Problems

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Solution:

1. \(X_1 \sim B(k, \tfrac12)\) so \(\E[X_1] = \frac{k}{2}\) Note that \(X_2 | X_1 = x_1 \sim B(x_1, \tfrac12)\) so \(\E[X_2 | X_1 = x_1) = \frac{x_1}{2}\) or \(\E[X_2 | X_1] = \frac12 X_1\). Therefore by the tower law, \(\E[\E[X_2|X_1]] = \E[\frac12 X_1] = \frac14k\) Notice also that \(\E[X_n] = \frac1{2^n} k\) and so \begin{align*} && \sum_{i=1}^\infty \E[X_i] &= \sum_{i=1}^{\infty} \frac1{2^i} k \\ &&&= \frac{\frac12 k}{1-\frac12} = k \end{align*}
2. Note that \(Y_1 \sim Geo(\tfrac12)-1\) which has generating function \(\E[t^{Y_1}] = \E[t^{G-1}] = \frac{\frac12 t}{1-(1-\frac12)t}\frac1{t} = \frac{\frac12}{1-\frac12t}\). Notice that \begin{align*} && \E \left [ t^Y \right] &= \E \left [ t^{\sum_{i=1}^kY_i} \right] \\ &&&= \prod_{i=1}^k \E[t^{Y_i}] \\ &&&= \frac{1}{(2-t)^k} \end{align*} Therefore \(\E[Y] = G'(1) = k(2-1)^{-(k+1)} = k\) \(\E[Y^2] = (tG'(t))'|_{t=1} = k(k+1)(2-1)^{-(k+2)}+k(2-1)^{-(k+1)} = k^2+2k\) so \(\var[Y] = k^2+2k - k^2 =2 k\). Finally \(\mathbb{P}(Y=r) = \binom{k+r-1}{k} \frac{1}{2^{r+k}}\)

[Note: this second distribution is a negative binomial distribution]

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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*}

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Solution: The the number of people being directed towards the buses (each cruise) is \(X \sim B(1024, \tfrac12) \approx N(512, 256) \approx 16Z + 512\). Therefore without an extra bus, the expected profit is \(\mathbb{E}[\min(X, 15 \times 32)]\). With the extra bus, the extra profit is \(\mathbb{E}[\min(X, 16 \times 32)]\), therefore the expected extra profit is: \(\mathbb{E}[\min(X, 16 \times 32)]-\mathbb{E}[\min(X, 15 \times 32)] = \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \) \begin{align*} \text{Expected extra profit} &= \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \\ &= \mathbb{E}[\min(16Z+512, 16 \times 32)-\min(16Z+512, 15 \times 32)] \\ &= 16\mathbb{E}[\min(Z+32, 32)-\min(Z+32, 30)] \\ &=16\int_{-\infty}^{\infty} \left (\min(Z+32, 32)-\min(Z+32, 30) \right)p_Z(z) \d z \\ &= 16 \left ( \int_{-2}^{0} (z+32-30) p_Z(z) \d z + \int_0^\infty (32-30)p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} (z+2) p_Z(z) \d z + \int_0^\infty 2p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} zp_Z(z) \d z + 2\int_{-2}^\infty p_Z(z) \d z \right) \\ &= 16 \left ( \int_{-2}^{0} z \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \d z + 2(1-\Phi(2)) \right) \\ &= 32(1-\Phi(2)) + \frac{16}{\sqrt{2\pi}} \left [ -e^{-\frac12z^2} \right]_{-2}^0 \\ &= 32(1-\Phi(2)) - \frac{16}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \end{align*} Across \(50\) different runs, this profit is \[ 1600(1-\Phi(2)) - \frac{800}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \]

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Solution: First notice that the the probability a page contains fewer than two errors is \(\mathbb{P}(X < 2)\) where \(X \sim Po(\lambda)\), ie \(\mathbb{P}(X<2) = e^{-\lambda} + \lambda e^{-\lambda} = (1+\lambda)e^{-\lambda}\). Therefore the number of pages \(Y\) with fewer than two errors out of our sample of \(n\) is \(Bin(n, p)\) where \(p\) is as before. ie \(\mathbb{P}(Y = k) = \binom{n}{k} p^kq^{n-k}\).

1. \(\,\) \begin{align*} && q &= 1- p = 1-(1+\lambda)e^{-\lambda} \\ &&&= 1 - (1+ \lambda)(1 - \lambda + \tfrac12 \lambda^2 + o(\lambda^3)) \\ &&&= 1 - 1+ \lambda - \lambda+\lambda^2 - \tfrac12 \lambda^2 + o(\lambda^3) \\ &&&= \tfrac12 \lambda^2 + o(\lambda^3) \end{align*}
2. \(\,\) \begin{align*} && \mathbb{P}(Y = n) &= p^n \\ &&&= (1+\lambda)^ne^{-\lambda n} \\ &&&= (1 + n \lambda + \frac{n(n-1)}{2} \lambda^2 + \cdots)(1 - \lambda n + \frac{\lambda^2 n^2}{2} + \cdots) \\ &&&= 1 + 0 \lambda + \left ( \frac{n(n-1)}{2} + \frac{n^2}{2} - n^2 \right) \lambda^2 + o(\lambda^3) \\ &&&= 1 - \frac{n}{2} \lambda^2 + o(\lambda^3) \end{align*} So if \(\frac{n}{2} \lambda \leq 1\) or \(n \leq \frac{2}{\lambda}\) \(\mathbb{P}(Y = n) \leq 1- \lambda\).
3. \(\,\) \begin{align*} && \mathbb{P}(Y > 1 | Y > 0) &= \frac{1-(q^n + npq^{n-1})}{1-q^n} \\ &&&= 1 - \frac{npq^{n-1}}{1-q^n} \\ &&&= 1 -n \frac{(1+ \lambda)e^{-\lambda} (\tfrac12 \lambda^2 + o(\lambda^3))^{n-1}}{1-(\tfrac12 \lambda^2 + o(\lambda^3))^n} \\ &&&= 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} \frac{(1+ \lambda)(1-\lambda + \lambda^2/2 - \cdots)(1+o(\lambda)^{n-1}}{1-(\tfrac12 \lambda^2 + o(\lambda^3))^n} \\ &&&= 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} (1 + o(\lambda)) \\ &&&\approx 1 - n \left (\frac{\lambda^2}{2} \right)^{n-1} \end{align*}

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Solution: Note that in the first \(10\)-short period the number of goals scored by each team is \(B(5, \p_i)\). For them to be equal they must both have scored the same number of goals, ie \begin{align*} && \alpha &= \sum_{i=0}^5 \mathbb{P}(\text{both teams score }5-i) \\ &&&= \sum_{i=0}^5 \binom{5}{i} (1-p_A)^ip_A^{5-i} \binom{5}{i} (1-p_B)^i p_B^{5-i} \\ &&&= \sum_{i=0}^5 \binom{5}{i} ^2(1-p_A)^i (1-p_B)^i p_A^{5-i} p_B^{5-i} \\ \end{align*} Suppose we make it to the end of the shoot out with scores tied. The probability that we finish each round is \(p_A(1-p_B) + p_B(1-p_A)\) (the probability \(A\) wins or \(B\) wins). This is \(p_A + p_B - 2p_Ap_B = \beta\)). Therefore the number of additional rounds is geometric with parameter \(\beta\) and the expected number of rounds is \(\frac{1}{\beta}\). Each round has two shots, and there is a probability \(\alpha\) of this occuring, ie \(\frac{2\alpha}{\beta}\). Added to the \(10\) guaranteed shots we get the desired result

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Solution: Notice that we can see the total number generated as \(X_n \sim B(2X_{n-1},p)\), since a Binomial is a sum of independent Bernoullis, and there are two Bernoullis per particle. \begin{align*} && \mathbb{P}(X_1=2 \mbox { and } X_2=2) &= \underbrace{p^2}_{\text{two generated in first iteration}} \cdot \underbrace{\binom{4}{2}p^2q^2}_{\text{two generated from the first two}} \\ &&&= 6p^4q^2 \end{align*} \begin{align*} && \mathbb{P})(X_1 = 2 |X_2 = 2) &= \frac{ \mathbb{P}(X_1=2 \mbox { and } X_2=2) }{ \mathbb{P}( X_2=2) } \\ &&&= \frac{6p^4q^2}{6p^4q^2+2pq \cdot p^2} \\ &&&= \frac{3pq}{3pq+1} \\ \Rightarrow && \frac{9}{25} &= \frac{3pq}{3pq+1} \\ \Rightarrow && 27pq + 9 &= 75pq \\ \Rightarrow && 9 &= 48pq \\ \Rightarrow && pq &= \frac{3}{16} \\ \Rightarrow && 0 &= p^2 - p + \frac3{16} \\ \Rightarrow && p &= \frac14, \frac34 \end{align*} By the same reasoning about the Bernoullis, we must have \(\E[X_n] = \E[\E[X_n | X_{n-1}]] = \E[2pX_{n-1}] = 2p \E[X_{n-1}]\) therefore \(\E[X_n] = (2p)^n\). If \(p = \frac14\) then \(\E[X_n] = \frac1{2^n} \to 0\) If \(p = \frac34\) then \(\E[X_n] = \left(\frac32 \right)^n \to \infty\)

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Solution: Let \(H_i\) be the random variable of how many heads the \(i\)th coin throws on a given day. Then \(H_i \sim B(M,p)\), and the probability that a given coin produces fewer than \(m\) heads is \(p_h = \P(H_i < m)\) Let \(C\) be the random variable the number of coins producing fewer than \(m\) heads, then \(C \sim B(L, p_h)\). The probability that more than \(l\) of the coins produce fewer than \(m\) heads is therefore \(\P(C > l)\). Finally, the probability that on exactly \(k\) days more than \(l\) of the coins will produce fewer than \(m\) heads is: \[ \binom{K}{k} \cdot \P(C > l)^k \cdot (1-\P(C > l))^{K-k} \] Let's start by assuming that all our Binomials can be approximated by a normal distribution. \(B(M,p) \approx N(Mp, Mp(1-p))\) and so: \begin{align*} p_h &= \P(H_i < m) \\ &\approx \P( \sqrt{Mp(1-p)}Z+Mp < m-\frac12) \\ &= \P \l Z < \frac{2m-2Mp-1}{2\sqrt{Mp(1-p)}} \r \\ &= \Phi\l\frac{2m-2Mp-1}{2\sqrt{Mp(1-p)}} \r \end{align*} \(B(L, p_h) \approx B \l L, \P \l Z < \frac{2m-2Mp-1}{2\sqrt{Mp(1-p)}} \r\r = B(L, \frac{h}{L}) \approx N(h, \frac{h(L-h)}{L})\) Therefore \begin{align*} \P(C > l) &= 1-\P(C \leq l) \\ &\approx 1- \P \l \sqrt{\frac{h(L-h)}{L}} Z + h \leq l+\frac12 \r \\ &= 1 - \P \l Z \leq \frac{2l-2h+1}{2\sqrt{\frac{h(L-h)}{L}}}\r \\ &= 1- \Phi\l \frac{2l-2h+1}{2\sqrt{\frac{h(L-h)}{L}}} \r \\ &= \Phi\l \frac{2h-2l-1}{2\sqrt{\frac{h(L-h)}{L}}} \r \end{align*} If we can approximate \(\sqrt{1-\frac{h}{L}}\) by \(1\) then we obtain the approximation in the question. Alternatively, \(B(L, \frac{h}{L}) \approx Po(h)\) and \(Po(h) \approx N(h,h)\) so we obtain: \begin{align*} \P(C > l) &= 1-\P(C \leq l) \\ &\approx 1 - \P(\sqrt{h} Z +h < l + \frac12) \\ &= 1 - \P \l Z < \frac{2l-2h+1}{2\sqrt{h}} \r \\ &= \Phi \l \frac{2h - 2l -1}{2\sqrt{h}}\r \end{align*} as required. [I think this is what the examiners expected]. Considering the case \(K=7\), \(k=2\), \(L=500\), \(l=4\), \(M=100\), \(m=48\) and \(p=0.6\), we have the first normal approximation depends on \(Mp\) and \(M(1-p)\) being large. They are \(60\) and \(40\) respectively, so this is likely a good approximation. The first approximation finds that \begin{align*} h &= 500 \cdot \Phi \l \frac{2 \cdot 48 - 2 \cdot 60 - 1}{2\sqrt{24}} \r \\ &= 500 \cdot \Phi \l \frac{2 \cdot 48 - 2 \cdot 60 - 1}{2\sqrt{24}} \r \\ &= 500 \cdot \Phi \l \frac{-25}{2 \sqrt{24}} \r \\ &\approx 500 \cdot \Phi (-2.5) \\ &= 500 \cdot 0.0062 \\ &\approx 3.1 \end{align*} The second binomial approximation will be good if \(500 \cdot \frac{3.1}{500} = 3.1\) is large, but this is quite small. Therefore, we shouldn't expect this to be a good approximation. However, since \(m = 48\) is far from the mean (in a normalised sense), we might expect the percentage error to be large. [Alternatively, using what I expect the desired approach] The approximation of \(B(L, \frac{h}{L}) \approx Po(h)\) is acceptable since \(n>50\) and \(h < 5\). The approximation of \(Po(h) \sim N(h,h)\) is not acceptable since \(h\) is small (in particular \(h < 15\)) Finally, we can compute all these values exactly using a modern calculator. \begin{array}{l|cc} & \text{correct} & \text{approx} \\ \hline p_h & 0.005760\ldots & 0.005362\ldots \\ \P(C > l) & 0.164522\ldots & 0.133319\ldots \\ \text{ans} & 0.231389\ldots & 0.182516\ldots \end{array} We can also see how the errors propagate, by doing the calculations assuming the previous steps are correct, and also including the Poisson step. \begin{array}{lccc} & \text{correct} & \text{approx} & \text{using approx } p_h \\ \hline p_h & 0.005760\ldots & 0.005362\ldots & - \\ \P(C > l)\quad [Po(h)] & 0.164522\ldots & 0.165044\ldots & 0.134293\ldots \\ \P(C > l)\quad [N(h,h)] & 0.164522\ldots & 0.169953\ldots & 0.133319\ldots \\ \P(C > l)\quad [N(h,h(1-\frac{h}{L})] & 0.164522\ldots & 0.169255\ldots & 0.132677\ldots \\ \text{ans} & 0.231389\ldots & 0.231389\ldots \end{array} By doing this, we discover that the largest errors are actually coming not from approximating the second approximation but from the small absolute (but large relative error) in the first approximation. This is, in fact, a coincidence; we can observe it by investigating the specific values being used. The first approximation looks as follows:

You might not be able to tell, but there's actually two plots on this chart. However, let's zoom in on the area we are worried about: We can see there are small differences, which could be large in percentage terms. (As we found when we computed them directly). First, we can immediately see that if we just look at the distribution of \(B(L, p_h)\) and \(B(L, p_{h_\text{approx}})\) we get quite different results, even before we do any approximations. If we plot the probability distribution of \(B(L, p_h)\) vs \(N(Lp_h, Lp_h(1-p_h))\) we find that it is not a great approximation. However, the CDF happens to be a very good approximation *just* for the value we care about. Very lucky, but not possible for someone sitting STEP to know at the time!

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Solution: \begin{align*} && \var[X+Y] &= \E\left [(X+Y-\E[X+Y])^2 \right] \\ &&&= \E \left [ (X - \E[X] + Y - \E[Y])^2 \right] \\ &&&= \E \left [(X - \E[X])^2 + (Y-\E[Y])^2 + 2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \E \left [(X - \E[X])^2 \right]+\E \left [(Y-\E[Y])^2 \right]+\E \left [2(X-\E[X])(Y-\E[Y]) \right] \\ &&&= \var[X] + \var[Y] + 2 \mathrm{Cov}(X,Y) \end{align*}

1. \(W+L+D = m\) where \(m\) is the number of games, which has variance \(0\). Therefore \(W+L+D\) has variance \(0\).
2. The probability of a decisive game is \(\frac23\) and \(W+L\) is the number of decisive games. Each game is independent so this meets the criteria for a binomial distribution.

Notice \(W+L \sim B(m, \tfrac23)\) and \(W, L, D \sim B(m, \tfrac13)\), in particular \(\var[W+L] = m \tfrac23 \tfrac13 = \tfrac29m\) and \(\var[W] = \var[D] = \var[D] = m \tfrac13 \tfrac13 = \tfrac29m\) \begin{align*} && \var[W+L] &= \var[W] + \var[L] + 2\mathrm{Cov}(W,L) \\ \Rightarrow && \mathrm{Cov}(W,L) &= -\tfrac19m \\ \Rightarrow && \frac{\mathrm{Cov}(W,L) }{\sqrt{\var[W]\var[L]}} &= -\frac12 \end{align*}

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Solution: Let \(X\) be the number of plates broken by the mathematician and \(Y\) by the other student. Then \(X \sim Po(\lambda), Y \sim Po(3\lambda)\) and \(X+Y \sim Po(4\lambda)\) \begin{align*} && \mathbb{P}(X = k | X+Y = n) &= \frac{\mathbb{P}(X = k, Y = n-k)}{\mathbb{P}(X+Y=n)} \\ &&&= \frac{e^{-\lambda} \lambda^k/k! \cdot e^{-3\lambda} (4\lambda)^{n-k}/(n-k)!}{e^{-4\lambda}(4\lambda)^n/n!} \\ &&&= \binom{n}{k} \left ( \frac{1}{4} \right)^k \left ( \frac{3}{5} \right)^{n-k} \end{align*} Therefore \(X | X+Y = n \sim Binomial(n, \tfrac14)\) \begin{align*} \mathbb{P}(X \geq 3 | X + Y = n) &= \binom{5}{3} \frac{3^2}{4^5} + \binom{5}{4} \frac{3}{4^5} + \binom{5}{5} \frac{1}{4^5} \\ &= \frac{1}{4^5} \left ( 90+ 15 + 1 \right) \\ &= \frac{106}{4^5} = \frac{53}{512} \approx \frac1{10} \end{align*}

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Solution: There are \(N-1\) gaps between the magnets which are independently gaps or not gaps. Therefore the total number of gaps is \(X \sim Binomial(N-1, \frac12)\) and \begin{align*} \mathbb{E}(X) &= \frac{N-1}{2} \\ \textrm{Var}(X) &= \frac{N-1}{4} \end{align*}