Problems

Solution:

1. \(\,\)
   

   + - ↺
   \begin{align*} && y & = x^{1/x} = \exp \left ( \tfrac1x \ln x \right ) \\ \Rightarrow && y' &= \exp \left ( \tfrac1x \ln x \right ) \cdot \left ( \frac{1}{x^2} - \frac{\ln x}{x^2} \right ) \\ &&&= \frac{\exp \left ( \tfrac1x \ln x \right ) }{x^2}(1 - \ln x) \\ y' =0: && x &= e \end{align*} Therefore the largest integer values will be \(2\) or \(3\). Comparing \((2^{\frac12})^6 = 8 < 9 = (3^{1/3})^6\) we see the maximum value of \(n^{1/n}\) where \(n\) is an integer is \(\sqrt[3]{3}\)
2. The number of tests is \(r\) plus however many groups fail times \(k\). The probability of a group failing is \(g = 1-(1-p)^k\) and the number of failing groups is \(\sim B(r, g)\) so the expected number of additional groups is \(rg\) and the expected total number of tests is \[ \frac{N}{k} + N(1-(1-p)^k) \]
3. \(\,\) \begin{align*} && N &\geq E = \frac{N}{k} + N(1-(1-p)^k) \\ \Rightarrow && 1 &\geq \frac{1}{k} + 1-(1-p)^k \\ \Rightarrow && (1-p)^k &\geq \frac1k \\ \Rightarrow && k \ln(1-p) &\geq - \ln k \\ \Rightarrow && \ln(1 - p) &\geq -\frac{1}{k} \ln k \geq -\frac13 \ln 3 \\ \Rightarrow && 1-p &\geq \frac{1}{\sqrt[3]{3}} \\ \Rightarrow && p &\leq 1-\frac{1}{\sqrt[3]{3}} \end{align*} (taking \(k=3\)) Claim: \(1 - \frac{1}{\sqrt[3]{3}} > \frac14\) Proof: This is equivalent to \(\sqrt[3]{3} > \frac43\) or \(81 > 4^3 = 64\) which is clearly true.
4. If \(pk\) is small then \((1-p)^k \approx 1 - pk\) and so we obtain \(N \left ( \frac1k +pk \right)\) as required. If \(p = 0.01\) and \(k = 10\) then \(\frac{1}{10} + 0.01 \cdot 10 = 0.2\) so the expected number of tests is \(\sim 20\%\) of \(N\)

Solution:

1. When \(N = 2n+1\), \(\mu = n +\frac12\) and so \begin{align*} && \delta &= \E[|X-\mu|] \\ &&&= \sum_{i=0}^n (n + \tfrac12 - i) \frac{1}{2^{2n+1}} \binom{2n+1}{i} + \sum_{i=n+1}^{2n+1} (i-n - \tfrac12) \frac{1}{2^{2n+1}} \binom{2n+1}{i} \\ &&&= 2\sum_{i=0}^n (n + \tfrac12 - i) \frac{1}{2^{2n+1}} \binom{2n+1}{i} \\ &&&= \frac{(2n +1)}{2^{2n+1}}\sum_{i=0}^n \binom{2n+1}i - \frac{2}{2^{2n+1}}\sum_{i=0}^n i \binom{2n+1}{i} \\ &&&= \frac{(2n +1)}{2^{2n+1}}\sum_{i=0}^n \binom{2n+1}i - \frac{2}{2^{2n+1}}\sum_{i=1}^n (2n+1) \binom{2n}{i-1} \\ &&&= \frac{(2n +1)}{2^{2n+1}}2^{2n} - \frac{2(2n+1)}{2^{2n+1}}\sum_{i=0}^{n-1} \binom{2n}{i} \\ &&&= \frac{(2n +1)}{2} - \frac{2(2n+1)}{2^{2n+1}} \frac12\left (2^{2n} - \binom{2n}{n} \right) \\ &&&= \frac{2n+1}{2^{2n+1}}\binom{2n}{n} \end{align*}

Solution:

1. Her probability of passing if she answers \(k \leq 2\) is \(0\), since she can attain at most \(4\) marks. If she attempts \(3\) questions, she needs to get all of them right, hence \(\mathbb{P}(\text{gets all }3\text{ correct}) = \frac{1}{n^3}\). If she attempts \(4\) questions, we can afford to get one wrong \begin{align*} && \mathbb{P}(\text{passes}|\text{attempts }4) &=\mathbb{P}(4/4) +\mathbb{P}(3/4) \\ &&&= \frac{1}{n^4} + 4\cdot\frac{1}{n^3} \cdot \frac{n-1}{n} \\ &&&= \frac{4n-3}{n^4} \end{align*} If she attempts \(5\) questions she can get \(5\) right (10), \(4\) right, \(1\) wrong (7), but \(3\) right will not work (\(6 - 2 = 4< 5\)), hence: \begin{align*} && \mathbb{P}(\text{passes}|\text{attempts }5) &=\mathbb{P}(5/5) +\mathbb{P}(4/5) \\ &&&= \frac{1}{n^5} + 5\cdot\frac{1}{n^4} \cdot \frac{n-1}{n} \\ &&&= \frac{5n-4}{n^5} \end{align*} If \(4n-3 > n \Leftrightarrow n \geq 2\) then \(4\) attempts is better than \(3\). If \(4n^2-3n > 5n-4 \Leftrightarrow 4n^2-8n+4 = 4(n-1)^2 > 0 \Leftrightarrow n \geq\) then \(4\) is better than \(5\), but \(n\) is \(\geq 2\) so, \(4\) is the best option.
2. \(\,\) \begin{align*} && \mathbb{P}(\text{passes}) &= \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot 0 + \frac16 \cdot \frac1{n^3} + \frac16 \frac{4n-3}{n^4} + \frac16 \frac{5n-4}{n^5} \\ &&&= \frac{n^2+4n^2-3n+5n-4}{6n^5} \\ &&&= \frac{5n^2+2n-4}{6n^5} \\ && \mathbb{P}(\text{answered }4|\text{passes}) &= \frac{\mathbb{P}(\text{answered }4\text{ and passes})}{ \mathbb{P}(\text{passes})} \\ &&&= \frac{\frac16 \cdot \frac{4n-3}{n^4}}{\frac{5n^2-2n-4}{6n^5} } \\ &&&= \frac{4n^2-3n}{5n^2+2n-4} \end{align*} Notice that the function takes all values for \(n\) between the roots of the denominator (which are either side of \(0\) and below \(3/4\). Therefore after \(3/4\) the function must be increase since otherwise we would have a quadratic equation with more than \(2\) roots.
3. \(\,\) \begin{align*} &&\mathbb{P}(C \text{ passes}) &= \binom{5}{3} \left ( \frac{n}{n+1} \right)^3 \left ( \frac{1}{n+1}\right)^2 \frac{1}{n^3} + \binom{5}{4} \left ( \frac{n}{n+1} \right)^4 \left ( \frac{1}{n+1}\right) \frac{4n-3}{n^4} +\\ &&&\quad \quad + \binom{5}{5} \left ( \frac{n}{n+1} \right)^5 \frac{5n-4}{n^5} \\ &&&= \frac{10}{(n+1)^5} + \frac{5(4n-3)}{(n+1)^5} + \frac{(5n-4)}{(n+1)^5} \\ &&&= \frac{10+20n-15+5n-4}{(n+1)^5}\\ &&&= \frac{25n-9}{(n+1)^5}\\ \end{align*}

Solution:

1. The probability no-one picks the winning number is \(\left ( 1 - \frac{1}{N}\right)^N \approx \frac1e\). \begin{align*} && \mathbb{E}(\text{profit}) &= Nc - (1-e^{-1})J \\ &&& < Nc -(1- \tfrac12 )J \\ &&& < Nc - \frac12 J \\ &&&= \frac{2Nc-J}{2} \end{align*} Therefore if \(J = 2Nc\) the expected profit is negative.
2. \(\,\) \begin{align*} && 1 &= \sum_{\text{all numbers}} \mathbb{P}(\text{pick }i) \\ &&&= \sum_{\text{popular numbers}} \mathbb{P}(\text{pick }i)+\sum_{\text{unpopular numbers}} \mathbb{P}(\text{pick }i) \\ &&&=\gamma N \frac{a}{N} + (1-\gamma)N \frac{b}{N} \\ &&&= \gamma a + (1-\gamma)b \end{align*} \begin{align*} && \mathbb{P}(\text{no-one picks winning number}) &= \mathbb{P}(\text{no-one picks winning number} | \text{winning number is popular})\mathbb{P})(\text{winning number is popular}) + \\ &&&\quad + \mathbb{P}(\text{no-one picks} | \text{unpopular})\mathbb{P}(\text{unpopular}) \\ &&&= \left (1 - \frac{a}{N} \right)^N \gamma + \left (1 - \frac{b}{N} \right)^N (1-\gamma) \\ &&&\approx \gamma e^{-a} + (1-\gamma)e^{-b} \\ \\ && \mathbb{E}(\text{profit}) &= Nc - (1-\gamma e^{-a} - (1-\gamma)e^{-b})J \\ &&&= Nc-J+J\gamma e^{-a} +J(1-\gamma)e^{-b} \end{align*} If \(\gamma = \frac18\) and \(a=9b\), then \(1=\frac18 a + \frac78b = 2b \Rightarrow b = \frac12, a = \frac92\) and \begin{align*} && \mathbb{E}(\text{profit}) &= Nc-J +J\tfrac18e^{-9/2}+J\tfrac78e^{-1/2} \\ &&&= Nc-J+\tfrac18Je^{-1/2}(e^{-4}+7) \end{align*} If we can show \(e^{-1/2}\frac{e^{-4}+7}{8} > \frac12\) we'd be done, so \begin{align*} && e^{-1/2}\frac{e^{-4}+7}{8} &> \frac12 \\ \Leftrightarrow && e^{-4}+7 &>4e^{1/2} \\ \Leftrightarrow && 49+14e^{-4}+e^{-8} &>16e \\ \end{align*} But clearly the LHS \(>49\) and the RHS \(<48\) so we're done

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(X+Y=r) &= \sum_{k=0}^r \mathbb{P}(X = k, Y = r-k) \\ &&&= \sum_{k=0}^r \mathbb{P}(X = k)\mathbb{P}( Y = r-k) \\ &&&= \sum_{k=0}^r \frac{e^{-\lambda T} (\lambda T)^k}{k!}\frac{e^{-\mu T} (\mu T)^{r-k}}{(r-k)!}\\ &&&= \frac{e^{-(\mu+\lambda)T}}{r!}\sum_{k=0}^r \binom{r}{k}(\lambda T)^k (\mu T)^{r-k}\\ &&&= \frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^r}{r!} \end{align*} Therefore \(X+Y \sim Po \left ( (\mu+\lambda)T \right)\)
2. \(\,\) \begin{align*} && \mathbb{P}(X = r | X+Y = k) &= \frac{\mathbb{P}(X=r, Y = k-r)}{\mathbb{P}(X+Y=k)} \\ &&&= \frac{\frac{e^{-\lambda T} (\lambda T)^r}{r!}\frac{e^{-\mu T} (\mu T)^{k-r}}{(k-r)!}}{\frac{e^{-(\mu+\lambda)T}((\mu+\lambda)T)^k}{k!}} \\ &&&= \binom{k}{r} \left ( \frac{\lambda}{\lambda + \mu} \right)^r \left ( \frac{\mu}{\lambda + \mu} \right)^{k-r} \end{align*} Therefore \(X|X+Y=k \sim B(k, \frac{\lambda}{\lambda + \mu})\)
3. \(P(X=1|X+Y = 1) = \frac{\lambda}{\lambda + \mu}\)
4. Let \(X_1, Y_1\) be the time to the first fish are caught by Adam and Eve, then \begin{align*} && \mathbb{P}(X_1, Y_1 > t) &= \mathbb{P}(X_1> t) \mathbb{P}( Y_1 > t) \\ &&&= e^{-\lambda t}e^{-\mu t} \\ &&&= e^{-(\lambda+\mu)t} \\ \Rightarrow && f_{\max(X_1,Y_1)}(t) &= (\lambda+\mu)e^{-(\lambda+\mu)} \end{align*} Therefore the expected time is \(\frac1{\mu+\lambda}\)

Solution:

1. There are several possibilities \begin{array}{c|c|c} \text{Alice} & \text{Bob} & P \\ \hline 0 & 1 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 2 & \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{3}{2^5} \\ 0 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ 1 & 2 & 2 \cdot \frac1{2^2} \cdot 3 \cdot \frac{1}{2^3} = \frac{6}{2^5} \\ 1 & 3 & 2\cdot \frac1{2^2} \cdot \frac{1}{2^3} = \frac{2}{2^5} \\ 2 & 3 & \frac1{2^2} \cdot \frac{1}{2^3} = \frac{1}{2^5} \\ \hline && \frac{1}{2^5}(3+3+1+6+2+1) = \frac{16}{2^5} = \frac12 \end{array}
2. There are several possibilities \begin{array}{c|c|c} A & B & \text{count} \\ \hline 0 & 1 & 4 \\ 0 & 2 & 6 \\ 0 & 3 & 4 \\ 0 & 4 & 1 \\ 1 & 2 & 3\cdot6 \\ 1 & 3 & 3\cdot4 \\ 1 & 4 & 3 \\ 2 & 3 & 3\cdot4 \\ 2 & 4 & 3 \\ 3 & 4 & 1 \\ \hline && 64 \end{array} Therefore the total probability is \(\frac12\)
3. \(\mathbb{P}(\text{Bob more than Alice}) = p_1 \cdot \underbrace{\frac12}_{\text{he wins by breaking the tie on his last flip}} + p_2\) If \(p_3\) is the probability that Alice gets more heads than Bob, then by symmetry \(p_3 = p_2\) and \(p_1 + p_2 + p_3 = 1\). Therefore \(p_1 + 2p_2 = 1\). ie \(\frac12 p_1 + p_2 = \frac12\) therefore the answer is always \(\frac12\) for all values of \(n\).

Solution:

1. \(X \sim B(16, \tfrac12)\), then \(X \approx N(8, 2^2)\), in particular \begin{align*} && \mathbb{P}(X = 8) &\approx \mathbb{P} \left ( 8 - \frac12 \leq 2Z + 8 \leq 8 + \frac12 \right) \\ &&&= \mathbb{P} \left (-\frac14 \leq Z \leq \frac14 \right) \\ &&&= \int_{-\frac14}^{\frac14} \frac{1}{\sqrt{2 \pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2\pi}} \int_{-\frac14}^{\frac14} 1\d x\\ &&&= \frac{1}{2 \sqrt{2\pi}} \end{align*}
2. Suppose \(X \sim B(2n, \frac12)\) then \(X \approx N(n, \frac{n}{2})\), and \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left ( n - \frac12 \leq \sqrt{\frac{n}{2}} Z + n \leq n + \frac12 \right) \\ &&&= \mathbb{P} \left ( - \frac1{\sqrt{2n}} \leq Z \leq \frac1{\sqrt{2n}}\right) \\ &&&= \int_{-\frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2 \pi}} e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{n\pi}}\\ \Rightarrow && \binom{2n}{n}\frac1{2^n} \frac{1}{2^n} & \approx \frac{1}{\sqrt{n \pi}} \\ \Rightarrow && (2n)! &\approx \frac{2^{2n}(n!)^2}{\sqrt{n\pi}} \end{align*}
3. \(X \sim Po(n)\), then \(X \approx N(n, (\sqrt{n})^2)\), therefore \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left (-\frac12 \leq \sqrt{n} Z \leq \frac12 \right) \\ &&&= \int_{-\frac{1}{2 \sqrt{n}}}^{\frac{1}{2 \sqrt{n}}} \frac{1}{\sqrt{2\pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && e^{-n} \frac{n^n}{n!} & \approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && n! &\approx \sqrt{2 \pi n} e^{-n}n^n \end{align*}

Solution:

1. Let \(N\) be the number of times the coin lands heads up, ie \(N \sim Binomial(100n, 0.2)\), then \(\mathbb{E}(N) = \mu = 20n, \mathrm{Var}(N) = \sigma^2 = 100n \cdot 0.2 \cdot 0.8 = 16n \Rightarrow \sigma = 4\sqrt{n}\). \begin{align*} && \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\ \Rightarrow && 1 - \mathbb{P}(|X - \mu| \leq k\sigma) &\leq \frac1{k^2} \\ \Rightarrow && 1 - \mathbb{P}(|X - 20n| \leq \sqrt{n} \cdot 4\sqrt{n}) &\leq \frac1{{\sqrt{n}}^2} \\ \Rightarrow && 1 - \mathbb{P}(16n \leq N \leq 24n) &\leq \frac{1}{n} \\ \Rightarrow && 1 - \frac1n &\leq \alpha \end{align*}
2. Suppose \(X \sim Pois(n)\), then \(\mathbb{E}(X) = n, \mathrm{Var}(X) = n\). Therefore \begin{align*} && \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\ \Rightarrow && 1-\mathbb{P}(|X - n| \leq \sqrt{n} \cdot \sqrt{n}) &> \frac{1}{\sqrt{n}^2} \\ \Rightarrow && 1 - \sum_{i=0}^{2n} \mathbb{P}(X = i) & \leq \frac{1}{n} \\ \Rightarrow && \sum_{i=0}^{2n} e^{-n} \frac{n^i}{i!} \geq 1 - \frac{1}{n} \\ \Rightarrow && \sum_{i=0}^{2n} \frac{n^i}{i!} \geq \left ( 1 - \frac1n \right)e^n \end{align*}

Solution:

1. \(\mathbb{P}(r \text{ need surgery}|n \text{ casualties}) = \binom{n}{r} \left ( \frac14\right)^r \left ( \frac34\right)^{n-r}\)
2. \(\,\) \begin{align*} && \mathbb{P}(r \text{ need surgery}) &= \sum_{n=r}^{\infty} \mathbb{P}(r \text{ need surgery} |n \text{ casualties}) \mathbb{P}(n \text{ casualties}) \\ &&&= \sum_{n=r}^{\infty} \binom{n}{r}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \sum_{n=r}^{\infty} \frac{n!}{(n-r)!r!}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{8^{n-r}}{(n-r)} \left ( \frac34\right)^{n-r} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{6^{n-r}}{(n-r)} \\ &&&= \frac{e^{-8}2^r}{r!} e^6 \\ &&&= \frac{e^{-2}2^r}{r!} \end{align*} Therefore the number requiring surgery is \(Po(2)\) with mean \(2\).
3. \(\,\) \begin{align*} && \mathbb{P}(X_1 = 8| X_1 + X_2 =12) &= \frac{\mathbb{P}(X_1 = 8,X_2 =4)} {\mathbb{P}(X_1+X_2 = 12)}\\ &&&= \frac{\frac{e^{-2}2^8}{8!} \cdot \frac{e^{-2}2^4}{4!}}{\frac{e^{-4}4^{12}}{12!}} \\ &&&= \frac{12!}{8!4!} \frac{1}{2^{12}} \\ &&&= \binom{12}4 \left ( \frac12 \right)^4\left ( \frac12 \right)^8 \\ &&&= \frac{495}{4096} \end{align*}

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.

Solution: Recall that for a random variable \(Z\) with pgf \(F(t)\) we have \(F(1) = 1\), \(\E[Z] = F'(1)\) and \(\E[Z^2] = F''(1) +F'(1)\) so \begin{align*} && \E[Y] &= G'(H(1))H'(1) \\ &&&= G'(1)H'(1) \\ &&&= \E[N]\E[X_i] \\ \\ && \E[Y^2] &= G''(H(1))(H'(1))^2+G'(H(1))H''(1) + G'(H(1))H'(1) \\ &&&= G''(1)(H'(1))^2+G'(1)H''(1) + G'(1)H'(1) \\ &&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N](\E[X_i^2]-\E[X_i]) + \E[N]\E[X_i] \\ &&&= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] \\ && \var[Y] &= (\E[N^2]-\E[N])(\E[X_i])^2 + \E[N]\E[X_i^2] - (\E[N])^2(\E[X_i])^2\\ &&&= (\var[N]+(\E[N])^2-\E[N])(\E[X_i])^2 + \E[N](\var[X_i]+\E[X_i]^2) - (\E[N])^2(\E[X_i])^2\\ &&&= \var[N](\E[X_i])^2 + \E[N]\var[X_i] \end{align*} Notice that \(N \sim Geo(\tfrac12)\) and \(Y = \sum_{i=1}^N X_i\) where \(X_i\) are Bernoulli. We have that \(G(t) = \frac{\frac12}{1-\frac12z}\) and \(H(t) = \frac12+\frac12p\) so the pgf of \(Y\) is \(G(H(t) = \frac{\frac12}{1 - \frac14-\frac14p} = \frac{2}{3-p}\). \begin{align*} && \E[X_i] &= \frac12\\ && \var[X_i] &= \frac14 \\ && \E[N] &= 2 \\ && \var[N] &= 2 \\ \\ && \E[Y] &= 2 \cdot \frac12 = 1 \\ && \var[Y] &= 2 \cdot \frac14 + 2 \frac14 = 1 \\ && \mathbb{P}(Y=r) &= \tfrac23 \left ( \tfrac13 \right)^r \end{align*}

Solution:

1. \(\mathbb{P}(X = r) = \binom{k}{r}p^rq^{k-r}\) where \(p = \frac{b}{n}, q = 1-p\). Therefore \begin{align*} && \frac{\mathbb{P}(X=r+1)}{\mathbb{P}(X=r)} &= \frac{\binom{k}{r+1}p^{r+1}q^{k-r-1}}{\binom{k}{r}p^rq^{k-r}} \\ &&&= \frac{(k-r)p}{(r+1)q} \\ &&&= \frac{(k-r)b}{(r+1)(n-b)} \end{align*} Comparing this to \(1\) we find: \begin{align*} && 1 & < \frac{\mathbb{P}(X=r+1)}{\mathbb{P}(X=r)} \\ \Leftrightarrow && 1 &< \frac{(k-r)b}{(r+1)(n-b)} \\ \Leftrightarrow && (r+1)(n-b) &<(k-r)b \\ \Leftrightarrow && rn& < (k+1)b - n \\ \Leftrightarrow && r &< \frac{(k+1)b}{n} - 1\\ \end{align*} If this equation is true, then \(\mathbb{P}(X=r+1)\) is larger, so \(r_{max} = \left \lfloor \frac{(k+1)b}{n} \right \rfloor\)
2. Let \(Y\) be the number of black balls in our sample, ie \(\mathbb{P}(Y = r) = \binom{b}{r}\binom{n-b}{k-r}/\binom{n}{k}\), so \begin{align*} && \frac{\mathbb{P}(Y = r+1)}{\mathbb{P}(Y=r)} &= \frac{\binom{b}{r+1}\binom{n-b}{k-(r+1)}/\binom{n}{k}}{\binom{b}{r}\binom{n-b}{k-r}/\binom{n}{k}} \\ &&&= \frac{b-r}{r+1} \frac{k-r}{n-b-k+r+1} \\ && 1 &< \frac{\mathbb{P}(Y = r+1)}{\mathbb{P}(Y=r)} \\ \Leftrightarrow && (r+1)(n-b-k+r+1) &< (b-r)(k-r) \\ \Leftrightarrow &&r(n-b-k+1)+(n-b-k+1) &< -r(b+k)+bk \\ \Leftrightarrow &&r(n+1) &< bk+b+k+1-n \\ \Leftrightarrow && r &< \frac{(b+1)(k+1)}{n+1} - \frac{n}{n+1} \end{align*} Therefore \(r = \left \lfloor \frac{ (b+1)(k+1)}{n+1}\right \rfloor\), it is not unique if \(n+1\) divides \((b+1)(k+1)\)