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2012 Paper 1 Q13
D: 1500.0 B: 1529.2
probability conditional expected value combinatorics counting digits inclusion-exclusion

I choose at random an integer in the range 10000 to 99999, all choices being equally likely. Given that my choice does not contain the digits 0, 6, 7, 8 or 9, show that the expected number of different digits in my choice is 3.3616.

Solution: We are choosing any \(5\) digit number from \(\{1,2,3,4,5\}\). There are \(5^5\) such numbers. \begin{align*} && \mathbb{E}(\text{different digits}) &= \frac1{5^5} \left (1 \cdot 5 + 2 \cdot \binom{5}{2}(2^5-2)+3 \cdot \binom{5}{3}(3^5-3 \cdot 2^5+3)+4 \cdot \binom{5}{4}(4^5 - 4 \cdot 3^5+6 \cdot 2^5-4) + 5 \cdot 5! \right) \\ &&&= \frac{2101}{625} = 3.3616 \end{align*}

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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.

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2008 Paper 2 Q13
D: 1600.0 B: 1516.0
probability conditional tree diagram combinatorics hypergeometric distribution optimisation

Bag \(P\) and bag \(Q\) each contain \(n\) counters, where \(n\ge2\). The counters are identical in shape and size, but coloured either black or white. First, \(k\) counters (\(0\le k\le n\)) are drawn at random from bag \(P\) and placed in bag \(Q\). Then, \(k\) counters are drawn at random from bag \(Q\) and placed in bag \(P\).

  1. If initially \(n-1\) counters in bag \(P\) are white and one is black, and all \(n\) counters in bag \(Q\) are white, find the probability in terms of \(n\) and \(k\) that the black counter ends up in bag \(P\). Find the value or values of \(k\) for which this probability is maximised.
  2. If initially \(n-1\) counters in bag \(P\) are white and one is black, and \(n-1\) counters in bag \(Q\) are white and one is black, find the probability in terms of \(n\) and \(k\) that the black counters end up in the same bag. Find the value or values of \(k\) for which this probability is maximised.

Solution:

  1. \(\,\) \begin{align*} \mathbb{P}(\text{black counter in }P) &= \mathbb{P}(\text{black counter moves twice})+\mathbb{P}(\text{black counter doesn't move}) \\ &= \mathbb{P}(\text{black counter moves out})\mathbb{P}(\text{black counter moves back}) + (1-\mathbb{P}(\text{black counter moves out})) \\ &= \frac{k}n\cdot \frac{k}{n+k}+\frac{n-k}{n} \\ &= \frac{k^2+n^2-k^2}{n(n+k)} \\ &= \frac{n^2}{n(n+k)} = \frac{n}{n+k} \end{align*} This is maximised if \(k\) is as small as possibe, ie \(k = 0\) (ie it doesn't leave it's bag)
  2. \(\,\) \begin{align*} && \mathbb{P}(\text{both counters in same bag}) &= \mathbb{P}(\text{both in }P)+ \mathbb{P}(\text{both in }Q) \\ &&&= \mathbb{P}(B_P \to Q \to P, B_Q \to P)+\mathbb{P}(B_P \text{ stays}, B_Q \to P)+\mathbb{P}(B_P \to Q, \text{both stay}) \\ &&&= \frac{k}{n} \cdot \frac{k(k-1)}{(n+k)(n+k-1)} + \frac{n-k}{n} \frac{k}{n+k} + \frac{k}{n} \frac{n(n-1)}{(n+k)(n+k-1)} \\ &&&= \frac{(k^3-k^2)+(n-k)k(n+k-1)+kn(n-1)}{n(n+k)(n+k-1)}\\ &&&= \frac{2kn(n-1)}{n(n+k)(n+k-1)}\\ &&&= \frac{2k(n-1)}{(n+k)(n+k-1)} \end{align*} \begin{align*} && \frac{P_{k+1}}{P_k} &= \frac{2(k+1)(n-1)}{(n+k+1)(n+k)} \frac{(n+k)(n+k-1)}{2k(n-1)} \\ &&&= \frac{(k+1)(n+k-1)}{k(n+k+1)} \\ &&& \geq 1 \\ \Leftrightarrow && (k+1)(n+k-1) &\geq k(n+k+1) \\ \Leftrightarrow && n-1 &\geq k \\ \end{align*} Therefore this probability is increasing while \(k \leq n-1\), ie it's maximised \(k = n-1\) or \(k=n\)

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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*}

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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*}

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2004 Paper 2 Q13
D: 1600.0 B: 1500.0
probability conditional discrete probability distributions expected value optimisation without replacement hypergeometric sequential decision

A bag contains \(b\) balls, \(r\) of them red and the rest white. In a game the player must remove balls one at a time from the bag (without replacement). She may remove as many balls as she wishes, but if she removes any red ball, she loses and gets no reward at all. If she does not remove a red ball, she is rewarded with \pounds 1 for each white ball she has removed. If she removes \(n\) white balls on her first \(n\) draws, calculate her expected gain on the next draw and show that %her expected total reward would be the same as before it is zero if \(\ds n = {b-r \over r+1}\,\). Hence, or otherwise, show that she will maximise her expected total reward if she aims to remove \(n\) balls, where \[ n = \mbox{ the integer part of } \ds {b + 1 \over r + 1}\;. \] With this value of \(n\), show that in the case \(r=1\) and \(b\) even, her expected total reward is \(\pounds {1 \over 4}b\,\), and find her expected total reward in the case \(r=1\) and \(b\) odd.

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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.

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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?

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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 \(p = 9\%\)

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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.

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1992 Paper 2 Q16
D: 1600.0 B: 1500.0
probability conditional Bayes' theorem binomial distribution combinatorics random walk

A taxi driver keeps a packet of toffees and a packet of mints in her taxi. From time to time she takes either a toffee (with probability \(p\)) or mint (with probability \(q=1-p\)). At the beginning of the week she has \(n\) toffees and \(m\) mints in the packets. On the \(N\)th occasion that she reaches for a sweet, she discovers (for the first time) that she has run out of that kind of sweet. What is the probability that she was reaching for a toffee?

Solution: \begin{align*} \mathbb{P}(\text{run out reading for toffee on } N\text{th occassion}) &= \binom{N-1}{n}p^nq^{N-1-n}p \end{align*} Since out of the first \(N-1\) times, we need to choose toffee \(n\) times, and then choose it again for the \(N\)th time. Therefore: \begin{align*} \mathbb{P}(\text{reaching for toffee} | \text{run out on }N\text{th occassion}) &= \frac{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion})}{\mathbb{P}(\text{reaching for toffee and run out on }N\text{th occassion}) + \mathbb{P}(\text{reaching for mint and run out on }N\text{th occassion})} \\ &= \frac{ \binom{N-1}{n}p^nq^{N-1-n}p}{ \binom{N-1}{n}p^nq^{N-1-n}p + \binom{N-1}{m}q^mp^{N-1-m}q} \\ &= \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} \l \frac{q}{p} \r^{m+ n+ 2-N}} \end{align*} Some conclusions we can draw from this are: As \(p \to 1, q \to 0\), the probability they were reaching for a Toffee tends to \(1\). (And vice versa). If \(p = q\), then the probability is: \begin{align*} \frac{ \binom{N-1}{n}}{ \binom{N-1}{n} + \binom{N-1}{m} } \end{align*} Since \(n+1 \leq N \leq n+m+1\) where \(n \geq m\) we can notice that: \begin{align*} \text{if } N = m + n + 1 && \binom{m+n+1 - 1}{n} &= \binom{m+n+1 - 1}{m} & \text{ so } \mathbb{P} = \frac12 \\ \text{if } N = n+k && \binom{n+k-1}{n} &< \binom{n+k-1}{m} & \text{ so } \mathbb{P} < \frac12 \\ \end{align*}

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1990 Paper 1 Q14
D: 1500.0 B: 1500.7
probability conditional expected value discrete probability distributions optimisation strategy sequential decision balls in bag

A bag contains 5 white balls, 3 red balls and 2 black balls. In the game of Blackball, a player draws a ball at random from the bag, looks at it and replaces it. If he has drawn a white ball, he scores one point, while for a red ball he scores two points, these scores being added to his total score before he drew the ball. If he has drawn a black ball, the game is over and his final score is zero. After drawing a red or white ball, he can either decide to stop, when his final score for the game is the total so far, or he may elect to draw another ball. The starting score is zero. Juggins' strategy is to continue drawing until either he draws a black ball (when of course he must stop, with final score zero), or until he has drawn three (non-black) balls, when he elects to stop. Find the probability that in any game he achieves a final score of zero by employing this strategy. Find also his expected final score. Muggins has so far scored \(N\) points, and is deciding whether to draw another ball. Find the expected score if another ball is drawn, and suggest a strategy to achieve the greatest possible average final score in each game.

Solution: The probability Juggin's has a non-zero score is the probability he never draws a black ball in his three goes. This is \((1-\frac15)^3 = \frac{64}{125}\). Let's consider the \(\frac{61}{125}\) probability world where he never draws a black ball. In this conditional probability space, he has \(\frac{5}{8}\) chances of pulling out white balls and \(\frac38\) or pulling out red. His expected score per pull is \(\frac58 \cdot 1 + \frac38 \cdot 2 = \frac{11}{8}\). Therefore his expected score in this universe is \(\frac{33}8\) and his expected score is \(\frac{33}{8} \cdot \frac{61}{125} = \frac{2013}{1000} = 2.013\) . The expected score after drawing another ball is \(( N + 1)\frac{5}{10} + (N+2) \frac{3}{10} + 0 \cdot \frac{2}{10} = \frac{8}{10}N + \frac{11}{10}\). A sensible strategy would be to only draw if \(\frac{8}{10}N + \frac{11}{10} > N \Rightarrow N < \frac{11}{2}\), ie keep drawing until \(N \geq 6\) or we bust out. [The expected score for this strategy is: \begin{array}{ccc} \text{score} & \text{route} & \text{count} & \text{prob} \\ \hline 6 & \text{6 1s} & 1 & \left ( \frac12 \right)^6 \\ 6 & \text{4 1s, 1 2} & 5 & 5 \cdot \left ( \frac12 \right)^4 \cdot \frac{3}{10} \\ 6 & \text{2 1s, 2 2s} & 6 & 6 \cdot \left ( \frac12 \right)^2 \cdot \left ( \frac{3}{10} \right)^2 \\ 6 & \text{3 2s} & 1 & 1 \cdot \left ( \frac{3}{10} \right)^3 \\ 7 & \text{5 1s, 1 2} & 1 &\left ( \frac12 \right)^5 \cdot \frac{3}{10} \\ 7 & \text{3 1s, 2 2s} & 4 & 4\cdot \left ( \frac12 \right)^3 \cdot \left ( \frac{3}{10} \right)^2 \\ 7 & \text{1 1, 3 2s} & 3 & 3\cdot \left ( \frac12 \right) \cdot \left ( \frac{3}{10} \right)^3 \\ \end{array} For an expected value of \(\frac{2171}{8000} \cdot 6 + \frac{759}{8000} \cdot 7 = \frac{18\,339}{8000} = 2.29 \quad (3\text{ s.f.})\)]

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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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1989 Paper 1 Q16
D: 1516.0 B: 1470.2
probability conditional game theory expected value optimisation strategy linear programming

A and B play a guessing game. Each simultaneously names one of the numbers \(1,2,3.\) If the numbers differ by 2, whoever guessed the smaller pays the opponent £\(2\). If the numbers differ by 1, whoever guessed the larger pays the opponent £\(1.\) Otherwise no money changes hands. Many rounds of the game are played.

  1. If A says he will always guess the same number \(N\), explain (for each value of \(N\)) how B can maximise his winnings.
  2. In an attempt to improve his play, A announces that he will guess each number at random with probability \(\frac{1}{3},\) guesses on different rounds being independent. To counter this, B secretly decides to guess \(j\) with probability \(b_{j}\) (\(j=1,2,3,\, b_{1}+b_{2}+b_{3}=1\)), guesses on different rounds being independent. Derive an expression for B's expected winnings on any round. How should the probabilities \(b_{j}\) be chosen so as to maximize this expression?
  3. A now announces that he will guess \(j\) with probability \(a_{j}\) (\(j=1,2,3,\, a_{1}+a_{2}+a_{3}=1\)). If B guesses \(j\) with probability \(b_{j}\) (\(j=1,2,3,\, b_{1}+b_{2}+b_{3}=1\)), obtain an expression for his expected winnings in the form \[ Xa_{1}+Ya_{2}+Za_{3}. \] Show that he can choose \(b_{1},b_{2}\) and \(b_{3}\) such that \(X,Y\) and \(Z\) are all non-negative. Deduce that, whatever values for \(a_{j}\) are chosen by A, B can ensure that in the long run he loses no money.

Solution:

  1. Suppose A always plays \(1\), then B should always play \(2\) and every time they will win 1. Suppose A always plays \(2\) then B should always play \(3\) and every time they will win 1. If A always plays \(3\) then B should always play \(1\) and every time they will win 2.
  2. \begin{array}{cccc} & b_1 & b_2 & b_3 \\ \frac13 & (0, \frac{b_1}{3}) & (1, \frac{b_2}{3}) & (-2, \frac{b_3}{3}) \\ \frac13 & (-1, \frac{b_1}{3}) & (0, \frac{b_2}{3}) & (1, \frac{b_3}{3}) \\ \frac13 & (2, \frac{b_1}{3}) & (-1, \frac{b_2}{3}) & (0, \frac{b_3}{3}) \\ \end{array} Therefore the expected value is: \(\frac{b_1}{3} - \frac{b_3}{3}\) and to maximise this he should always guess \(1\) (ie \(b_1 = 1, b_2 = 0, b_3 = 0\).)
  3. \begin{array}{cccc} & b_1 & b_2 & b_3 \\ a_1 & (0, a_1b_1) & (1, a_1b_2) & (-2, a_1b_3) \\ a_2 & (-1, a_2b_1) & (0, a_2b_2) & (1, a_2b_3) \\ a_3 & (2, a_3b_1) & (-1, a_3b_2) & (0, a_3b_3) \\ \end{array} Therefore the expected value is: \((b_2-2b_3)a_1 + (b_3-b_1)a_2 + (2b_1-b_2)a_3\) We need \(b_2 \geq 2b_3, b_3 \geq b_1, 2b_1 \geq b_2\) so \(b_1 \leq b_3 \leq \frac12 b_2 \leq b_1\) so we could take \(b_1 = b_3 = \frac12 b_2\) or \(b_1 = b_3 = \frac14, b_2 = \frac12\) and all values would be \(0\). Therefore by choosing these values \(B\) can guarantee his expected value is \(0\) and therefore shouldn't expect to lose money in the long run.

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1987 Paper 1 Q16
D: 1500.0 B: 1500.0
probability conditional Bayes' theorem normal approximation binomial distribution voting inverse normal

The parliament of Laputa consists of 60 Preservatives and 40 Progressives. Preservatives never change their mind, always voting the same way on any given issue. Progressives vote at random on any given issue.

  1. A randomly selected member is known to have voted the same way twice on a given issue. Find the probability that the member will vote the same way a third time on that issue.
  2. Following a policy change, a proportion \(\alpha\) of Preservatives now consistently votes against Preservative policy. The Preservative leader decides that an election must be called if the value of \(\alpha\) is such that, at any vote on an item of Preservative policy, the chance of a simple majority would be less than 80\%. By making a suitable normal approximation, estimate the least value of \(\alpha\) which will result in an election being called.

Solution:

  1. The vote is will now be \(60(1-\alpha)\) for, \(60\alpha\) against and \(X \sim B(40, \frac12)\) at random between those. For a majority, they need \(60(1-\alpha) + X > 50\), ie \(\P(X > 60\alpha - 10) \geq 0.8\). Using a normal approximation to the binomial, we need \(X \approx N(20, 10)\), so \begin{align*} \P(X > 60 \alpha - 10) &= 1- \P(X \leq 60 \alpha - 10) \\ &\approx 1 - \P(\sqrt{10}Z+20 \leq 60\alpha - 10.5) \\ &\approx 1 - \P(Z \leq \frac{60\alpha - 30.5}{\sqrt{10}}) \end{align*} If we want this to be less than \(0.2\) we need \( \frac{60\alpha - 30.5}{\sqrt{10}} < -0.8416 \Rightarrow \alpha < 0.4639\). This would correspond to 27 or fewer exiles or 33 or more remaining preservatives. [Actual computations using Binomial distribution shows we should expect at least 17 to randomly join 20% of the time, so 34 preservatives are required]

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