Problems

---

Solution: \begin{align*} && \E[\text{salary}] &= B\sum_{k=0}^N \frac{1}{1+k}\binom{N}{k}p^k(1-p)^{N-k} \\ \\ && (q+x)^N &= \sum_{k=0}^N \binom{N}{k}x^kq^{N-k} \\ \Rightarrow && \int_0^p(q+x)^N \d x &= \sum_{k=0}^N \binom{N}{k} \frac{p^{k+1}}{k+1}q^{N-k} \\ && \frac{(q+p)^{N+1}-q^{N+1}}{N+1} &= \frac{p}{B} \E[\text{salary}] \\ \Rightarrow && \E[\text{salary}] &= B\frac{1-q^{N+1}}{p(N+1)} \end{align*} Therefore if \(Ap(N+1) < B(1-(1-p)^{N+1})\) the expected value of the new salary is higher. (Whether or not the new salary is worth it in a risk adjusted sense is for the birds). We could model \(X\) by a Poisson random variable if \(N\) is large and \(Np = \lambda \) is small. Suppose \(X \approx Po(\lambda)\) then \begin{align*} \E \left [\frac{B}{1+X} \right] &= B\sum_{k=0}^\infty \frac{1}{1+k}\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \frac{B}{\lambda} \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} \\ &= \frac{B}{\lambda}e^{-\lambda}(e^{\lambda}-1) \\ &= \frac{B(1-e^{-\lambda})}{\lambda} = B \frac{1-e^{-Np}}{Np} \end{align*}

---

Solution: \begin{align*} && \mathbb{P}(k \text{ heads} | n\text{ tosses}) &= \binom{n}k 2^{-n} \\ && \mathbb{P}(1 \text{ head} | n\text{ tosses}) &= n2^{-n} \\ \Rightarrow && \frac{ \mathbb{P}(1 \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(1 \text{ head} | n\text{ tosses}) } &= \frac{n+1}{2n} \end{align*} Which is less than \(1\) unless \(n \geq 1\). Therefore the MLE is \(n = 1\) or \(n= 2\). \begin{align*} \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}}{2 \binom{n}{k}} \\ &= \frac{(n+1)!(n-k)!}{2n!(n+1-k)!} \\ &= \frac{n+1}{2(n+1-k)} \end{align*} This is less than or equal to \(1\) if \(n+1 = 2(n+1-k) \Leftrightarrow n= 2k-1\), therefore the MLEs are \(2k-1\) and \(2k\). If the coin is biased, we have \begin{align*} && \frac{ \mathbb{P}(k \text{ head} | n+1\text{ tosses}) }{ \mathbb{P}(k \text{ head} | n\text{ tosses}) } &= \frac{\binom{n+1}{k}p^kq^{n+1-k}}{\binom{n}{k}p^kq^{n-k}} \\ &&&= \frac{n+1}{(n+1-k)}q \\ \\ && 1 & \geq \frac{n+1}{(n+1-k)}q \\ \Leftrightarrow && (n+1)(1-q) &\geq k \\ \Leftrightarrow && n+1 & \geq \frac{k}{p} \end{align*} Therefore the probability is increasing until \(n+1 \geq \frac{k}{p}\). If \(\frac{k}p\) is an integer the MLEs are \(\frac{k}{p}-1\) and \(\frac{k}p\), otherwise it is \(\lfloor \frac{k}{p} \rfloor\) and the MLE is unique.

---

Solution: \begin{align*} && (1+t)^{n}(1+t)^{n}&=(1+t)^{2n} \\ [t^n]: &&\sum_{k=0}^n \underbrace{\binom{n}{k}}_{t^k\text{ from left bracket}} \underbrace{\binom{n}{n-k}}_{t^{n-k}\text{ from right bracket}} &= \binom{2n}{n} \end{align*}

From each point, we can get to the diagonal ahead of us, so each move only takes us one diagonal closer together. Therefore we can only meet on the diagonal. The number of routes we can meet at is \begin{align*} && R &= \sum_{k=0}^n \underbrace{\binom{n}{k}}_{\text{I go up } k}\underbrace{\binom{n}{n-k}}_{\text{she goes down }n-k} \\ &&&= \binom{2n}{n} \end{align*} Therefore the probability is \(\displaystyle \frac1{2^{2n}} \binom{2n}n\). If I leave \(2k\) minutes late, then we will be attempting meet on a diagonal which is \(2k\) closer to me. The probability this occurs is \begin{align*} && \frac{1}{2^{2n}}\sum_{j=0}^{n-k}\binom{n-k}{j}\binom{n+k}{n-j} &= \frac{1}{2^{2n}}\binom{2n}{n} \end{align*} (by considering the coefficient of \(t^n\) in \((1+t)^{n+k}(1+t)^{n-k} =(1+t)^{2n}\)) This probability is unchanged, because you can consider the two paths as one path by one random person, conditional on them meeting and the delay doesn't change anything. If \(n = 1\) and I leave late, the only way we meet is if we end up walking towards each other down the same street (not at an intersection). This means I need to walk towards the intersection she reaches after the first minute \(\frac12\) and she needs to walk towards me \(\frac12\) so we have probability \(\frac14\)

---

Solution: After \(5\) steps we can get to: \begin{array}{c|c} x & \text{ways} \\ \hline 5 & 1 \text { - go positive every time}\\ 4 & 0 \\ 3 & \binom{5}{1} \text { - go positive every time but 1} \\ 2 &0 \\ \hline & 6 \end{array} Therefore there are \(\frac{6}{2^5} = \frac{3}{16}\) ways to get to \(\{2,3,4,5\}\) After \(6\) steps we can get to: \begin{array}{c|c} x & \text{ways} \\ \hline 5 & 0 \\ 4 & \binom{6}{1} \text { - go positive every time but 1}\\ 3 & 0 \\ 2 & \binom{6}{2} - \text{ - go positive every time but 2} \\ \hline & 21 \end{array} Therefore there are \(\frac{21}{2^6} = \frac{21}{64}\) ways to get to \(\{2,3,4,5\}\) After \(2k\) steps we can reach \(2\) or \(4\). To get to \(2\) we must take \(k+1\) positive steps and \(k-1\) negative steps, ie \(\binom{2k}{k-1}\). To get to \(4\) we must take \(k+2\) positive steps and \(k-2\) negative steps, ie \(\binom{2k}{k-2}\) Therefore there are \(\binom{2k+1}{k-1}\) routes, ie a probability of \(\frac{1}{2^{2k}} \binom{2k+1}{k-1}\) After \(2k+1\) steps we can reach \(3\) or \(5\). To get to \(3\) we must take \(k+2\) positive steps and \(k-1\) negative steps, ie \(\binom{2k+1}{k-1}\). To get to \(5\) we must take \(k+3\) positive steps and \(k-2\) negative steps, ie \(\binom{2k+1}{k-2}\) Therefore there are \(\binom{2k+2}{k-1}\) routes, ie a probability of \(\frac{1}{2^{2k+1}} \binom{2k+2}{k-1}\) To find the maximum of \(P(s)\) notice that \begin{align*} && \frac{P(2k+1)}{P(2k)} &= \frac12 \frac{\binom{2k+2}{k-1}}{\binom{2k+1}{k-1}} \\ &&&= \frac12 \frac{(2k+2)!(k-1)!(k+2)!}{(2k+1)!(k-1)!(k+3)!} \\ &&&= \frac12 \frac{2k+2}{k+3} = \frac{k+1}{k+3} < 1 \end{align*} So we should only look at the even terms. \begin{align*} && \frac{P(2k+2)}{P(2k)} &= \frac14 \frac{\binom{2k+3}{k}}{\binom{2k+1}{k-1}} \\ &&&= \frac14 \frac{(2k+3)!(k-1)!(k+2)!}{(2k+1)!k!(k+3)!} \\ &&&= \frac14 \frac{(2k+3)(2k+2)}{k(k+3)} \\ &&&= \frac{(2k+3)(k+1)}{2k(k+3)} \geq 1 \\ \Leftrightarrow && (2k+3)(k+1) &\geq 2k(k+3) \\ \Leftrightarrow && 2k^2+5k+3 &\geq 2k^2+6k \\ \Leftrightarrow && 3 &\geq k \\ \end{align*} Therefore the maximum is when \(s = 2\cdot 3\) or \(s = 2\cdot 4\) which we computed earlier to be \(\frac{21}{64}\)

---

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*}

---

Solution:

1. \(\binom{m+n}{m} p^m (1-p)^n\)
2. \(\,\) \begin{align*} \mathbb{P}(\text{failed to spot }n\text{ misprints}|\text{spotted }m\text{ misprints}) &= \frac{\mathbb{P}(\text{failed to spot }n\text{ misprints and spotted }m\text{ misprints}) }{\mathbb{P}(\text{spotted }m\text{ misprints})} \\ &= \frac{\binom{m+n}{n}p^m(1-p)^n e^{-\lambda} \lambda^{m+n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}p^m(1-p)^k e^{-\lambda} \lambda^{m+k}/(n+k)!} \\ &= \frac{\binom{m+n}{n}(1-p)^n \lambda^{n}/(n+m)!}{\sum_{k=0}^{\infty} \binom{m+k}{k}(1-p)^k \lambda^{k}/(n+k)!} \\ &= \frac{(1-p)^n \lambda^{n}/n!}{\sum_{k=0}^{\infty} (1-p)^k \lambda^{k}/k!} \\ &= \frac{(1-p)^n\lambda^n}{n!} e^{-(1-p)\lambda} \end{align*} Alternatively, given the missed misprints and spotted misprints are independent, we can view them as both following \(Po(p\lambda)\) and \(Po((1-p)\lambda)\) and so we obtain exactly this result, without calculation.

---

Solution:

1. It cannot overflow on the first day, since it is already \(V\) below the top. The only way it can overflow is if it rains both days. This will occur with probability \(p^2\). The probability it overflows therefore is the probability that bad timing hampers us, ie \(V - u(1+t_2) > 0\) where \(t_2\) is the timing of the rain on day 2 (as a fraction of a day). Ie \(t_2 < \frac{V}{u}-1\). Therefore \begin{align*} && Q &= p^2 \left (\frac{V}{u} - 1 \right) \\ \Rightarrow && u &= \frac{Vp^2}{p^2+Q} \end{align*}
2. The probability the reservoir overflows during this \(18\) days is \(\mathbb{P}(\text{rains more than }k\text{ times})\). The number of times it rains (\(X\)) is \(B(18, \tfrac13)\), since \(18 \cdot \tfrac13 = 6 > 5\) a normal approximation is reasonable, ie \(X \approx N(6, 4)\). We wish to find \(k\) such that \(\mathbb{P}( X > k + 0.5) < \tfrac1{10}\) therefore \(k \approx 1.28 \cdot 2 + 6 - 0.5 \approx 8.1\) so they should set \(k\) to \(9\)
3. In this case we have \(B(36, \tfrac16)\) approximated by \(B(6, 5)\) which has a larger standard deviation, therefore we need to choose a larger value for \(k\). [It turns out to actually be the same, but there's no reason to be able to expect students without a calculator to establish this]

---

Solution: \begin{align*} \mathbb{P}(\text{fever tree grows}) &= \mathbb{P}(0 \leq L \leq 1) + \mathbb{P}(2 \leq L \leq 3) \\ &= \int_0^1 \frac12 -\frac18 x \d x + \int_2^3 \frac12 - \frac18 x \d x \\ &= \left [\frac12 x - \frac1{16}x^2 \right]_0^1+ \left [\frac12 x - \frac1{16}x^2 \right]_2^3 \\ &= \frac12 - \frac1{16}+\frac32-\frac9{16} - 1 + \frac{4}{16} \\ &= \frac58 \end{align*} The expected number of fever trees is just \(96 \cdot \frac58 = 60\).

1. \(f_Y(t)\) must match the distribution for \(L\), but limited to the points we care about, therefore it should be: $f_Y(t) = \begin{cases} ( \frac45 - \frac15t ) & \text{if } t \in [0,1]\cup[2,3] \\ 0 & \text{otherwise} \end{cases}$
   

   + - ↺
   \begin{align*} \mathbb{E}(Y) &= \frac12 \cdot \frac15 (4 - \frac12)+\frac52 \cdot (1 - \frac15 (4 - \frac12)) \\ &= \frac12 \cdot \frac7{10} + \frac52 \cdot \frac3{10} \\ &= \frac{22}{20} \\ &= \frac{11}{10} \end{align*}
2. Given the seeds are blown independently and the wind hasn't changed, it is reasonable to model the number of fever trees as \(B(96, \frac{5}{8})\), it is also acceptable to approximate this using a Normal distribution, ie \(N(60, 22.5)\), \(23\) is \(\frac{23-60}{\sqrt{22.5}}\) is a very negative number, so he should be extremely suspicious.

---

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 \(A\).

---

Solution: \begin{align*} \mathbb{P}(\text{exactly }k\text{ faults after test}) &= \mathbb{P}(k\text{ faults in non-tested, 0 in batch})+\mathbb{P}(k-1\text{ faults in non-tested, 1 in batch}) \\ &=\binom{N}{k}(1-p)^{N-k}p^k\binom{n}{0}(1-p)^n+\binom{N}{k-1}(1-p)^{N-k+1}p^{k-1}\binom{n}{1}(1-p)^{n-1}p \\ &= (1-p)^{N-k+n}p^k \cdot \left ( \binom{N}{k}+n\binom{N}{k-1} \right) \\ &= (1-p)^{N-k+n}p^k \cdot \left (\frac{N!}{k!(N-k)!}+\frac{N!n}{(k-1)!(N-k+1)!}\right) \\ &= (1-p)^{N-k+n}p^k \frac{N!}{k!(N-k+1)!} \cdot \left ((N-k+1)+nk \right) \\ &= \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!k!}p^{k}\left(1-p\right)^{N+n-k} \end{align*} When \(k = N+1\) we get: \begin{align*} \frac{(N+1)n N!}{(N+1)!} p^{N+1}(1-p)^{N+n-k} &= np^{N+1}(1-p)^{N+n-k} \end{align*} and the probability is: \begin{align*} \mathbb{P}(\text{exactly }N+1\text{ faults after test}) &= \mathbb{P}(N\text{ faults in non-tested, 1 in batch}) \\ &= \binom{N}{N}p^N \cdot \binom{n}{1}p(1-p)^{N-1} \\ &= np^{N+1}(1-p)^{N+n-k} \end{align*} So the formula does work for \(k = N+1\). \begin{align*} \mathbb{E}(faults) &= \sum_{k=0}^{N+1} k \cdot \mathbb{P}(\text{exactly }k\text{ faults after test}) \\ &= \sum_{k=0}^{N+1} k \cdot \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!k!}p^{k}\left(1-p\right)^{N+n-k} \\ &= \sum_{k=1}^{N+1} \frac{\left[N+1+k\left(n-1\right)\right]N!}{\left(N-k+1\right)!(k-1)!}p^{k}\left(1-p\right)^{N+n-k} \\ &= \sum_{k=1}^{N+1} \left[N+1+k\left(n-1\right)\right] p(1-p)^{n-1}\binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1} \\ &= p(1-p)^{n-1} \cdot \left ( (N+1+n-1)\sum_{k=1}^{N+1} \binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1}+ (n-1)\sum_{k=1}^{N+1} (k-1)\binom{N}{k-1}p^{k-1}\left(1-p\right)^{N-k+1} \right) \\ &= p(1-p)^{n-1} \left ((N+1+n-1) + (n-1)pN \right) \\ &= \left[N+n+pN(n-1)\right]p(1-p)^{n-1} \end{align*}

---

Solution: Let \(X \sim N(0,1)\), and $\displaystyle X_{M}=\begin{cases} X & \text{if }X < M,\\ M & \text{if }X\geqslant M. \end{cases} $. Then we can calculate: \begin{align*} \mathbb{E}[X_M] &= \int_{-\infty}^M xf_X(x)\,dx + M\mathbb{P}(X \geq M) \\ &= \int_{-\infty}^M x \frac1{\sqrt{2\pi}}e^{-\frac12x^2}\,dx + M\mathbb{P}(X \geq M) \\ &= \left [ -\frac{1}{\sqrt{2\pi}}e^{-\frac12x^2} \right ]_{-\infty}^M + M (1-\mathbb{P}(X < M)) \\ &= -\phi(M) + M(1-\Phi(M)) \end{align*} Let \(B \sim B\left (1024, \frac12 \right)\) be the number of potential bus passengers. Then \(B \approx N(512, 256) = N(512, 16^2)\) which is a good approximation since both \(np\) and \(nq\) are large. The question is asking us, how much additional profit would the bus company get if they ran an additional bus. Currently each week they is (on average) \(512\) passengers worth of demand, but they can only supply \(496\) seats, so we should expect that there is demand for another bus. The question is how much that demand is worth. Using the first part of the question, we can see that their profit is something like a `capped normal', \(X_M\), except we are scaled and with a different cap. So we are interested in $\displaystyle Y_{M}=\begin{cases} B & \mbox{if }B< M,\\ M & \mbox{if }B\geqslant M. \end{cases}\(, but since \)B \approx N\left (512,16^2\right)$ this is similar to \begin{align*} Y_{M}&=\begin{cases} 16X+512 & \mbox{if }16X+512< M,\\ M & \mbox{if }16X+512\geq M. \end{cases} \\ &= \begin{cases} 16X+512 & \mbox{if }X< \frac{M-512}{16},\\ M & \mbox{if }X \geq \frac{M-512}{16}. \end{cases} \\ &= 16X_{\frac{M-512}{16}} + 512\end{align*} We are interested in \(\mathbb{E}[Y_{16\times31}]\) and \(\mathbb{E}[Y_{17\times31}]\), which are \(16\mathbb{E}[X_{-1}]+512\) and \(16\mathbb{E}[Y_{\frac{15}{16}}]+512\) Since \(\frac{15}{16} \approx 1\), lets look at \(16(\mathbb{E}[X_1] - \mathbb{E}[X_{-1}])\) \begin{align*} \mathbb{E}[X_1] - \mathbb{E}[X_{-1}] &= \left ( -\phi(1) + 1-\Phi(1)\right) - \left ( - \phi(-1) -(1 - \Phi(-1)) \right ) \\ &= -\phi(1) + \phi(-1) + 1-\Phi(1) + 1 - \Phi(-1) \\ &= 1 - \Phi(1) + \Phi(1) \\ &= 1 \end{align*} Therefore the extra \(31\) will fill roughly \(16\) of them. (This is a slight overestimate, which is worth bearing in mind). A better approximation might be that \(\mathbb{E}[X_t] - \mathbb{E}[X_{-1}] = \frac{t +1}{2}\) for \(t \approx 1\), (since we want something increasing). This would give us an approximation of \(15.5\), which is very close to the `true' answer. Therefore, over \(50\) bus runs, we should earn roughly \(800\) vloska extra from an additional bus. (Again an overestimate, and with an uncertain pay-off, they should consider offering maybe \(600\)). Since this is the future, we can quite easily calculate the exact values using the binomial distribution on a computer. This gives the true value as \(15.833\), and so they should pay up to \(791\)

---

Solution: \begin{array}{c|c|c|c} k & \P(k \text{ goals}) & \P(\geq 2k+1 \text{ fouls}) & \P(k \text{ goals and } \geq 2k+1 \text{ fouls}) \\ \hline 0 & \frac{3-|2|}{9} = \frac19 & \frac{9}{10} & \frac{9}{90}\\ 1 & \frac{3-|2-1|}{9} = \frac29 & \frac{7}{10} & \frac{14}{90} \\ 2 & \frac{3-|2-2|}{9} = \frac39 & \frac{5}{10} & \frac{15}{90} \\ 3 & \frac{3-|2-3|}{9} = \frac29 & \frac{3}{10} & \frac{6}{90} \\ 4 & \frac{3-|2-4|}{9} = \frac19 & \frac{1}{10} & \frac{1}{90} \\ \hline &&& \frac{9+14+15+6+1}{90} = \frac12 \end{array} The probability a team scores more than half its points from fouls is \(\frac12\). Letting \(X\) be the number of times a team committed \(9\) fouls, then \(X \sim B(300, p)\). Consider two hypotheses: \(H_0: p = \frac1{10}\) \(H_1: p < \frac1{10}\) Under \(H_0\), we are interested in \(\P(X \leq 9)\). Since \(300 \frac{1}{10} > 5\) it is appropriate to use a normal approximation, \(N(30, 27)\). Therefore, \begin{align*} && \P(X \leq 9) &\approx \P(3\sqrt{3}Z + 30 \leq 9.5) \\ &&&= \P( Z \leq \frac{9.5-30}{3\sqrt{3}}) \\ &&&= \P(Z \leq \frac{-20.5}{3\sqrt{3}}) \\ &&&< \P(Z \leq -\frac{7}{2}) \end{align*} Which is very small. Therefore there is good evidence to believe there has been a change in the number of fouls.

---

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]

---

Solution:

1. \begin{align*} && p_n &= \mathrm{P}(A\leqslant k\mbox{ and }B\geqslant1-k) \\ &&&= \mathrm{P}(A\leqslant k) +\P(B\geqslant1-k) - \mathrm{P}(A\leqslant k\mbox{ or }B\geqslant1-k)\\ &&&= 1-\mathrm{P}(A\geq k) +1-\P(B \leq 1-k) - \l 1- \mathrm{P}(A\geq k\mbox{ and }B\leq 1-k)\r\\ &&&= 1 - \P(X_i \geq k) - \P(X_i \leq 1-k) + \P(k \leq X_i \leq 1-k) \\ &&&= 1 - k^n - (1-k)^n + (1-2k)^n \end{align*} Therefore as \(n \to \infty\) \(p_n \to 1\), since \(k, (1-k), (1-2k)\) are all between \(0\) and \(1\) and so their powers will tend to \(0\).
2. Let \(N = 4\,000\,000\). The probability exactly one person wins is \(Np(1-p)^{N-1}\). The probability exactly two people win is \(\binom{N}{2} p^2 (1-p)^{N-2}\). We wish to maximise the sum of these probabilities. To find this maximum, differentiate wrt \(p\). \begin{align*} \frac{\d}{\d p} : && \small N(1-p)^{N-1}-N(N-1)p(1-p)^{N-2} + N(N-1)p(1-p)^{N-2} - \frac12 N(N-1)(N-2)p^2(1-p)^{N-3} \\ &&= N(1-p)^{N-3} \l (1-p)^2 - \frac12(N-1)(N-2)p^2\r \\ \Rightarrow && \frac{(1-p)}{p} = \sqrt{\frac{(N-1)(N-2)}{2}} \\ \Rightarrow && p = \frac{1}{1+ \sqrt{\frac{(N-1)(N-2)}{2}}} \end{align*} This will be a maximum, since this is an increasing function at \(p=0\) and decreasing at \(p=1\) and there's only one stationary point. Note that \(p > \frac{\sqrt{2}}{(N-2)}\) and \(p < \frac{\sqrt{2}}{N-1+\sqrt{2}} < \frac{\sqrt{2}}{N}\) and so: \begin{align*} Np(1-p)^{N-1} &< \sqrt{2}(1-\frac{\sqrt{2}}{N-2})^{N-1} \\ &\approx \sqrt{2} e^{-\sqrt{2}} \end{align*} \begin{align*} \frac{N(N-1)}{2}p^2(1-p)^{N-2} &<(1-\frac{\sqrt{2}}{N-2})^{N-1} \\ &\approx e^{-\sqrt{2}} \end{align*} Alternatively, we can use a Poisson approximation. The number of winners is \(B(N, p)\) where we are hoping \(np\) is small but not zero. Therefore it's reasonable to approximation \(B(N,p)\) by \(Po(Np)\). (Call this value \(\lambda\)). Then we wish to maximise: \begin{align*} && p &= e^{-\lambda} \l \lambda + \frac{\lambda^2}{2} \r \\ &&&= e^{-\lambda} \lambda \l 1+ \frac{\lambda}{2} \r \\ \Rightarrow && \ln p &= -\lambda + \ln \lambda + \ln(1+\frac12 \lambda) \\ \frac{\d}{\d \lambda}: && \frac{p'}{p} &= -1 + \frac{1}{\lambda} + \frac{1}{2+\lambda} \\ &&&= \frac{-(2+\lambda)\lambda+2+2\lambda}{\lambda(2+\lambda)} \\ &&&= \frac{2-\lambda^2}{\lambda(2+\lambda)} \\ \Rightarrow && \lambda &= \sqrt{2} \end{align*} \begin{align*} \frac{\sqrt{2}+1}{e^{\sqrt{2}}} &< \frac{\sqrt{2}+1}{1+\sqrt{2}+1+\frac{1}{3}\sqrt{2}+\frac{1}{6}} \\ &= \frac{30\sqrt{2}-18}{41} \end{align*} Either way, we find we want to estimate \(e^{-\sqrt{2}}(1+\sqrt{2})\)