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2018 Paper 1 Q11
D: 1500.0 B: 1513.7

A bag contains three coins. The probabilities of their showing heads when tossed are \(p_1\), \(p_2\) and \(p_3\).

  1. A coin is taken at random from the bag and tossed. What is the probability that it shows a head?
  2. A coin is taken at random from the bag (containing three coins) and tossed; the coin is returned to the bag and again a coin is taken at random from the bag and tossed. Let \(N_1\) be the random variable whose value is the number of heads shown on the two tosses. Find the expectation of \(N_1\) in terms of \(p\), where \(p = \frac{1}{3}(p_1+p_2+p_3)\,\), and show that \(\var(N_1) =2p(1-p)\,\).
  3. Two of the coins are taken at random from the bag (containing three coins) and tossed. Let \(N_2\) be the random variable whose value is the number of heads showing on the two coins. Find \(\E(N_2)\) and \(\var(N_2)\).
  4. Show that \(\var(N_2)\le \var(N_1)\), with equality if and only if \(p_1=p_2=p_3\,\).

Solution:

  1. \(\mathbb{P}(\text{head}) = \mathbb{P}(\text{head}|1)\mathbb{P}(\text{coin 1}) + \mathbb{P}(\text{head}|2)\mathbb{P}(\text{coin 2})+\mathbb{P}(\text{head}|3)\mathbb{P}(\text{coin 3}) = \frac13(p_1+p_2+p_3)\)
  2. \(N_1 = X_1 + X_2\) where \(X_i \sim Bernoulli(p)\), therefore \(\mathbb{E}(N_1) = 2p\) and \(\textrm{Var}(N_1) = \textrm{Var}(X_1)+ \textrm{Var}(X_2) = p(1-p)+p(1-p) = 2p(1-p)\)
  3. Let \(Y_i\) be the indicator for the \(i\)th coin is heads. Then \(\mathbb{E}(Y_i) = p\) and so \(\mathbb{E}(N_2) = 2p\). \begin{align*} && \textrm{Var}(N_2) &= \mathbb{E}(N_2^2) - [\mathbb{E}(N_2)]^2\\ &&&= 2^2 \cdot \left (\frac13 \left (p_1p_2+p_2p_3+p_3p_1 \right) \right) + 1 \cdot \left (\frac13 \left (p_1 (1-p_2) + (1-p_1)p_2 + p_2(1-p_3) +(1-p_2)p_3 + p_3(1-p_1) + (1-p_3)p_1 \right) \right) - [\mathbb{E}(N_2)]^2 \\ &&&= \frac43\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac13 \left ( 2(p_1+p_2+p_3) - 2(p_1p_2+p_2p_3+p_3p_1)\right)-[\mathbb{E}(N_2)]^2 \\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac23 \left ( p_1+p_2+p_3 \right)-[\mathbb{E}(N_2)]^2\\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) + \frac23 \left ( p_1+p_2+p_3 \right)-\left[\frac23(p_1+p_2+p_3)\right]^2\\ &&&= \frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) +2p(1-2p)\\ \end{align*}
  4. \(\,\) \begin{align*} && \textrm{Var}(N_1) - \textrm{Var}(N_2) &= 2p(1-p) - \left (\frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) +2p(1-2p) \right) \\ &&&= 2p^2-\frac23\left (p_1p_2+p_2p_3+p_3p_1 \right) \\ &&&= \frac23 \left ( \frac13(p_1+p_2+p_3)^2 -\left (p_1p_2+p_2p_3+p_3p_1 \right)\right)\\ &&&= \frac29 \left (p_1^2+p_2^2+p_3^2 -(p_1p_2+p_2p_3+p_3p_1) \right)\\ &&&= \frac19 \left ((p_1-p_2)^2+(p_2-p_3)^2+(p_3-p_1)^2 \right) &\geq 0 \end{align*} with equality iff \(p_1 = p_2 = p_3\)

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2018 Paper 1 Q12
D: 1500.0 B: 1500.0

A multiple-choice test consists of five questions. For each question, \(n\) answers are given (\(n\ge2\)) only one of which is correct and candidates either attempt the question by choosing one of the \(n\) given answers or do not attempt it. For each question attempted, candidates receive two marks for the correct answer and lose one mark for an incorrect answer. No marks are gained or lost for questions that are not attempted. The pass mark is five. Candidates A, B and C don't understand any of the questions so, for any question which they attempt, they each choose one of the \(n\) given answers at random, independently of their choices for any other question.

  1. Candidate A chooses in advance to attempt exactly \(k\) of the five questions, where \(k=0, 1, 2, 3, 4\) or \(5\). Show that, in order to have the greatest probability of passing the test, she should choose \(k=4\,\).
  2. Candidate B chooses at random the number of questions he will attempt, the six possibilities being equally likely. Given that Candidate B passed the test find, in terms of \(n\), the probability that he attempted exactly four questions. [Not on original test: Show that this probability is an increasing function of \(n\).]
  3. For each of the five questions Candidate C decides whether to attempt the question by tossing a biased coin. The coin has a probability of \(\frac n{n+1}\) of showing a head, and she attempts the question if it shows a head. Find the probability, in terms of \(n\), that Candidate C passes the test.

Solution:

  1. Her probability of passing if she answers \(k \leq 2\) is \(0\), since she can attain at most \(4\) marks. If she attempts \(3\) questions, she needs to get all of them right, hence \(\mathbb{P}(\text{gets all }3\text{ correct}) = \frac{1}{n^3}\). If she attempts \(4\) questions, we can afford to get one wrong \begin{align*} && \mathbb{P}(\text{passes}|\text{attempts }4) &=\mathbb{P}(4/4) +\mathbb{P}(3/4) \\ &&&= \frac{1}{n^4} + 4\cdot\frac{1}{n^3} \cdot \frac{n-1}{n} \\ &&&= \frac{4n-3}{n^4} \end{align*} If she attempts \(5\) questions she can get \(5\) right (10), \(4\) right, \(1\) wrong (7), but \(3\) right will not work (\(6 - 2 = 4< 5\)), hence: \begin{align*} && \mathbb{P}(\text{passes}|\text{attempts }5) &=\mathbb{P}(5/5) +\mathbb{P}(4/5) \\ &&&= \frac{1}{n^5} + 5\cdot\frac{1}{n^4} \cdot \frac{n-1}{n} \\ &&&= \frac{5n-4}{n^5} \end{align*} If \(4n-3 > n \Leftrightarrow n \geq 2\) then \(4\) attempts is better than \(3\). If \(4n^2-3n > 5n-4 \Leftrightarrow 4n^2-8n+4 = 4(n-1)^2 > 0 \Leftrightarrow n \geq\) then \(4\) is better than \(5\), but \(n\) is \(\geq 2\) so, \(4\) is the best option.
  2. \(\,\) \begin{align*} && \mathbb{P}(\text{passes}) &= \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot \frac1{n^3} + \frac16 \frac{4n-3}{n^4} + \frac16 \frac{5n-4}{n^5} \\ &&&= \frac{n^2+4n^2-3n+5n-4}{6n^5} \\ &&&= \frac{5n^2+2n-4}{6n^5} \\ && \mathbb{P}(\text{answered }4|\text{passes}) &= \frac{\mathbb{P}(\text{answered }4\text{ and passes})}{ \mathbb{P}(\text{passes})} \\ &&&= \frac{\frac16 \cdot \frac{4n-3}{n^4}}{\frac{5n^2-2n-4}{6n^5} } \\ &&&= \frac{4n^2-3n}{5n^2+2n-4} \end{align*} Notice that the function takes all values for \(n\) between the roots of the denominator (which are either side of \(0\) and below \(3/4\). Therefore after \(3/4\) the function must be increase since otherwise we would have a quadratic equation with more than \(2\) roots.
  3. \(\,\) \begin{align*} &&\mathbb{P}(C \text{ passes}) &= \binom{5}{3} \left ( \frac{n}{n+1} \right)^3 \left ( \frac{1}{n+1}\right)^2 \frac{1}{n^3} + \binom{5}{4} \left ( \frac{n}{n+1} \right)^4 \left ( \frac{1}{n+1}\right) \frac{4n-3}{n^4} +\\ &&&\quad \quad + \binom{5}{5} \left ( \frac{n}{n+1} \right)^5 \frac{5n-4}{n^5} \\ &&&= \frac{10}{(n+1)^5} + \frac{5(4n-3)}{(n+1)^5} + \frac{(5n-4)}{(n+1)^5} \\ &&&= \frac{10+20n-15+5n-4}{(n+1)^5}\\ &&&= \frac{25n-9}{(n+1)^5}\\ \end{align*}

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