1}{1+ \sqrt{\frac{(N-1)(N-2)}{2}}} \end{align*} This will be a maximum, since this is an increasing function at \(p=0\) and decreasing at \(p=1\) and there's only one stationary point. Note that \(p > \frac{\sqrt{2}}{(N-2)}\) and \(p < \frac{\sqrt{2}}{N-1+\sqrt{2}} < \frac{\sqrt{2}}{N}\) and so: \begin{align*} Np(1-p)^{N-1} &< \sqrt{2}(1-\frac{\sqrt{2}}{N-2})^{N-1} \\ &\approx \sqrt{2} e^{-\sqrt{2}} \end{align*} \begin{align*} \frac{N(N-1)}{2}p^2(1-p)^{N-2} &<(1-\frac{\sqrt{2}}{N-2})^{N-1} \\ &\approx e^{-\sqrt{2}} \end{align*} Alternatively, we can use a Poisson approximation. The number of winners is \(B(N, p)\) where we are hoping \(np\) is small but not zero. Therefore it's reasonable to approximation \(B(N,p)\) by \(Po(Np)\). (Call this value \(\lambda\)). Then we wish to maximise: \begin{align*} && p &= e^{-\lambda} \l \lambda + \frac{\lambda^2}{2} \r \\ &&&= e^{-\lambda} \lambda \l 1+ \frac{\lambda}{2} \r \\ \Rightarrow && \ln p &= -\lambda + \ln \lambda + \ln(1+\frac12 \lambda) \\ \frac{\d}{\d \lambda}: && \frac{p'}{p} &= -1 + \frac{1}{\lambda} + \frac{1}{2+\lambda} \\ &&&= \frac{-(2+\lambda)\lambda+2+2\lambda}{\lambda(2+\lambda)} \\ &&&= \frac{2-\lambda^2}{\lambda(2+\lambda)} \\ \Rightarrow && \lambda &= \sqrt{2} \end{align*} \begin{align*} \frac{\sqrt{2}+1}{e^{\sqrt{2}}} &< \frac{\sqrt{2}+1}{1+\sqrt{2}+1+\frac{1}{3}\sqrt{2}+\frac{1}{6}} \\ &= \frac{30\sqrt{2}-18}{41} \end{align*} Either way, we find we want to estimate \(e^{-\sqrt{2}}(1+\sqrt{2})\)

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