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2019 Paper 3 Q11
D: 1500.0 B: 1500.0
Poisson distribution probability thinning property expected value geometric series optimisation conditional probability calculus optimisation

The number of customers arriving at a builders' merchants each day follows a Poisson distribution with mean \(\lambda\). Each customer is offered some free sand. The probability of any given customer taking the free sand is \(p\).

  1. Show that the number of customers each day who take sand follows a Poisson distribution with mean \(p\lambda\).
  2. The merchant has a mass \(S\) of sand at the beginning of the day. Each customer who takes the free sand gets a proportion \(k\) of the remaining sand, where \(0 \leq k < 1\). Show that by the end of the day the expected mass of sand taken is $$\left(1 - e^{-kp\lambda}\right)S.$$
  3. At the beginning of the day, the merchant's bag of sand contains a large number of grains, exactly one of which is made from solid gold. At the end of the day, the merchant's assistant takes a proportion \(k\) of the remaining sand. Find the probability that the assistant takes the golden grain. Comment on the case \(k = 0\) and on the limit \(k \to 1\). In the case \(p\lambda > 1\) find the value of \(k\) which maximises the probability that the assistant takes the golden grain.

Solution:

  1. Let \(X\) be the number of people arriving on a given day, and \(Y\) be the number taking sand, then \begin{align*} && \mathbb{P}(Y = k) &= \sum_{x=k}^{\infty} \mathbb{P}(x \text{ arrive and }k\text{ of them take sand}) \\ &&&= \sum_{x=k}^{\infty} \mathbb{P}(X=x)\mathbb{P}(k \text{ out of }x\text{ of them take sand})\\ &&&= \sum_{x=k}^{\infty} e^{-\lambda} \frac{\lambda^x}{x!}\binom{x}{k}p^k(1-p)^{x-k}\\ &&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \sum_{x=k}^{\infty} \frac{((1-p)\lambda)^x}{k!(x-k)!} \\ &&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \frac{((1-p)\lambda)^k}{k!} \sum_{x=0}^{\infty} \frac{((1-p)\lambda)^x}{x!} \\ &&&= e^{-\lambda} \left ( \frac{p}{1-p} \right)^k \frac{((1-p)\lambda)^k}{k!}e^{(1-p)\lambda)} \\ &&&= e^{-p\lambda} \frac{(p\lambda)^k}{k!} \end{align*} which is precisely a Poisson with parameter \(p\lambda\). Alternatively, \(Y = B_1 + B_2 + \cdots + B_X\) where \(B_i \sim Bernoulli(p)\) so \(G_Y(t) = G_X(G_B(t)) = G_X(1-p+pt) = e^{-\lambda(1-(1-p+pt))} = e^{-p\lambda(1-t)}\) so \(Y \sim Po(\lambda)\) Alternatively, alternatively, let \(Z\) be the number of people not taking sand, so \begin{align*} && \mathbb{P}(Y = y, Z= z) &= \mathbb{P}(X=y+z) \cdot \binom{y+z}{y} p^y(1-p)^z \\ &&&= e^{-\lambda} \frac{\lambda^{y+z}}{(y+z)!} \frac{(y+z)!}{y!z!} p^y(1-p)^z \\ &&&=\left ( e^{-p\lambda} \frac{(p\lambda)^y}{y!} \right) \cdot \left ( e^{-(1-p)\lambda} \frac{((1-p)\lambda)^z}{z!}\right) \end{align*} So clearly \(Y\) and \(Z\) are both (independent!) Poisson with parameters \(p\lambda \) and \((1-p)\lambda\)
  2. The amount taken is \(Sk + S(1-k)k + \cdots +Sk(1-k)^{Y-1} = Sk\cdot \frac{1-(1-k)^Y}{k} = S(1-(1-k)^Y)\) so \begin{align*} \E[\text{taken sand}] &= \E \left [ S(1-(1-k)^Y)\right] \\ &= S-S\E\left [(1-k)^Y \right] \\ &= S - SG_Y(1-k)\\ &=S - Se^{-p\lambda(1-(1-k))} \tag{pgf for Poisson} \\ &= S\left (1-e^{-kp\lambda} \right) \end{align*}
  3. The fraction of grains the assistant takes home is: \((1-k)^Yk\), which has expected value \(ke^{-kp\lambda}\). This the the probability he takes home the golden grain. When \(k = 0\) the probability is \(0\) which makes sense (no-one takes home any sand, including the merchant's assistant). As \(k \to 1\) we get \(e^{-p\lambda}\) which is the probability that no-one gets any sand other than him. \begin{align*} && \frac{\d }{\d k} \left ( ke^{-kp\lambda} \right) &= e^{-kp\lambda} - (p\lambda)ke^{-kp\lambda} \\ &&&= e^{-kp\lambda}(1 - (p\lambda)k) \end{align*} Therefore maximised at \(k = \frac{1}{p\lambda}\). (Clearly this is a maximum just by sketching the function)

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1995 Paper 1 Q13
D: 1500.0 B: 1484.0
Poisson distribution thinning property probability generating function conditional probability compound distribution order statistics summation

A scientist is checking a sequence of microscope slides for cancerous cells, marking each cancerous cell that she detects with a red dye. The number of cancerous cells on a slide is random and has a Poisson distribution with mean \(\mu.\) The probability that the scientist spots any one cancerous cell is \(p\), and is independent of the probability that she spots any other one.

  1. Show that the number of cancerous cells which she marks on a single slide has a Poisson distribution of mean \(p\mu.\)
  2. Show that the probability \(Q\) that the second cancerous cell which she marks is on the \(k\)th slide is given by \[ Q=\mathrm{e}^{-\mu p(k-1)}\left\{ (1+k\mu p)(1-\mathrm{e}^{-\mu p})-\mu p\right\} . \]

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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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