Problems

Solution:

1. \(\,\) \begin{align*} \mathbb{P}(\text{stop after r}) &= \mathbb{P}(\text{both stop at r}) + 2\mathbb{P}(\text{first stops before r second stops at r})\\ &= (q^2)^{r-1} p^2 + 2\cdot q^{r-1} p\cdot(1-q^{r-1}) \\ &= q^{r-1}p\left (2-2q^{r-1}+pq^{r-1} \right) \\ &= q^{r-1}p\left (2-q^{r-1}(1+p+q-p) \right) \\ &= q^{r-1}p\left (2-q^{r-1}-q^r\right) \\ \end{align*} \begin{align*} \E[\text{throws}] &= \sum_{r=1}^{\infty} r \mathbb{P}(\text{stop after r}) \\ &= \sum_{r=1}^{\infty} r q^{r-1}p\left (2-q^{r-1}-q^r\right) \\ &= \sum_{r=1}^{\infty} 2r q^{r-1}p-\sum_{r=1}^{\infty}r pq^{2r-2}-\sum_{r=1}^{\infty}r q^{2r-1}p \\ &=2p \sum_{r=1}^{\infty} r q^{r-1}-pq^{-2}\sum_{r=1}^{\infty}r q^{2r}-pq^{-1}\sum_{r=1}^{\infty}r q^{2r} \\ &= 2p(1-q)^{-2} - pq^{-2}q^2(1-q^2)^{-2}-pq^{-1}q^2(1-q^2)^{-2} \\ &= 2pp^{-2} -p(1+q)(1-q^2)^{-2} \\ &= 2p^{-1}-p(1+q)(1+q)^{-2}p^{-2} \\ &= 2p^{-1}-p^{-1}(1+q)^{-1} \\ &= \frac{2(1+q)-1}{p(1+q)} \\ &= \frac{1+2q}{p(1+q)} \\ &= \frac{3-2p}{p(2-p)} \end{align*}
2. \(\,\) \begin{align*} && m &= \frac{3-2p}{p(2-p)} \\ \Rightarrow && 2mp-mp^2 &= 3-2p \\ \Rightarrow && 0 &= mp^2-(2m+2)p + 3 \\ \Rightarrow && p &= \frac{2m+2 \pm \sqrt{(2m+2)^2-12m}}{2m} \\ &&&= \frac{m+1- \sqrt{m^2-m + 1}}{m} \\ \end{align*} If we are looking for an approximation, we could say \(p^2 \approx 0\) and \(p \approx \frac{3}{2(m+1)}\)

Solution: Given \(0 < r \ll n\), then \(\frac{r}{n}\) is small and so, \(e^x \approx 1+x\), therefore: \(\displaystyle e^{-r/n} \approx 1 - \frac{r}{n} = \frac{n-r}{n}\). Line everyone in the room up in some order. The first person is always going to have a birthday we haven't seen before. The probability the second person has a new birthday is \(\displaystyle 1 - \frac{1}{365}\) since they can't be born on the same day as the first person. The third person has a \(\displaystyle 1 - \frac{2}{365}\) probability of having a birthday we've not seen before, since they can't share a birthday with either of the first two people. Similarly the \(k\)th person has a \(\displaystyle 1 - \frac{k-1}{365}\) chance of having a unique birthday. \begin{align*} \prod_{i=1}^k \mathbb{P}(\text{the } i \text{th person has a new birthday}) &= \prod_{i=1}^k \l 1 - \frac{i-1}{365}\r \\ &\approx \prod_{i=1}^k \exp \l -\frac{i-1}{365}\r \\ &= \exp\l - \sum_{i=1}^k\frac{i-1}{365}\r \\ &= \exp\l - \frac{k(k-1)}{2\cdot365}\r \\ &= e^{-k(k-1)/730} \end{align*} But this the probability no-one shares a birthday, so the answer we are looking for is \(1-\) this, ie \(1 - e^{-k(k-1)/730}\) Suppose \(1 - e^{-k(k-1)/730} = \frac12\), then \begin{align*} && 1 - e^{-k(k-1)/730} &= \frac12 \\ \Rightarrow && e^{-k(k-1)/730} &= \frac12 \\ \Rightarrow && -k(k-1)/730 &= -\ln 2 \\ \Rightarrow && k(k-1)/730 &\approx \frac{253}{365} \\ \Rightarrow && k(k-1) &\approx 506 \end{align*} Therefore since \(22 \cdot 23 = 506\), we should expect the number to be approximately \(23\). Since \(e^{-r/n} > \frac{n-r}{n}\) we should expect this to be an overestimate, therefore \(23\) should suffice.

Solution:

1. 

   + - ↺
2. Since \(f(x)\) is continuous, \(a -bx\) joins \((2k, \ln k)\) and \((4k ,0)\). ie it has a gradient of \(\frac{-\ln k}{2k}\) and is zero at \(4k\), hence \(\displaystyle b = -\frac{\ln k}{2k}, a = 2\ln k\). The \(3\) sections have areas \(\int_1^k \ln x \d x = k \ln k -k +1\), \(k \ln k, k \ln k\). Therefore \begin{align*} &&1&= 3k\ln k - k +1 \\ \Rightarrow &&0 &= k(3\ln k - 1) \\ \Rightarrow &&\ln k &= \frac13 \\ \Rightarrow &&k &= e^{1/3} \\ && a &= \frac23 \\ && b&= -\frac16e^{-1/3} \end{align*}
3. Clearly \(1 > k \ln k > \frac{1}{3}\), therefore the median must lie between \(k\) and \(2k\). So we need, \(\frac12\) to be the area of the rectangle + the triangle, ie: \begin{align*} && \frac12 &= k \ln k + (2k-M) \ln k \\ &&&= \frac13 k + \frac13 (2k - M) \\ \Rightarrow && M &= 3k - \frac32 \\ \Rightarrow && M &= 3e^{1/3} - \frac32 \end{align*}

Solution:

1. This is the probability of having the sequence \(\underbrace{BB\cdots B}_{r-1 \text{ times}}R\) which has probability \(\displaystyle \left ( \frac{n}{n+1} \right)^{r-1}\frac{1}{n+1}\). Maximising this, is equivalent to maximising \(\log\) of it, ie \begin{align*} && y &= (r-1) \ln n - r \ln (n+1) \\ \Rightarrow && \frac{\d y}{\d n} &= \frac{r-1}{n} - \frac{r}{n+1} \\ &&&= \frac{(r-1)(n+1)-rn}{n(n+1)} \\ &&&= \frac{r-n-1}{n(n+1)} \end{align*} Therefore this is maximised when \(n = r-1\)

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*}

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 \[ 1600(1-\Phi(2)) - \frac{800}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \]

Solution:

1. \(\,\)
   

   + - ↺
   The overlap can be at most 0.4, which would mean \(p =0.7-0.4 = 0.3\) It must be at least 0.1, which would mean \(p =0.7-0.1 = 0.6\) so \(0.3 \leq p \leq 0.6\)
2. 

   + - ↺
   Notice that: \begin{align*} && 1 &= 0.05 + 0.7 -(hc+hr+0.1) + \\ &&&\quad\quad 0.8 - (hc+cr + 0.1) + \\ &&&\quad \quad \quad 0.4 - (hr+cr+0.1) +\\ &&&\quad \quad \quad \quad hc+hr+cr+0.1 \\ && &= 0.05 +0.7+0.8+0.4 - (hc+hr+cr)-2\cdot 0.1 \\ \Rightarrow && hc+hr+cr &=0.05 +0.7 + 0.8 + 0.4 - 0.2-1 \\ \Rightarrow && \mathbb{P}(\text{exactly 2}) &= 0.75 \end{align*} Notice \(q = \frac{hc}{0.8}\) Notice that we must have: \(hc, hr cr \geq 0\) as well as \(hc+hr+cr = 0.75\) \begin{align*} && \mathbb{P}(\text{only hat}) &= 0.7 -(hc+hr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.6 \\ && \mathbb{P}(\text{only cloak}) &= 0.8 - (hc+cr + 0.1)\geq 0 \\ \Rightarrow &&hc+cr & \leq 0.7 \\ && \mathbb{P}(\text{only ring}) &= 0.4 - (hr+cr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.3 \\ \end{align*} To find the minimum for \(hc\) we want to maximise \(hr+cr = 0.3\), so \(hc = 0.75 - 0.3 = 0.45\). To find the maximum for \(hc\) we want to minimise \(hr\) and \(cr\) \(cr \leq 0.7 - hc\) and \(hr \leq 0.6 - hc\) so \(0.75 \leq hc + (0.6 - hc) + (0.7 - hc) = 1.3-hc\) so \(hc \leq 1.3 - 0.75 = 0.55\) Therefore the range for \(q\) is \(\frac{.45}{.8}\) to \(\frac{.55}{.8}\) or \(\frac9{16} \leq q \leq \frac{11}{16}\)

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*}
2. 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*}

Solution:

1. We must have the total probability summing to \(1\), therefore \(1 =m + k\) (as \(x \to \infty\)) therefore \(k = 1-m\).
2. \(\,\) \begin{align*} && \E[X] &= \mathbb{P}(X=-1) \cdot (-1) + \int_0^{\infty} kx e^{-x} \d x \\ &&&= -m + (1-m) = 1-2m \end{align*}
3. \(\,\) \begin{align*} && \var[X] &= \E[X^2]-\E[X]^2 \\ &&&= \mathbb{P}(X=-1)\cdot(-1)^2 + (1-m)\int_0^{\infty} x^2e^{-x} \d x - (1-2m)^2 \\ &&&= m + (1-m)(1+1^2) - (1-2m)^2 \\ &&&= 3-4m - 1+4m -4m^2 \\ &&&= 2(1-m^2) \end{align*} To find the median \(q\), we need \begin{align*} && \frac12 &= \mathbb{P}(X \leq q) \\ &&&= m + (1-m)(1-e^{-q}) \\ \Rightarrow && e^{-q} &= 1-\frac{\frac12-m}{1-m} \\ &&&= \frac{1-m - \frac12+m}{1-m} \\ &&&= \frac{1}{2(1-m)} \\ \Rightarrow && q &= \ln 2(1-m) \end{align*}
4. \(\,\) \begin{align*} && \E\left [|X|^{\frac12}\right] &= \mathbb{P}(X=-1) \cdot 1 + \int_0^{\infty} \sqrt{x} (1-m)e^{-x} \d x \\ &&&= m + (1-m)\int_0^\infty \sqrt{x} e^{-x} \d x \\ u^2 = x, \d x = 2u \d u : &&&= m + (1-m) \int_{u=0}^{u=\infty} u e^{-u^2} \cdot 2u \d u \\ &&&= m + 2(1-m) \int_0^{\infty} u^2 e^{-u^2} \d u \\ &&&= m + (1-m)\frac{\sqrt{\pi}}2 \end{align*}

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*}

Solution: \begin{align*} \P(X \leq 0.8) &= \P(X_1 \leq 0.8,X_2 \leq 0.8,X_3 \leq 0.8) \\ &= 0.8^3 \\ &= 0.512 \end{align*} \begin{align*} && \P(X < c) &= c^3 \\ \Rightarrow && f_X(x) &= 3x^2 \\ \Rightarrow && \E[X] &= \int_0^1 x \cdot (3x^2) \, dx \\ && &= \left [ \frac{3}{4}x^4 \right]_0^1 \\ &&&= \frac{3}{4} \end{align*} \(X\) is distributed the maximum of \(N\) numbers on \([0,a]\). \begin{align*} H_0 : & x= 1 \\ H_1 : & x < 1 \end{align*} \begin{align*} &&\P(X < c) &= c^N \\ &&&= \frac1{20} \\ \Rightarrow && N &= -\frac{\log(20)}{\log(c)} \end{align*} where \(c = 0.8\), we have \begin{align*} N &= \frac{\log(20)}{\log(5/4)} \\ &= \frac{\log(5)+\log(4)}{\log(5)-\log(4)} \\ &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} \end{align*} \begin{align*} && 2^{10} &\approx 10^{3} \\ && 10\log(2) &\approx 3 (\log(5) + \log(2)) \\ && 7\log(2) &\approx 3 \log(5) \\ && \frac{\log(5)}{2\log(2)} &\approx \frac{7}{6} \end{align*} \begin{align*} &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} &= \frac{\frac{7}{6} + 1}{\frac{7}{6} -1} \\ &= 13 \end{align*} Since \(2^{10} > 10^3\) then \(N=14\) is the value we seek. \(\P(X < 0.8 | a= 0.8) = 1\) \(\P(X < 0.8 | a= 0.9, N=14) = \frac{8^{14}}{9^{14}}\)