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2010 Paper 1 Q12
D: 1500.0 B: 1508.1
probability discrete random variable expectation geometric distribution tail sum formula inequality optimisation coupon collector

A discrete random variable \(X\) takes only positive integer values. Define \(\E(X)\) for this case, and show that \[\E(X) =\sum^{\infty}_{n=1}\P\left(X\ge n \right).\] I am collecting toy penguins from cereal boxes. Each box contains either one daddy penguin or one mummy penguin. The probability that a given box contains a daddy penguin is \(p\) and the probability that a given box contains a mummy penguin is \(q\), where \(p\ne0\), \(q\ne0\) and \(p+q=1\,\). Let \(X\) be the number of boxes that I need to open to get at least one of each kind of penguin. Show that \(\P(X\ge 4)= p^{3}+q^{3}\), and that \[ \E(X)=\frac{1}{pq}-1.\, \] Hence show that \(\E(X)\ge 3\,\).

Solution: \[ \E[X] := \sum_{n=1}^{\infty} n \mathbb{P}(X=n) \] \begin{align*} && \sum^{\infty}_{n=1}\mathbb{P}\left(X\ge n \right) &= \sum^{\infty}_{n=1}\sum_{k=n}^\infty \mathbb{P}(X=k) \\ &&&= \sum_{k=1}^\infty k \cdot \mathbb{P}(X=k) \\ &&&= \E[X] \end{align*} \begin{align*} &&\mathbb{P}(X \geq 4) &= \mathbb{P}(\text{first 3 are daddies}) +\mathbb{P}(\text{first 3 are mummies}) \\ &&&= p^3 + q^3 \\ \Rightarrow && \E[X] &= \sum_{n=1}^{\infty} \mathbb{P}\left(X\ge n \right) \\ &&&= 1+\sum_{n=2}^{\infty} \left ( p^{n-1} + q^{n-1}\right) \\ &&&= 1+\frac{p}{1-p} + \frac{q}{1-q} \\ &&&= 1+\frac{p}q + \frac{q}p \\ &&&= 1+\frac{p^2+q^2}{pq} \\ &&&= 1+\frac{(p+q)^2-2pq}{pq} \\ &&&= \frac{1}{pq} -1 \\ &&& \underbrace{\geq}_{AM-GM} \frac{1}{4}-1 = 3 \end{align*}

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2010 Paper 1 Q13
D: 1484.0 B: 1516.0
Poisson distribution exponential waiting time quadratic in exponential probability independent random variables inequality roots of equations

The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean \(\lambda\) texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is \(p\), show that \[ p\e^{2\lambda}-\e^{\lambda}+1=0. \] Given that \(4p<1\), show that there are two positive values of \(\lambda\) that satisfy this equation. The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means \(\lambda_{1}\) and \(\lambda_{2}\) texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also \(p\), find an expression for \(\lambda_{1}+\lambda_{2}\) in terms of \(p\). Find the probability, in terms of \(p\), that she waits between 1 and 2 hours in the morning to receive her first text.

Solution: Let \(X_t\) be the number of texts he recieves before \(t\) hours. So \(X_t \sim P(t\lambda)\) \begin{align*} &&\mathbb{P}(X_1 = 0 \, \cap \, X_2 > 0) &= e^{-\lambda} \cdot \left ( 1-e^{-\lambda}\right) = p \\ \Rightarrow && e^{2\lambda}p &= e^{\lambda} - 1 \\ \Rightarrow && 0 &= pe^{2\lambda}-e^{\lambda} + 1 \\ \Rightarrow && e^{\lambda} &= \frac{1 \pm \sqrt{1-4p}}{2p} \end{align*} Which clearly has two positive roots if \(4p < 1\). We need to show both roots are \(>1\). So considering the smaller one we are looking at: \begin{align*} && \frac{1-\sqrt{1-4p}}{2p} & > 1 \\ \Leftrightarrow && 1-\sqrt{1-4p} &> 2p \\ \Leftrightarrow && 1-2p&> \sqrt{1-4p} \\ \Leftrightarrow && (1-2p)^2&> 1-4p \\ \Leftrightarrow && 1-4p+4p^2&> 1-4p \\ \end{align*} which is clearly true. We must have \(e^{\lambda_1}\cdot e^{\lambda_2} = \frac{1}{p}\), so \(\lambda_1 + \lambda_2 = -\ln p\) by considering the product of the roots in our quadratic. (Vieta). Therefore the probability she waits between 1 and 2 hours in the morning is \(e^{-(\lambda_1 + \lambda_2)} \cdot ( 1- e^{-(\lambda_1+\lambda_2)}) = p \cdot (1-p)\)

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2010 Paper 2 Q12
D: 1600.0 B: 1486.8
probability cumulative distribution functions piecewise PDF expectation median inequality continuous random variable

The continuous random variable \(X\) has probability density function \(\f(x)\), where \[ \f(x) = \begin{cases} a & \text {for } 0\le x < k \\ b & \text{for } k \le x \le 1\\ 0 & \text{otherwise}, \end{cases} \] where \(a > b > 0\) and \(0 < k < 1\). Show that \(a > 1\) and \(b < 1\).

  1. Show that \[ \E(X) = \frac{1-2b+ab}{2(a-b)}\,. \]
  2. Show that the median, \(M\), of \(X\) is given by \(\displaystyle M=\frac 1 {2a}\) if \(a+b\ge 2ab\) and obtain an expression for the median if \(a+b\le 2ab\).
  3. Show that \(M < \E(X)\,\).

Solution: \begin{align*} && 1 &= \int_0^1 f(x) \d x \\ &&&= ak + b(1-k) \\ &&&= b + (a-b)k \\ \Rightarrow && k &= \frac{1-b}{a-b} \\ \Rightarrow && b & < 1 \tag{\(0 < k, \,a > b\)} \\ && k &> 1 \\ \Rightarrow && a-b & > 1-b \\ \Rightarrow && a > 1 \end{align*}

  1. \(\,\) \begin{align*} && \E[X] &= \int_0^1 x \cdot f(x) \d x \\ &&&= \int_0^k ax \d x + \int_k^1 b x \d x \\ &&&= a \frac{k^2}{2} + b \frac{1-k^2}{2} \\ &&&= \frac12b + (a-b) \frac{(1-b)^2}{2(a-b)^2} \\ &&&= \frac{(1-b)^2+b(a-b)}{2(a-b)} \\ &&&= \frac{1-2b+ab}{2(a-b)} \end{align*}
  2. \(\,\) The median \(M\) satisfies \[\frac12 = \int_0^M f(x) \d x \] If \(ka = \frac{a-ab}{a-b} \leq \frac12 \Leftrightarrow 2a-2ab \leq a-b \Leftrightarrow a+b \leq 2ab\) then \(M > k\) otherwise \(M < k\). In the latter case: \begin{align*} && \frac12 &= Ma \\ \Rightarrow && M &= \frac{1}{2M} \end{align*} In the former case \begin{align*} && \frac12 &= ka + (M-k)b \\ &&&= k(a-b) + Mb \\ &&&= 1-b + M b \\ \Rightarrow && M &= 1-\frac1{2b} \end{align*}

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2010 Paper 2 Q13
D: 1600.0 B: 1502.2
probability tree diagram optimisation inequality combinatorics strategy comparison complementary counting

Rosalind wants to join the Stepney Chess Club. In order to be accepted, she must play a challenge match consisting of several games against Pardeep (the Club champion) and Quentin (the Club secretary), in which she must win at least one game against each of Pardeep and Quentin. From past experience, she knows that the probability of her winning a single game against Pardeep is \(p\) and the probability of her winning a single game against Quentin is \(q\), where \(0 < p < q < 1\).

  1. The challenge match consists of three games. Before the match begins, Rosalind must choose either to play Pardeep twice and Quentin once or to play Quentin twice and Pardeep once. Show that she should choose to play Pardeep twice.
  2. In order to ease the entry requirements, it is decided instead that the challenge match will consist of four games. Now, before the match begins, Rosalind must choose whether to play Pardeep three times and Quentin once (strategy 1), or to play Pardeep twice and Quentin twice (strategy 2) or to play Pardeep once and Quentin three times (strategy 3). Show that, if \(q-p > \frac 12\), Rosalind should choose strategy 1. If \(q-p<\frac12\), give examples of values of \(p\) and \(q\) to show that strategy 2 can be better or worse than strategy 1.

Solution:

  1. If she plays \(P\) twice her probability is \(q(p^2+2p(1-p)) = qp(2-p)\). If she plays \(Q\) twice her probability is \(pq(2-q)\). Since \(p < q\) she should play \(P\) twice.
  2. Under strategy 1, her probability is \(q(p^3+3p^2(1-p)+3p(1-p)^2) = qp(p^2+3p-3p^2+3-6p+3p^2) = qp(3-3p+p^2)\) Under strategy 2 her probability is \((p^2+2p(1-p))(q^2+2q(1-q)) = pq(2-p)(2-q)\). Under strategy 3 her probability is \(qp(3-3q+q^2)\) \begin{align*} && q - p &> \frac12 \\ \Rightarrow && (2-p)(2-q) & < (2-p)(\frac32 - p) \\ &&&= 3 - \frac72p + p^2 \\ &&&< 3- 3p + p^2 \end{align*} Therefore Strategy 1 dominates if \(q-p > \frac12\). If \(p = \frac14, q = \frac12\) then \((2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}\) and \(3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}\) so strategy 2 dominates. Notice that strategy 1 always dominates strategy 3 since \(f(x) = 3-3x+x^2\) is decreasing for \(x < 1.5\). If \(p = \frac14, q = \frac12\) then \((2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}\) and \(3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}\) so strategy 2 dominates. For strategy 1 to dominate, we need \(3-3p+p^2 > (2-q)(2-p)\) or \(\frac{3-3p+p^2}{2-p} > 2-q\). When \(p = \frac12\) this is \(\frac{3-\frac32 + \frac14}{2 - \frac12} = \frac{\frac{7}{4}}{\frac{3}{2}} = \frac76 = 2-\frac{5}{6}\) so take any value of \(q\) larger than \(\frac56\).

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2010 Paper 3 Q12
D: 1700.0 B: 1500.0
geometric distribution infinite series arithmetico-geometric series expected value probability alternating turns proof summing series

The infinite series \(S\) is given by \[ S = 1 + (1 + d)r + (1 + 2d)r^2 + \cdots + (1+nd)r^n +\cdots\; ,\] for \(\vert r \vert <1\,\). By considering \(S - rS\), or otherwise, prove that \[ S = \frac 1{1-r} + \frac {rd}{(1-r)^2} \,.\] Arthur and Boadicea shoot arrows at a target. The probability that an arrow shot by Arthur hits the target is \(a\); the probability that an arrow shot by Boadicea hits the target is \(b\). Each shot is independent of all others. Prove that the expected number of shots it takes Arthur to hit the target is \(1/a\). Arthur and Boadicea now have a contest. They take alternate shots, with Arthur going first. The winner is the one who hits the target first. The probability that Arthur wins the contest is \(\alpha\) and the probability that Boadicea wins is \(\beta\). Show that \[ \alpha = \frac a {1-a'b'}\,, \] where \(a' = 1-a\) and \(b'=1-b\), and find \(\beta\). Show that the expected number of shots in the contest is \(\displaystyle \frac \alpha a + \frac \beta b\,.\)

Solution: Notice that \begin{align*} && S - rS &= 1 + dr + dr^2 + \cdots \\ &&&= 1 + dr(1 + r+r^2+ \cdots) \\ &&&= 1 + \frac{rd}{1-r} \\ \Rightarrow && S &= \frac{1}{1-r} + \frac{rd}{(1-r)^2} \end{align*} The number of shots Arthur takes is \(\textrm{Geo}(a)\), so it's expectation is \(1/a\). The probability Arthur wins is: \begin{align*} \alpha &= a + a'b'a + (a'b')^2a + \cdots \\ &= a(1+a'b' + \cdots) \\ &= \frac{a}{1-a'b'} \\ \\ \beta &= a'b + a'b'a'b + \cdots \\ &= a'b(1+b'a' + (b'a')^2 + \cdots ) \\ &= \frac{a'b}{1-a'b'} \end{align*} The expected number of shots in the contest is: \begin{align*} E &= a + 2a'b + 3a'b'a + 4a'b'a'b + \cdots \\ &= a(1 + 3a'b' + 5(a'b')^2 + \cdots) + 2a'b(1 + 2(a'b') + 3(a'b')^2 + \cdots) \\ &= a \left ( \frac{1}{1-a'b'} + \frac{2a'b'}{(1-a'b')^2} \right) + 2a'b \left ( \frac{1}{1-a'b'} + \frac{a'b'}{(1-a'b')^2}\right) \\ &= \frac{a}{1-a'b'} \left (1 + \frac{2a'b'}{(1-a'b')} \right) + 2\frac{a'b}{1-a'b'} \left ( 1 + \frac{a'b'}{(1-a'b')}\right) \\ &= \alpha \frac{1+a'b'}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ &= \alpha \frac{1+1-a-b+ab}{1-a'b'} + \beta \frac{2}{1-a'b'} \\ \end{align*}

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2009 Paper 1 Q12
D: 1500.0 B: 1501.5
probability tree diagram inequality AM-GM inequality combinatorics algebraic proof symmetric functions

Prove that, for any real numbers \(x\) and \(y\), \(x^2+y^2\ge2xy\,\).

  1. Carol has two bags of sweets. The first bag contains \(a\) red sweets and \(b\) blue sweets, whereas the second bag contains \(b\) red sweets and \(a\) blue sweets, where \(a\) and \(b\) are positive integers. Carol shakes the bags and picks one sweet from each bag without looking. Prove that the probability that the sweets are of the same colour cannot exceed the probability that they are of different colours.
  2. Simon has three bags of sweets. The first bag contains \(a\) red sweets, \(b\) white sweets and \(c\) yellow sweets, where \(a\), \(b\) and \(c\) are positive integers. The second bag contains \(b\) red sweets, \(c\) white sweets and \(a\) yellow sweets. The third bag contains \(c\) red sweets, \(a\) white sweets and \(b\) yellow sweets. Simon shakes the bags and picks one sweet from each bag without looking. Show that the probability that exactly two of the sweets are of the same colour is \[ \frac {3(a^2b+b^2c+c^2a+ab^2 + bc^2 +ca^2)}{(a+b+c)^3}\,, \] and find the probability that the sweets are all of the same colour. Deduce that the probability that exactly two of the sweets are of the same colour is at least 6 times the probability that the sweets are all of the same colour.

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2009 Paper 1 Q13
D: 1500.0 B: 1504.1
probability discrete probability distributions expectation combinatorics arrangements counting arguments permutations

I seat \(n\) boys and \(3\) girls in a line at random, so that each order of the \(n+3\) children is as likely to occur as any other. Let \(K\) be the maximum number of consecutive girls in the line so, for example, \(K=1\) if there is at least one boy between each pair of girls.

  1. Find \(\P(K=3)\).
  2. Show that \[\P(K=1)= \frac{n(n-1)}{(n+2)(n+3)}\,. \]
  3. Find \(\E(K)\).

Solution:

  1. If all the girls are say together there are \(n+1\) ways to place the block of 3 girls. There are \(\binom{n+3}{3}\) ways to choose where to place the girls in total, therefore: \begin{align*} && \mathbb{P}(K =3) &= \frac{n+1}{\binom{n+3}3} \\ &&&= \frac{6(n+1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{6}{(n+3)(n+2)} \end{align*}
  2. If \(K= 1\) then all of the girls are separated. We can place three girls and two boys separating them, then we are allocating \(N-2\) boys to \(4\) gaps, ie \(\binom{N-2+3}{3} = \binom{N+1}{3}\). \begin{align*} && \mathbb{P}(K=3) &= \frac{\binom{n+1}{3}}{\binom{n+3}{3}} \\ &&&= \frac{(n+1)n(n-1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{n(n-1)}{(n+3)(n+2)} \end{align*}
  3. \(\,\) \begin{align*} \mathbb{E}(K) &= \sum_{k=1}^3 k \mathbb{P}(K=k) \\ &= \frac{6}{(n+3)(n+2)} + 2 \left (1 - \frac{6}{(n+3)(n+2)} - \frac{n(n-1)}{(n+3)(n+2)} \right) + 3\frac{n(n-1)}{(n+3)(n+2)} \\ &= 2+\frac{6-12+n(n-1)}{(n+3)(n+2)} \\ &= 2 + \frac{n^2-n-6}{(n+2)(n+3)}\\ &= 2 + \frac{(n-3)(n+2)}{(n+2)(n+3)} \\ &= 2 + \frac{n-3}{n+3} \\ &= \frac{2n}{n+3} \end{align*}

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2009 Paper 2 Q13
D: 1600.0 B: 1500.0
probability tree diagram binomial distribution expected value polynomial equation engine failure cost optimisation

Satellites are launched using two different types of rocket: the Andover and the Basingstoke. The Andover has four engines and the Basingstoke has six. Each engine has a probability~\(p\) of failing during any given launch. After the launch, the rockets are retrieved and repaired by replacing some or all of the engines. The cost of replacing each engine is \(K\). For the Andover, if more than one engine fails, all four engines are replaced. Otherwise, only the failed engine (if there is one) is replaced. Show that the expected repair cost for a single launch using the Andover is \[ 4Kp(1+q+q^2-2q^3) \ \ \ \ \ \ \ \ \ \ \ \ \ (q=1-p) \tag{*} \] For the Basingstoke, if more than two engines fail, all six engines are replaced. Otherwise only the failed engines (if there are any) are replaced. Find, in a form similar to \((*)\), the expected repair cost for a single launch using the Basingstoke. Find the values of \(p\) for which the expected repair cost for the Andover is \(\frac23\) of the expected repair cost for the Basingstoke.

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2008 Paper 1 Q12
D: 1516.0 B: 1484.0
probability discrete random variables order statistics expectation median summation maximum of random variables approximation

In this question, you may use without proof the results: \[ \sum_{r=1}^n r = \tfrac12 n(n+1) \qquad\text{and}\qquad \sum_{r=1}^n r^2 = \tfrac1 6 n(n+1)(2n+1)\,. \] The independent random variables \(X_1\) and \(X_2\) each take values \(1\), \(2\), \(\ldots\), \(N\), each value being equally likely. The random variable \(X\) is defined by \[ X= \begin{cases} X_1 & \text { if } X_1\ge X_2\\ X_2 & \text { if } X_2\ge X_1\;. \end{cases} \]

  1. Show that \(\P(X=r) = \dfrac{2r-1}{N^2}\,\) for \(r=1\), \(2\), \(\ldots\), \(N\).
  2. Find an expression for the expectation, \(\mu\), of \(X\) and show that \(\mu=67.165\) in the case \(N=100\).
  3. The median, \(m\), of \(X\) is defined to be the integer such that \(\P(X\ge m) \ge \frac 12\) and \(\P(X\le m)\ge \frac12\). Find an expression for \(m\) in terms of \(N\) and give an explicit value for \(m\) in the case \(N=100\).
  4. Show that when \(N\) is very large, \[ \frac \mu m \approx \frac {2\sqrt2}3\,. \]

Solution: \begin{align*} \P(X = r) &= \P(X_1 = r, X_2 \leq r) + \P(X_2 = r, X_1 < r) \\ &= \P(X_1 = r) \P(X_2 \leq r) + \P(X_2 = r)\P( X_1 < r) \\ &= \frac{1}{N} \frac{r}{N} + \frac{1}{N} \frac{r-1}{N} \\ &= \frac{2r-1}{N^2} \end{align*} \begin{align*} \E(X) &= \sum_{r=1}^N r \P(X = r) \\ &= \sum_{r=1}^N \frac{2r^2 - r}{N^2} \\ &= \frac{1}{N^2} \l \frac{N(N+1)(2N+1)}{3} - \frac{N(N+1)}{2} \r \\ &= \frac{N+1}{N} \l \frac{4N-1}{6} \r \end{align*} When \(N = 100\), this is equal to \(\frac{101 \cdot 399}{6 \cdot 100} = \frac{101 \cdot 133}{200} = 67.165\) \begin{align*} &&\frac12 &\leq \P(X \leq m) \\ &&&=\sum_{r=1}^m \P(X=r) \\ &&&=\sum_{r=1}^m \frac{2r-1}{N^2} \\ &&&= \frac{1}{N^2} \l m(m+1) - m \r \\ &&&= \frac{m^2}{N^2} \\ \Rightarrow && m^2 &\geq \frac{N^2}{2} \\ \Rightarrow && m &\geq \frac{N}{\sqrt{2}} \\ \Rightarrow && m &= \left \lceil \frac{N}{\sqrt{2}} \right \rceil \end{align*} When \(N = 100\), \(100/\sqrt{2} = \sqrt{2}50\). \(\sqrt{2} > 1.4 \Rightarrow 50\sqrt{2} > 70\) \(\sqrt{2} < 1.42 \Rightarrow 50 \sqrt{2} < 71\), therefore \(\displaystyle \left \lceil \frac{100}{\sqrt{2}} \right \rceil = 71\) \begin{align*} \lim_{N \to \infty} \frac{\frac{(N+1)(4N-1)}{6N}}{ \left \lceil\frac{N}{\sqrt{2}} \right \rceil} &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l \frac{4N^2 +3N - 1}{2N^2} \r \tag{since the floor will be irrelevant}\\ &= \lim_{N \to \infty} \frac{\sqrt{2}}{3}\l 2 + \frac{3}{2N} - \frac{1}{N^2} \r \\ &= \lim_{N \to \infty} \frac{2\sqrt{2}}{3} \end{align*}

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2008 Paper 1 Q13
D: 1500.0 B: 1452.7
probability counting circular permutations inclusion-exclusion combinatorics married couples round table

Three married couples sit down at a round table at which there are six chairs. All of the possible seating arrangements of the six people are equally likely.

  1. Show that the probability that each husband sits next to his wife is \(\frac{2}{15}\).
  2. Find the probability that exactly two husbands sit next to their wives.
  3. Find the probability that no husband sits next to his wife.

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2008 Paper 2 Q12
D: 1600.0 B: 1500.0
probability tree diagrams expectation independent events conditional probability quadratic equation

In the High Court of Farnia, the outcome of each case is determined by three judges: the ass, the beaver and the centaur. Each judge decides its verdict independently. Being simple creatures, they make their decisions entirely at random. Past verdicts show that the ass gives a guilty verdict with probability \(p\), the beaver gives a guilty verdict with probability \(p/3\) and the centaur gives a guilty verdict with probability \(p^2\). Let \(X\) be the number of guilty verdicts given by the three judges in a case. Given that \(\E(X)= 4/3\), find the value of \(p\). The probability that a defendant brought to trial is guilty is \(t\). The King pronounces that the defendant is guilty if at least two of the judges give a guilty verdict; otherwise, he pronounces the defendant not guilty. Find the value of \(t\) such that the probability that the King pronounces correctly is \(1/2\).

Solution: \begin{align*} && \mathbb{E}(X) &= p + \frac{p}{3} + p^2 = \frac43p+p^2 \\ \Rightarrow && \frac43 &= \frac43p+p^2 \\ \Rightarrow && 0 &= 3p^2+4p-4 \\ &&&= (3p-2)(p+2) \\ \Rightarrow && p &= \frac23, -2 \end{align*} Since \(p \in [0,1]\) we must have \(p = \frac23\). \begin{align*} && \mathbb{P}(\text{correct verdict}) &= t p+ (1-t) (1-p) \\ &&&= t(2p-1)+(1-p)\\ \Rightarrow && \frac12 &= t(2p-1)+(1-p) \\ \Rightarrow && t &= \frac{\frac12-(1-p)}{2p-1} \\ &&&= \frac{2p-1}{2(2p-1)} = \frac12 \end{align*} (so it doesn't depend at all on what the judges are doing, the only way to be fair is if the trials happen at random!)

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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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2008 Paper 3 Q13
D: 1700.0 B: 1500.0
probability expected value variance indicator random variables linearity of expectation harmonic series combinatorics discrete probability distributions

A box contains \(n\) pieces of string, each of which has two ends. I select two string ends at random and tie them together. This creates either a ring (if the two ends are from the same string) or a longer piece of string. I repeat the process of tying together string ends chosen at random until there are none left. Find the expected number of rings created at the first step and hence obtain an expression for the expected number of rings created by the end of the process. Find also an expression for the variance of the number of rings created. Given that \(\ln 20 \approx 3\) and that \(1+ \frac12 + \cdots + \frac 1n \approx \ln n\) for large \(n\), determine approximately the expected number of rings created in the case \(n=40\,000\).

Solution: Let \(X_i\) be the indicator variable a loop is formed when there are \(i\) strings in the bag, so \(\mathbb{P}(X_i = 1) = \frac{1}{2i-1}\). Therefore \begin{align*} && Y_n &= X_n + Y_{n-1} \\ && Y_n &= X_n + \cdots + X_1 \\ \Rightarrow && \E[Y_n] &= \frac{1}{2n-1} + \frac{1}{2n-3} + \cdots + \frac{1}{1} \\ && \var[Y_n] &= \sum_{i=1}^n \frac{1}{2i-1} \frac{2i-2}{2i-1} \\ &&&= 2\sum_{i=1}^n \frac{i-1}{(2i-1)^2} \end{align*} \begin{align*} && \E[Y_{n}] &= 1 + \frac13 + \cdots + \frac{1}{2n-1} \\ &&&= 1 + \frac12 + \cdots + \frac1{2n} - \frac12\left (1 + \frac12 + \cdots + \frac1n \right) \\ &&&\approx \ln 2n -\frac12 \ln n \\ &&&= \ln 2 \sqrt{n} \\ \\ && \E[Y_{40\,000}] &= \ln 2 \sqrt{40\,000} \\ &&&= \ln 400 \\ &&&= 2 \ln 20 \approx 6 \end{align*}

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2007 Paper 1 Q12
D: 1500.0 B: 1484.0
probability conditional probability without replacement law of total probability symmetry combinatorics transfer problems

  1. A bag contains \(N\) sweets (where \(N \ge 2\)), of which \(a\) are red. Two sweets are drawn from the bag without replacement. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.
  2. There are two bags, each containing \(N\) sweets (where \(N \ge 2\)). The first bag contains \(a\) red sweets, and the second bag contains \(b\) red sweets. There is also a biased coin, showing Heads with probability \(p\) and Tails with probability \(q\), where \(p+q = 1\). The coin is tossed. If it shows Heads then a sweet is chosen from the first bag and transferred to the second bag; if it shows Tails then a sweet is chosen from the second bag and transferred to the first bag. The coin is then tossed a second time: if it shows Heads then a sweet is chosen from the first bag, and if it shows Tails then a sweet is chosen from the second bag. Show that the probability that the first sweet is red is equal to the probability that the second sweet is red.

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2007 Paper 1 Q13
D: 1500.0 B: 1469.5
probability conditional probability combinatorics without replacement counting discrete probability

A bag contains eleven small discs, which are identical except that six of the discs are blank and five of the discs are numbered, using the numbers 1, 2, 3, 4 and 5. The bag is shaken, and four discs are taken one at a time without replacement. Calculate the probability that:

  1. all four discs taken are numbered;
  2. all four discs taken are numbered, given that the disc numbered ``3'' is taken first;
  3. exactly two numbered discs are taken, given that the disc numbered ``3'' is taken first;
  4. exactly two numbered discs are taken, given that the disc numbered ``3'' is taken;
  5. exactly two numbered discs are taken, given that a numbered disc is taken first;
  6. exactly two numbered discs are taken, given that a numbered disc is taken.

Solution: There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.

  1. There are we are choose \(4\) objects in order from \(5\) (ie \({^5\P_4}\)) to obtain valid draws, this is out of a total of picking \(4\) objects from \(11\) (\({^{11}\P_4}\)). Ie the probability is: \(\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}\) Alternatively, there are \(\binom{5}{4}\) ways to choose four numbered discs, out of \(\binom{11}{4}\) ways to choose four discs. ie \(\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}\)
  2. \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\ &= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\ &= \frac{4! \cdot 7! \cdot 11}{11!} \\ &= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\ &= \frac{1}{30} \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac{1}{30} \end{align*} Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in \(\binom{4}{3}\) ways, and there are \(11 \cdot \binom{10}{3}\) was overall. Another alternative using Bayes rule: \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\ &= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\ &= \frac1{30} \end{align*}
  3. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\ &= \frac12 \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac12 \end{align*}
  4. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\ &= \frac{\frac{2}{11}}{\frac4{11}} \\ &= \frac12 \end{align*} Using Bayes rule: \(\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}\) \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\ &= \frac{\frac4{10}{5 / 11}}{4/11} \\ &= \frac{1}{2} \end{align*}
  5. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\ &= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\ &= \frac{1}{2} \end{align*}
  6. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\ &= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\ &= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\ &= \frac{\frac{5}{11}}{\frac{21}{22}} \\ &= \frac{10}{21} \neq \frac12 \end{align*}

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