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2019 Paper 3 Q11
D: 1500.0 B: 1500.0

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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2019 Paper 3 Q12
D: 1500.0 B: 1485.6

The set \(S\) is the set of all integers from 1 to \(n\). The set \(T\) is the set of all distinct subsets of \(S\), including the empty set \(\emptyset\) and \(S\) itself. Show that \(T\) contains exactly \(2^n\) sets. The sets \(A_1, A_2, \ldots, A_m\), which are not necessarily distinct, are chosen randomly and independently from \(T\), and for each \(k\) \((1 \leq k \leq m)\), the set \(A_k\) is equally likely to be any of the sets in \(T\).

  1. Write down the value of \(P(1 \in A_1)\).
  2. By considering each integer separately, show that \(P(A_1 \cap A_2 = \emptyset) = \left(\frac{3}{4}\right)^n\). Find \(P(A_1 \cap A_2 \cap A_3 = \emptyset)\) and \(P(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset)\).
  3. Find \(P(A_1 \subseteq A_2)\), \(P(A_1 \subseteq A_2 \subseteq A_3)\) and \(P(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m)\).

Solution: For every element in \(S\) we can choose whether or not it appears in a subset of \(S\), therefore there are \(2^n\) choices so \(2^n\) distinct subsets.

  1. \(\mathbb{P}(1 \in A_1) = \frac12\) (since \(1\) is in exactly half the subsets)
  2. \(\,\) \begin{align*} && \mathbb{P}(A_1 \cap A_2 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2) \forall i) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12\right) \\ &&&= \left (\frac34 \right)^n \end{align*}
  3. \(\,\) \begin{align*} && \mathbb{P}(A_1 \cap A_2 \cap A_3 = \emptyset) &= \mathbb{P}(i \not \in (A_1 \cap A_2 \cap A_3) \forall i) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1 \cap A_2 \cap A_3) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\mathbb{P}(i \in A_1)\mathbb{P}(i \in \cap A_2))\mathbb{P}(i \in \cap A_3) \right) \\ &&&= \prod_{i=1}^n \left ( 1-\frac12 \cdot \frac12 \cdot \frac12\right) \\ &&&= \left (\frac78 \right)^n \end{align*} Similarly, \(\displaystyle \mathbb{P}(A_1 \cap A_2 \cap \cdots \cap A_m = \emptyset) = \left ( \frac{2^m-1}{2^m} \right)^n\)
  4. \(\,\) \begin{align*} && \mathbb{P}(A_1 \subseteq A_2) &= \mathbb{P}(A_1 \cap A_2^c = \emptyset) \\ &&&= \left (\frac34 \right)^n \\ \\ && \mathbb{P}(A_1 \subseteq A_2 \subseteq A_3) &= \prod_{i=1}^n \mathbb{P}(\text{once }i\text{ appears it keeps appearing}) \\ &&&= \prod_{i=1}^n \frac{\#\{(0,0,0), (0,0,1), (0,1,1), (1,1,1) \}}{2^3} \\ &&&= \prod_{i=1}^n \frac{4}{8} \\ &&&= \frac{1}{2^n} \\ \\ && \mathbb{P}(A_1 \subseteq A_2 \subseteq \cdots \subseteq A_m) &= \prod_{i=1}^n \frac{m+1}{2^m} \\ &&&= \left ( \frac{m+1}{2^m} \right)^n \end{align*}

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