Problems

Filters
Clear Filters

28 problems found

2018 Paper 1 Q12
D: 1500.0 B: 1500.0
conditional probability Bayes' theorem binomial distribution optimisation multiple choice test probability random strategy

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*}

View
2012 Paper 1 Q12
D: 1484.0 B: 1516.0
probability conditional probability Bayes' theorem continuous probability distribution PDF integration fire extinguisher testing law of total probability

Fire extinguishers may become faulty at any time after manufacture and are tested annually on the anniversary of manufacture. The time \(T\) years after manufacture until a fire extinguisher becomes faulty is modelled by the continuous probability density function \[ f(t) = \begin{cases} \frac{2t}{(1+t^2)^2}& \text{for \(t\ge0\)}\,,\\[4mm] \ \ \ \ 0& \text{otherwise}. \end{cases} \] A faulty fire extinguisher will fail an annual test with probability \(p\), in which case it is destroyed immediately. A non-faulty fire extinguisher will always pass the test. All of the annual tests are independent. Show that the probability that a randomly chosen fire extinguisher will be destroyed exactly three years after its manufacture is \(p(5p^2-13p +9)/10\). Find the probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture.

Solution: The probability it becomes faulty in each year is: \begin{align*} \mathbb{P}(\text{faulty in Y}1) &= \int_0^1 \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_0^1 \\ &= 1 - \frac{1}{2} = \frac{1}{2} \\ \mathbb{P}(\text{faulty in Y}2) &= \frac{1}{2} - \frac{1}{5} = \frac{3}{10} \\ \mathbb{P}(\text{faulty in Y}3) &= \frac{1}{5} - \frac{1}{10} = \frac{1}{10} \end{align*} The probability of failing for the first time after exactly \(3\) years is: \begin{align*} \mathbb{P}(\text{faulty in Y1, }PPF) &+ \mathbb{P}(\text{faulty in Y2, }PF) + + \mathbb{P}(\text{faulty in Y3, }F) \\ &= \frac12 (1-p)^2p + \frac3{10}(1-p)p + \frac1{10}p \\ &= \frac{p}{10} \l 5(1-p)^2 + 3(1-p) + 1 \r \\ &= \frac{p}{10} \l 5 - 10p + 5p^2 + 3 -3p +1 \r \\ &= \frac{p}{10} \l 9 - 13p + 5p^2 \r \end{align*} as required. The probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture is: \begin{align*} \mathbb{P}(\text{faulty 18 months after} | \text{fails after 3 tries}) &= \frac{\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})}{\mathbb{P}(\text{fails after exactly 3 tries})} \end{align*} We can compute \(\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})\) by looking at \(2\) cases, fails between \(12\) months and \(18\) years, and between \(0\) years and \(1\) year. \begin{align*} \mathbb{P}(\text{faulty between 1y and 18m}) &= \int_{1}^{\frac32} \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_{1}^{\frac32} \\ &= \frac12 - \frac{4}{13} = \frac{5}{26} \\ \end{align*} So the probability is: \begin{align*} \mathbb{P} &= \frac{\frac{5}{26}(1-p)p + \frac12(1-p)^2p}{\frac{p}{10} \l 9 - 13p + 5p^2 \r} \\ &= \frac{\frac{25}{13}(1-p) + 5(1-p)^2}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 5 + 13(1-p) \r}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 18 - 13p \r}{9 - 13p + 5p^2} \\ \end{align*}

View
2012 Paper 2 Q12
D: 1600.0 B: 1500.7
probability conditional Bayes' theorem tree diagram binomial distribution mode inequality optimisation

A modern villa has complicated lighting controls. In order for the light in the swimming pool to be on, a particular switch in the hallway must be on and a particular switch in the kitchen must be on. There are four identical switches in the hallway and four identical switches in the kitchen. Guests cannot tell whether the switches are on or off, or what they control. Each Monday morning a guest arrives, and the switches in the hallway are either all on or all off. The probability that they are all on is \(p\) and the probability that they are all off is \(1-p\). The switches in the kitchen are each on or off, independently, with probability \(\frac12\).

  1. On the first Monday, a guest presses one switch in the hallway at random and one switch in the kitchen at random. Find the probability that the swimming pool light is on at the end of this process. Show that the probability that the guest has pressed the swimming pool light switch in the hallway, given that the light is on at the end of the process, is \(\displaystyle \frac{1-p}{1+2p}\).
  2. On each of seven Mondays, guests go through the above process independently of each other, and each time the swimming pool light is found to be on at the end of the process. Given that the most likely number of days on which the swimming pool light switch in the hallway was pressed is 3, show that \(\frac14 < p < \frac{5}{14}\).

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(\text{hall switch on}) &= \underbrace{p \cdot \frac34 }_{\text{already on and not flipped}}+ \underbrace{(1-p) \cdot \frac14}_{\text{not on and flipped}} \\ &&&= \frac14 +\frac12 p\\ && \mathbb{P}(\text{kitchen on}) &= \frac12 \\ \Rightarrow && \mathbb{P}(\text{pool is on}) &= \frac18 + \frac14p \end{align*} \begin{align*} && \mathbb{P}(\text{flipped hall switch} | \text{pool on}) &= \frac{\mathbb{P}(\text{flipped hall and pool on})}{\mathbb{P}(\text{pool on})} \\ &&&= \frac{(1-p)\frac14 \cdot \frac 12}{\frac18 + \frac14 p} \\ &&&= \frac{1-p}{1+2p} \end{align*}
  2. The number of days the swimming pool light was pressed is \(X = B\left (7, \frac{1-p}{1+2p} \right)\), and we have that \(\mathbb{P}(X = 2) < \mathbb{P}(X = 3) > \mathbb{P}(X=4)\) (since the binomial is unimodal). Let \(q = \frac{1-p}{1+2p} \) \begin{align*} && \mathbb{P}(X = 2) &< \mathbb{P}(X = 3) \\ \Rightarrow && \binom{7}{2} q^2(1-q)^5 &< \binom{7}{3}q^3(1-q)^4 \\ \Rightarrow && 21(1-q) &< 35q \\ \Rightarrow && 21 &< 56q \\ \Rightarrow && \frac{3}{8} &< \frac{1-p}{1+2p} \\ \Rightarrow && 3+6p &< 8-8p \\ \Rightarrow && 14p &< 5\\ \Rightarrow && p &< \frac5{14} \\ \\ && \mathbb{P}(X = 3) &> \mathbb{P}(X = 4) \\ \Rightarrow && \binom{7}{3} q^3(1-q)^4 &> \binom{7}{4}q^4(1-q)^3 \\ \Rightarrow &&(1-q)&> q \\ \Rightarrow && \frac12 &> q \\ \Rightarrow && \frac12 &> \frac{1-p}{1+2p} \\ \Rightarrow && 1+2p &> 2-2p \\ \Rightarrow && 4p &> 1\\ \Rightarrow && p &> \frac1{4} \end{align*} Therefore \(\frac14 < p < \frac{5}{14}\) as required.

View
2006 Paper 2 Q12
D: 1600.0 B: 1516.0
probability binomial distribution Poisson distribution conditional probability Bayes' theorem cricket approximation

A cricket team has only three bowlers, Arthur, Betty and Cuba, each of whom bowls 30 balls in any match. Past performance reveals that, on average, Arthur takes one wicket for every 36 balls bowled, Betty takes one wicket for every 25 balls bowled, and Cuba takes one wicket for every 41 balls bowled.

  1. In one match, the team took exactly one wicket, but the name of the bowler was not recorded. Using a binomial model, find the probability that Arthur was the bowler.
  2. Show that the average number of wickets taken by the team in a match is approximately 3. Give with brief justification a suitable model for the number of wickets taken by the team in a match and show that the probability of the team taking at least five wickets in a given match is approximately \(\frac15\). [You may use the approximation \(\e^3 = 20\).]

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}(\text{Arthur took wicket and exactly one wicket}) &= \binom{30}{1} \frac{1}{36} \left ( \frac{35}{36} \right)^{29} \binom{30}{0} \left ( \frac{24}{25} \right)^{30} \binom{30}{0} \left ( \frac{40}{41} \right)^{30}\\ &&&= \frac{30 \cdot 35^{29} \cdot 24^{30} \cdot 40^{30}}{36^{30} \cdot 25^{30} \cdot {41}^{30}}\\ &&&= \frac{1}{35} N\\ && \mathbb{P}(\text{B took wicket and exactly one wicket}) &= \binom{30}{0}\left ( \frac{35}{36} \right)^{30} \binom{30}{1} \frac{1}{25} \left ( \frac{24}{25} \right)^{29} \binom{30}{0} \left ( \frac{40}{41} \right)^{30}\\ &&&= \frac{1}{24} N \\ && \mathbb{P}(\text{C took wicket and exactly one wicket}) &= \binom{30}{0}\left ( \frac{35}{36} \right)^{30} \binom{30}{0}\left ( \frac{24}{25} \right)^{30} \binom{30}{1} \frac{1}{41} \left ( \frac{40}{41} \right)^{29}\\ &&&= \frac{1}{40} N \\ && \mathbb{P}(\text{Arthur took wicket} | \text{exactly one wicket}) &= \frac{ \mathbb{P}(\text{Arthur took wicket and exactly one wicket}) }{ \mathbb{P}(\text{exactly one wicket}) } \\ &&&= \frac{ \frac{1}{35} N}{\frac1{35} N + \frac{1}{24}N + \frac{1}{40} N} \\ &&&= \frac{3}{10} \end{align*} Alternatively, we could look at: \begin{align*} && \mathbb{P}(X_A = 1 | X_A + X_B + X_C =1) &= \frac{\mathbb{P}(X_A = 1, X_B = 0,X_C = 0)}{\mathbb{P}(X_A = 1, X_B = 0,X_C = 0)+\mathbb{P}(X_A = 0, X_B = 1,X_C = 0)+\mathbb{P}(X_A = 0, X_B = 0,X_C = 1)} \\ &&&= \frac{\frac{\mathbb{P}(X_A = 1)}{\mathbb{P}(X_A=0)}}{\frac{\mathbb{P}(X_A = 1)}{\mathbb{P}(X_A=0)}+\frac{\mathbb{P}(X_B = 1)}{\mathbb{P}(X_B=0)}+\frac{\mathbb{P}(X_C = 1)}{\mathbb{P}(X_C=0)}} \end{align*} and we can calculate these relatively likelihoods in a similar way to above.
  2. \(\,\) \begin{align*} && \mathbb{E}(\text{number of wickets}) &= \mathbb{E} \left ( \sum_{i=1}^{90} \mathbb{1}_{i\text{th ball is a wicket}} \right) \\ &&&= \sum_{i=1}^{90} \mathbb{E} \left (\mathbb{1}_{i\text{th ball is a wicket}} \right) \\ &&&= 30 \cdot \frac{1}{36} + 30 \cdot \frac{1}{25} + 30 \cdot \frac{1}{41} \\ &&&\approx 1 + 1 + 1 = 3 \end{align*} We might model the number of wickets taken as \(Po(\lambda)\), where \(\lambda\) is the average number of wickets taken. We can think of this roughly as the Poisson approximation to the binomial where \(N\) is large and \(Np\) is small. Assuming we use \(Po(3)\) we have \begin{align*} && \mathbb{P}(\text{at least 5 wickets}) &= 1-\mathbb{P}(\text{4 or fewer wickets}) \\ &&&= 1- e^{-3} \left (1 + \frac{3}{1} + \frac{3^2}{2} + \frac{3^3}{6} + \frac{3^4}{24} \right) \\ &&&= 1 - \frac{1}{20} \left ( 1 + 3 + \frac{9}{2} + \frac{9}{2} + \frac{27}{8} \right) \\ &&&= 1 - \frac{1}{20} \left (13 + 3\tfrac38 \right) \\ &&&\approx 1 - \frac{16}{20} = \frac15 \end{align*}

View
2005 Paper 1 Q13
D: 1500.0 B: 1516.0
probability conditional Bayes' theorem normal distribution symmetric distribution mean and standard deviation law of total probability

The random variable \(X\) has mean \(\mu\) and standard deviation \(\sigma\). The distribution of \(X\) is symmetrical about \(\mu\) and satisfies: \[\P \l X \le \mu + \sigma \r = a \mbox{ and } \P \l X \le \mu + \tfrac{1}{ 2}\sigma \r = b\,,\] where \(a\) and \(b\) are fixed numbers. Do not assume that \(X\) is Normally distributed.

  1. Determine expressions (in terms of \(a\) and \(b\)) for \[ \P \l \mu-\tfrac12 \sigma \le X \le \mu + \sigma \r \mbox{ and } \P \l X \le \mu +\tfrac12 \sigma \; \vert \; X \ge \mu - \tfrac12 \sigma \r.\]
  2. My local supermarket sells cartons of skimmed milk and cartons of full-fat milk: \(60\%\) of the cartons it sells contain skimmed milk, and the rest contain full-fat milk. The volume of skimmed milk in a carton is modelled by \(X\) ml, with \(\mu = 500\) and \(\sigma =10\,\). The volume of full-fat milk in a carton is modelled by \(X\) ml, with \(\mu = 495\) and \(\sigma = 10\,\).
    1. Today, I bought one carton of milk, chosen at random, from this supermarket. When I get home, I find that it contains less than 505 ml. Determine an expression (in terms of \(a\) and \(b\)) for the probability that this carton of milk contains more than 500 ml.
    2. Over the years, I have bought a very large number of cartons of milk, all chosen at random, from this supermarket. \(70\%\) of the cartons I have bought have contained at most 505 ml of milk. Of all the cartons that have contained at least 495 ml of milk, one third of them have contained full-fat milk. Use this information to estimate the values of \(a\) and \(b\).

Solution:

  1. \(\,\) \begin{align*} && \mathbb{P}\left (\mu - \tfrac12 \sigma \leq X \right) &= \mathbb{P}\left (X \leq \mu + \tfrac12 \sigma \right) \tag{by symmetry} \\ &&&= b \\ \Rightarrow && \mathbb{P} \left (\mu - \tfrac12 \sigma \leq X \leq \mu + \sigma \right) &= a - (1-b) = a+b - 1\\ \\ && \mathbb{P} \left ( X \le \mu +\tfrac12 \sigma \vert X \ge \mu - \tfrac12 \sigma \right ) &= \frac{ \mathbb{P} \left (\mu - \tfrac12 \sigma \leq X \leq \mu + \tfrac12 \sigma \right)}{\mathbb{P} \left ( X \ge \mu - \tfrac12 \sigma \right )} \\ &&&= \frac{b-(1-b)}{1-(1-b)} \\ &&&= \frac{2b-1}{b} \end{align*}
    1. Let \(Y\) be the volume of milk in the carton I bring home, we are interested in: \begin{align*} && \mathbb{P}(Y \geq 500 | Y \leq 505) &= \frac{\mathbb{P}(500 \leq Y \leq 505)}{\mathbb{P}(Y \leq 505)} \\ &&&=\frac{\mathbb{P}(500 \leq Y \leq 505|\text{skimmed})\mathbb{P}(\text{skimmed})+\mathbb{P}(500 \leq Y \leq 505|\text{full fat})\mathbb{P}(\text{full fat})}{\mathbb{P}(Y \leq 505|\text{skimmed})\mathbb{P}(\text{skimmed})+\mathbb{P}(Y \leq 505|\text{full fat})\mathbb{P}(\text{full fat})} \\ &&&= \frac{\frac35 \cdot \mathbb{P}(\mu \leq X \leq \mu + \tfrac12 \sigma) + \frac25 \cdot \mathbb{P}(\mu+\tfrac12 \sigma \leq X \leq \mu +\sigma)}{\frac35 \cdot \mathbb{P}(X \leq \mu + \tfrac12 \sigma) + \frac25 \cdot \mathbb{P}(X \leq \mu +\sigma)} \\ &&&= \frac{\frac35 \cdot(b-\tfrac12) + \frac25 \cdot (a-b)}{\frac35 \cdot b + \frac25 \cdot a} \\ &&&= \frac{b+2a-\frac32}{3b+2a} \\ &&&= \frac{4a+2b-3}{4a+6b} \end{align*}
    2. \(70\%\) of cartons have contained at most 505 ml, so: \begin{align*} && \tfrac7{10} &= \mathbb{P}(Y \leq 505) \\ &&&= \mathbb{P}(Y \leq 505 | \text{ skimmed}) \mathbb{P}(\text{skimmed}) + \mathbb{P}(Y \leq 505 | \text{ full fat}) \mathbb{P}(\text{full fat}) \\ &&&= \mathbb{P}(X \leq \mu + \tfrac12 \sigma) \cdot \tfrac35 + \mathbb{P}(X\leq \mu + \sigma ) \cdot \tfrac25 \\ \Rightarrow && 7 &= 6b+ 4a \end{align*} \(\tfrac13\) of cartons containing 495 ml contained full fat milk: \begin{align*} && \tfrac13 &= \mathbb{P}(\text{full fat} | Y \geq 495) \\ &&&= \frac{\mathbb{P}(\text{full fat and} Y \geq 495) }{\mathbb{P}(Y \geq 495)} \\ &&&= \frac{\mathbb{P}(X \geq \mu)\frac25}{\mathbb{P}(X \geq \mu)\cdot \frac25+\mathbb{P}(X \geq \mu-\tfrac12 \sigma)\cdot \frac35} \\ &&&= \frac{\frac15}{\frac12 \cdot \frac25 + b\frac35}\\ &&&= \frac{1}{1+ 3b }\\ \Rightarrow && 3b+1 &= 3 \\ \Rightarrow && b &= \frac23 \\ && a &= \frac34 \end{align*}

View
2005 Paper 2 Q12
D: 1600.0 B: 1500.0
conditional probability Bayes' theorem independence coin toss truthful answering law of total probability posterior probability

The twins Anna and Bella share a computer and never sign their e-mails. When I e-mail them, only the twin currently online responds. The probability that it is Anna who is online is \(p\) and she answers each question I ask her truthfully with probability \(a\), independently of all her other answers, even if a question is repeated. The probability that it is Bella who is online is~\(q\), where \(q=1-p\), and she answers each question truthfully with probability \(b\), independently of all her other answers, even if a question is repeated.

  1. I send the twins the e-mail: `Toss a fair coin and answer the following question. Did the coin come down heads?'. I receive the answer `yes'. Show that the probability that the coin did come down heads is \(\frac{1}{2}\) if and only if \(2(ap+bq)=1\).
  2. I send the twins the e-mail: `Toss a fair coin and answer the following question. Did the coin come down heads?'. I receive the answer `yes'. I then send the e-mail: `Did the coin come down heads?' and I receive the answer `no'. Show that the probability (taking into account these answers) that the coin did come down heads is \(\frac{1}{2}\,\).
  3. I send the twins the e-mail: `Toss a fair coin and answer the following question. Did the coin come down heads?'. I receive the answer `yes'. I then send the e-mail: `Did the coin come down heads?' and I receive the answer `yes'. Show that, if \(2(ap+bq)=1\), the probability (taking into account these answers) that the coin did come down heads is \(\frac{1}{2}\,\).

View
2004 Paper 1 Q12
D: 1500.0 B: 1529.3
probability conditional Bayes' theorem proportions estimation manufacturing defects law of total probability

In a certain factory, microchips are made by two machines. Machine A makes a proportion \(\lambda\) of the chips, where \(0 < \lambda < 1\), and machine B makes the rest. A proportion \(p\) of the chips made by machine A are perfect, and a proportion \(q\) of those made by machine B are perfect, where \(0 < p < 1\) and \(0 < q < 1\). The chips are sorted into two groups: group 1 contains those that are perfect and group 2 contains those that are imperfect. In a large random sample taken from group 1, it is found that \(\frac 2 5\) were made by machine A. Show that \(\lambda\) can estimated as \[ {2q \over 3p + 2q}\;. \] Subsequently, it is discovered that the sorting process is faulty: there is a probability of \(\frac 14\) that a perfect chip is assigned to group 2 and a probability of \(\frac 14\) that an imperfect chip is assigned to group 1. Taking into account this additional information, obtain a new estimate of \(\lambda\,\).

Solution: \begin{align*} && \frac25 &= \frac{\lambda p}{\lambda p + (1-\lambda) q} \\ \Rightarrow && 2(1-\lambda)q &= 3\lambda p \\ \Rightarrow && \lambda(3p+2q) &= 2q \\ \Rightarrow && \lambda &= \frac{2q}{3p+2q} \end{align*} \begin{align*} && \frac25 &= \frac{\lambda (p + \frac14(1-p))}{\lambda (p + \frac14(1-p))+(1-\lambda) (q + \frac14(1-q))} \\ &&&= \frac{\lambda(\frac34p + \frac14)}{\lambda(\frac34p + \frac14)+(1-\lambda)(\frac34q + \frac14)} \\ \Rightarrow && \lambda &= \frac{2(\frac34q+\frac14)}{3(\frac34p + \frac14)+2(\frac34q+\frac14)} \\ &&&= \frac{\frac32q + \frac12}{\frac94p + \frac32q + \frac54} \\ &&&= \frac{6q+2}{9p+6q+5} \end{align*}

View
2002 Paper 2 Q14
D: 1600.0 B: 1484.1
probability Poisson distribution conditional Bayes' theorem expectation summation weighted average concentric regions

A densely populated circular island is divided into \(N\) concentric regions \(R_1\), \(R_2\), \(\ldots\,\), \(R_N\), such that the inner and outer radii of \(R_n\) are \(n-1\) km and \(n\) km, respectively. The average number of road accidents that occur in any one day in \(R_n\) is \(2-n/N\,\), independently of the number of accidents in any other region. Each day an observer selects a region at random, with a probability that is proportional to the area of the region, and records the number of road accidents, \(X\), that occur in it. Show that, in the long term, the average number of recorded accidents per day will be \[ 2-\frac16\left(1+\frac1N\right)\left(4-\frac1N\right)\;. \] [Note: \(\sum\limits_{n=1}^N n^2 = \frac16 N(N+1)(2N+1) \;\).] Show also that \[ \P(X=k) = \frac{\e^{-2}N^{-k-2}}{k!}\sum_{n=1}^N (2n-1)(2N-n)^k\e^{n/N} \;. \] Suppose now that \(N=3\) and that, on a particular day, two accidents were recorded. Show that the probability that \(R_2\) had been selected is \[ \frac{48}{48 + 45\e^{1/3} +25 \e^{-1/3}}\;. \]

Solution: The area of \(R_n\) is \(\pi(n^2 - (n-1)^2) = (2n-1)\pi\). The area of the whole region is \(\pi N^2\). \begin{align*} && \E[X] &= \E[\E[X | \text{choose region }n]] \\ &&&= \sum_{n=1}^N \left (2 - \frac{n}{N} \right) \cdot \frac{(2n-1)\pi}{N^2 \pi} \\ &&&= \sum_{n=1}^N \left (2\cdot \frac{(2n-1)\pi}{N^2 \pi} - \frac{n}{N}\cdot \frac{(2n-1)\pi}{N^2 \pi} \right) \\ &&&= 2 - \frac{1}{N^3} \sum_{n=1}^N (2n^2-n) \\ &&&= 2 - \frac{1}{N^3} \left (\frac{2N(N+1)(2N+1)}{6} - \frac{N(N+1)}{2} \right) \\ &&&= 2 - \frac{N+1}{6N^2} \left (2(2N+1)-3 \right) \\ &&&= 2 - \frac{N+1}{6N^2} (4N - 1) \\ &&&= 2 - \frac16 \left (1 + \frac1N \right) \left (4 - \frac1N \right) \end{align*} Modelling each region as \(Po(2 - n/N)\) we have \begin{align*} \mathbb{P}(X = k ) &= \sum_{n=1}^N \exp(-2 + n/N) \frac{(2-n/N)^k}{k!} \frac{2n-1}{N^2} \\ &= \frac{e^{-2}N^{-k-2}}{k!} \sum_{n=1}^N e^{n/N} (2N-n)^k(2n-1) \end{align*} as desired. Supposing \(N=3\) and two accidents then \begin{align*} \mathbb{P}(R_2 | X = 2) &= \frac{\frac{3}{9} e^{-4/3}\frac{(\frac43)^2}{2!}}{\mathbb{P}(X=2)} \\ &= \frac{\frac{3}{9} e^{-4/3} \frac{(\frac43)^2}{2!}}{\frac{1}{9} e^{-5/3} \frac{(\frac53)^2}{2!} + \frac{3}{9} e^{-4/3} \frac{(\frac43)^2}{2!} + \frac{5}{9} e^{-2/3} \frac{(\frac33)^2}{2!}} \\ &= \frac{3 \cdot 16}{25e^{-1/3} + 3 \cdot 16 + 5 \cdot 9e^{1/3}} \\ &= \frac{48}{25e^{-1/3} + 48 + 45e^{1/3}} \end{align*} as required.

View
2000 Paper 1 Q13
D: 1484.0 B: 1484.7
probability conditional probability Bayes' theorem genetics binomial distribution combinatorics

Every person carries two genes which can each be either of type \(A\) or of type \(B\). It is known that \(81\%\) of the population are \(AA\) (i.e. both genes are of type \(A\)), \(18\%\) are \(AB\) (i.e. there is one gene of type \(A\) and one of type \(B\)) and \(1\%\) are \(BB\). A child inherits one gene from each of its parents. If one parent is \(AA\), the child inherits a gene of type \(A\) from that parent; if the parent is \(BB\), the child inherits a gene of type \(B\) from that parent; if the parent is \(AB\), the inherited gene is equally likely to be \(A\) or \(B\).

  1. Given that two \(AB\) parents have four children, show that the probability that two of them are \(AA\) and two of them are \(BB\) is \(3/128\).
  2. My mother is \(AB\) and I am \(AA\). Find the probability that my father is \(AB\).

View
1999 Paper 1 Q14
D: 1500.0 B: 1516.0
probability geometric probability expected value conditional probability density function discrete score dart target Bayes' theorem

When I throw a dart at a target, the probability that it lands a distance \(X\) from the centre is a random variable with density function \[ \mathrm{f}(x)=\begin{cases} 2x & \text{ if }0\leqslant x\leqslant1;\\ 0 & \text{ otherwise.} \end{cases} \] I score points according to the position of the dart as follows: %

%
%
%\newline\hspace*{10mm} if~\(0\le X< \frac14\), my score is 4; %\newline\hspace*{10mm} if~\(\frac14\le X< \frac12\), my score is 3; %\newline\hspace*{10mm} if \(\frac12\le X< \frac34\), my score is 2; %\newline\hspace*{10mm} if \(\frac34\le X\le 1\), my score is 1.
  1. Show that my expected score from one dart is 15/8.
  2. I play a game with the following rules. I start off with a total score 0, and each time~I throw a dart my score on that throw is added to my total. Then: \newline \hspace*{10mm} if my new total is greater than 3, I have lost and the game ends; \newline \hspace*{10mm} if my new total is 3, I have won and the game ends; \newline \hspace*{10mm} if my new total is less than 3, I throw again. Show that, if I have won such a game, the probability that I threw the dart three times is 343/2231.

View
1999 Paper 2 Q12
D: 1600.0 B: 1484.0
conditional probability Bayes' theorem maximum likelihood estimation optimisation manufacturers defective items

It is known that there are three manufacturers \(A, B, C,\) who can produce micro chip MB666. The probability that a randomly selected MB666 is produced by \(A\) is \(2p\), and the corresponding probabilities for \(B\) and \(C\) are \(p\) and \(1 - 3p\), respectively, where \({{0} \le p \le {1 \over 3}}.\) It is also known that \(70\%\) of MB666 micro chips from \(A\) are sound and that the corresponding percentages for \(B\) and \(C\) are \(80\%\) and \(90\%\), respectively. Find in terms of \(p\), the conditional probability, \(\P(A {\vert} S)\), that if a randomly selected MB666 chip is found to be sound then it came from \(A\), and also the conditional probability, \(\P(C {\vert} S)\), that if it is sound then it came from \(C\). A quality inspector took a random sample of one MB666 micro chip and found it to be sound. She then traced its place of manufacture to be \(A\), and so estimated \(p\) by calculating the value of \(p\) that corresponds to the greatest value of \(\P(A {\vert} S)\). A second quality inspector also a took random sample of one MB666 chip and found it to be sound. Later he traced its place of manufacture to be \(C\) and so estimated \(p\) by applying the procedure of his colleague to \(\P(C {\vert} S)\). Determine the values of the two estimates and comment briefly on the results obtained.

View
1999 Paper 3 Q13
D: 1700.0 B: 1484.0
probability conditional Bayes' theorem continuous random variable density function binomial distribution expected value geometric probability

The cakes in our canteen each contain exactly four currants, each currant being randomly placed in the cake. I take a proportion \(X\) of a cake where \(X\) is a random variable with density function \[{\mathrm f}(x)=Ax\] for \(0\leqslant x\leqslant 1\) where \(A\) is a constant.

  1. What is the expected number of currants in my portion?
  2. If I find all four currants in my portion, what is the probability that I took more than half the cake?

View
1998 Paper 1 Q13
D: 1484.0 B: 1532.0
conditional probability Bayes' theorem sampling with replacement combinatorial probability geometric series

I have a bag initially containing \(r\) red fruit pastilles (my favourites) and \(b\) fruit pastilles of other colours. From time to time I shake the bag thoroughly and remove a pastille at random. (It may be assumed that all pastilles have an equal chance of being selected.) If the pastille is red I eat it but otherwise I replace it in the bag. After \(n\) such drawings, I find that I have only eaten one pastille. Show that the probability that I ate it on my last drawing is \[\frac{(r+b-1)^{n-1}}{(r+b)^{n}-(r+b-1)^{n}}.\]

View
1998 Paper 2 Q12
D: 1600.0 B: 1560.2
probability conditional Bayes' theorem false positive diagnostic test rare disease

The diagnostic test AL has a probability 0.9 of giving a positive result when applied to a person suffering from the rare disease mathematitis. It also has a probability 1/11 of giving a false positive result when applied to a non-sufferer. It is known that only \(1\%\) of the population suffer from the disease. Given that the test AL is positive when applied to Frankie, who is chosen at random from the population, what is the probability that Frankie is a sufferer? In an attempt to identify sufferers more accurately, a second diagnostic test STEP is given to those for whom the test AL gave a positive result. The probablility of STEP giving a positive result on a sufferer is 0.9, and the probability that it gives a false positive result on a non-sufferer is \(p\). Half of those for whom AL was positive and on whom STEP then also gives a positive result are sufferers. Find \(p\).

Solution: \begin{align*} \mathbb{P}(M | P_{AL}) &= \frac{\mathbb{P}(M \cap P_{AL})}{\mathbb{P}(P_{AL})} \\ &= \frac{\frac{1}{100} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} + \frac{99}{100} \frac{1}{11}} \\ &= \frac{\frac{9}{10}}{\frac{9}{10} + \frac{9}{1}} \\ &= \frac{9}{99} = \frac{1}{11} \\ \end{align*} \begin{align*} && \frac12 &= \mathbb{P}(M | P_{STEP}, P_{AL}) \\ &&&= \frac{\frac{1}{100} \frac{9}{10} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} \frac{9}{10} + \frac{99}{100} \frac{1}{11}p} \\ &&&= \frac{81}{81+900p} \\ \Rightarrow && p &= \frac{81}{900} = \frac{9}{100} \end{align*} Therefore \(p = 9\%\)

View
1997 Paper 1 Q13
D: 1500.0 B: 1547.3
probability conditional Bayes' theorem decision analysis expected value optimisation

Mr Blond returns to his flat to find it in complete darkness. He knows that this means that one of four assassins Mr 1, Mr 2, Mr 3 or Mr 4 has set a trap for him. His trained instinct tells him that the probability that Mr \(i\) has set the trap is \(i/10\). His knowledge of their habits tells him that Mr \(i\) uses a deadly trained silent anaconda with probability \((i+1)/10\), a bomb with probability \(i/10\) and a vicious attack canary with probability \((9-2i)/10\) \([i=1,2,3,4]\). He now listens carefully and, hearing no singing, concludes correctly that no canary is involved. If he switches on the light and the trap is a bomb he has probability \(1/2\) of being killed but if the trap is an anaconda he has probability \(2/3\) of survival. If he does not switch on the light and the trap is a bomb he is certain to survive but, if the trap is an anaconda, he has a probability \(1/2\) of being killed. His professional pride means that he must enter the flat. Advise Mr Blond, giving reasons for your advice.

Solution: \begin{array}{c|c|c|c} & A & B & C \\ \hline 1 & \frac{1}{10} \cdot \frac{2}{10} & \frac{1}{10} \cdot \frac{1}{10} & \frac{1}{10} \cdot \frac{7}{10} \\ 2 & \frac{2}{10} \cdot \frac{3}{10} &\frac{2}{10} \cdot \frac{2}{10} &\frac{2}{10} \cdot \frac{5}{10} \\ 3 & \frac{3}{10} \cdot \frac{4}{10} &\frac{3}{10} \cdot \frac{3}{10} &\frac{3}{10} \cdot \frac{3}{10} \\ 4 & \frac{4}{10} \cdot \frac{5}{10} &\frac{4}{10} \cdot \frac{4}{10} &\frac{4}{10} \cdot \frac{1}{10} \\ \hline & \frac{2+6+12+20}{100} & \frac{1 + 4 + 9 + 16}{100} & \frac{7 + 10 + 9 + 4}{100} \end{array} Therefore \(\mathbb{P}(A) = \frac{4}{10}, \mathbb{P}(B) = \frac{3}{10}, \mathbb{P}(C) = \frac{3}{10}\), in particular, \begin{align*} \mathbb{P}(A | \text{not }C) &= \frac{4}{7} \\ \mathbb{P}(B | \text{not }C) &= \frac{3}{7} \\ \end{align*} If he switches the light on, his probability of survival is \(\frac47 \cdot \frac23 + \frac37 \cdot \frac12 = \frac{25}{42}\), if he doesn't his probability is \(\frac12 \cdot \frac47 +\frac37= \frac{5}{7} = \frac{30}{42}\) therefore he shouldn't switch the light on.

View