Problems

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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

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Solution: The \(1\)st slice of bread can only be the first slice in a sandwich or a slice of toast. Therefore \(s_1 = 0\) \begin{align*} && t_r &= \underbrace{s_{r-1}}_{r-1\text{th is the end of a sandwich}} \cdot \underbrace{p}_{\text{and we make toast}} + \underbrace{t_{r-1}}_{r-1\text{th is toast}} \cdot \underbrace{p}_{\text{and we make toast}} \\ &&&= (s_{r-1}+t_{r-1})p \\ \\ && s_r &= 1-\mathbb{P}(\text{previous slice is not the first of a sandwich}) \\ &&&= 1-(s_{r-1} + t_{r-1}) \\ \\ \Rightarrow && s_r &= 1 - \frac{t_r}{p} \\ \Rightarrow && t_r &= p - ps_r \\ \Rightarrow && s_r &= 1 - s_{r-1} - (p-ps_{r-1}) \\ &&&= 1 -p -(1-p)s_{r-1} \\ &&&= q(1-s_{r-1}) \end{align*} Therefore since \(s_r + qs_{r-1} = q\) we should look for a solution of the form \(s_r = A(-q)^r + B\). The particular solution will have \((1+q)B = q \Rightarrow B = \frac{q}{1+q}\), the initial condition will have \(s_1 = \frac{q}{1+q} +A(-q) = 0 \Rightarrow q = \frac{1}{1+q}\), so we must have \begin{align*} && s_r &= \frac{q+(-q)^r}{1+q}\\ \Rightarrow && t_r &= p(1-s_r) \\ &&&= p \frac{1+q-q-(-q)^r}{1+q} \\ &&&= \frac{(1-q)(1-(-q)^r)}{1+q} \\ && s_n &= 1-\frac{q+(-q)^{n-1}}{1+q} - \frac{p(1-(-q)^{n-1})}{1+q} \\ &&&= 1-\frac{1+(1-p)(-q)^{n-1}}{1+q}\\ &&&= 1-\frac{1-(-q)^n}{1+q}\\ &&&= \frac{q+(-q)^n}{1+q}\\ && t_n &=1-s_n \\ &&&=\frac{1-(-q)^n}{1+q} \end{align*}

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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 \(\frac1{\mu+\lambda}\)

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Solution:

1. The probability they pull the key out on the \(k\)th attempt will be \(\left ( \frac{n-1}{n} \right)^{k-1} \frac1n\), so we want: \begin{align*} \E[G_1] &= \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \frac1n \\ &= \frac{1}n \sum_{k=1}^{\infty} k \cdot \left ( \frac{n-1}{n} \right)^{k-1} \\ &= \frac1n \frac{1}{\left (1 - \frac{n-1}{n} \right)^2} \\ &= \frac{1}{n} \frac{n^2}{1^2} = n \end{align*}
2. In version 2, the probability the correct key comes out at the \(k\)th attempt is \(\frac1n\) (assume we take out all the keys, then the correct key is equally likely to appear in all of the space). Therefore \(\E[G_2] = \frac1n (1 + 2 + \cdots + n) = \frac{n+1}{2}\)
3. The probability the key comes out on the correct attempt is: \begin{align*} && \mathbb{P}(G_3 = k) &= \frac{n-1}{n} \cdot \frac{n}{n+1} \cdot \frac{n+1}{n+2} \cdots \frac{n+k-3}{n+k-2} \cdot \frac{1}{n+k-1} \\ &&&= \frac{n-1}{(n+k-2)(n+k-1)} \\ \\ &&k \cdot \mathbb{P}(G_3 = k) &= \frac{k(n-1)}{(n+k-2)(n+k-1)} \\ &&&= \frac{(n-1)(2-n)}{n+k-2} + \frac{(n-1)^2}{n+k-1} \\ &&&= \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n-k+2} \\ \Rightarrow && \E[G_3] &= \sum_{k=1}^{\infty} k \cdot \mathbb{P}(G_3 = k) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} + \frac{n-1}{n+k-2} \right) \\ &&&= \sum_{k=1}^{\infty} \left ( \frac{(n-1)^2}{n+k-1} - \frac{(n-1)^2}{n+k-2} \right) +\underbrace{\sum_{k=1}^{\infty} \frac{n-1}{n-k+2}}_{\to \infty} \\ \end{align*}

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(X = x) &= \sum_{y=1}^n \mathbb{P}(X=x,Y=y) \\ &&&= \sum_{y=1}^n k(x+y) \\ &&&= nkx + k\frac{n(n+1)}2 \\ \\ && 1 &= \sum_{x=1}^n \mathbb{P}(X=x) \\ &&&= nk\frac{n(n+1)}{2} + kn\frac{n(n+1)}2 \\ &&&= kn^2(n+1) \\ \Rightarrow && k &= \frac{1}{n^2(n+1)} \\ \Rightarrow && \mathbb{P}(X = x) &= \frac{nx}{n^2(n+1)} + \frac{n(n+1)}{2n^2(n+1)} \\ &&&= \frac{n+1+2x}{2n(n+1)} \\ \\ && \mathbb{P}(X=x)\mathbb{P}(Y=y) &= \frac{(n+1)^2+2(n+1)(x+y)+4xy}{4n^2(n+1)^2} \\ &&&\neq \frac{x+y}{n^2(n+1)} \end{align*} Therefore \(X\) and \(Y\) are not independent.
2. \(\,\) \begin{align*} && \E[X] &= \sum_{x=1}^n x \mathbb{P}(X=x) \\ &&&= \sum_{x=1}^n x \mathbb{P}(X=x)\\ &&&= \sum_{x=1}^n x \frac{n+1+2x}{2n(n+1)} \\ &&&= \frac{1}{2n(n+1)} \left ( (n+1) \sum x + 2\sum x^2\right)\\ &&&= \frac{1}{2n(n+1)} \left ( \frac{n(n+1)^2}{2} + \frac{n(n+1)(2n+1)}{3} \right) \\ &&&= \frac{1}{2} \left ( \frac{n+1}{2} + \frac{2n+1}{3} \right)\\ &&&= \frac{1}{2} \left ( \frac{7n+5}{6} \right)\\ &&&= \frac{7n+5}{12} \\ \\ && \textrm{Cov}(X,Y) &= \mathbb{E}\left[XY\right] - \E[X] \E[Y] \\ &&&= \sum_{x=1}^n \sum_{y=1}^n xy \frac{x+y}{n^2(n+1)} - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \sum \sum (x^2 y+xy^2) - \E[X]^2 \\ &&&= \frac{1}{n^2(n+1)} \left (\sum y \right )\left (\sum x^2\right ) - \E[X]^2 \\ &&&=\frac{(n+1)(2n+1)}{12} - \left ( \frac{7n+5}{12}\right)^2 \\ &&&= \frac1{144} \left (12(2n^2+3n+1) - (49n^2+70n+25) \right)\\ &&&= \frac{1}{144} \left (-25n^2-34n-13 \right) \\ &&& < 0 \end{align*} since \(\Delta = 34^2 - 4 \cdot 25 \cdot 13 = 4(17^2-25 \times 13) = -4 \cdot 36 < 0\)

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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 \(n\).

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Solution: \begin{align*} && \mathbb{P}(A \cup B \cup C) &= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cup B) \cap C) \tag{applying with \(A\cup B\) and \(C\)} \\ &&&= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \tag{applying with \(A\) and \(B\)}\\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \left ( \mathbb{P}(A \cap C) +\mathbb{P}(B \cap C) - \mathbb{P}( (A \cap C) \cap (B \cap C) )\right) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) +\mathbb{P}(C)- \mathbb{P}(A\cap B)- \mathbb{P}(A \cap C) -\mathbb{P}(B \cap C)+\mathbb{P}( A \cap B \cap C) \end{align*} \[ \mathbb{P}(A_1 \cup A_2 \cup A_3 \cup A_4) = \sum_i \mathbb{P}(A_i) - \sum_{i \neq j} \mathbb{P}(A_i \cap A_j) + \sum_{i \neq j \neq j} \mathbb{P}(A_i \cap A_j \cap A_k) - \mathbb{P}(A_1 \cap A_2 \cap A_3 \cap A_4) \]

1. \(\mathbb{P}(E_i) = \frac{1}{n}\)
2. \(\mathbb{P}(E_i \cap E_j) = \frac{1}{n} \cdot \frac{1}{n-1} = \frac{1}{n(n-1)}\)
3. \(\mathbb{P})(E_i \cap E_j \cap E_k) = \frac{1}{n(n-1)(n-2)}\)

First notice that the probability that \(k\) (or more) cards are in the correct place is \(\frac{(n-k)!}{n!}\) (place the other \(n-k\) cards in any order. We are interested in: \begin{align*} \mathbb{P} \left ( \bigcup_{i=1}^n E_i \right) &= \sum_{i} \mathbb{P}(E_i) - \sum_{i \neq j} \mathbb{P}(E_i \cap E_j) + \sum_{i \neq j \neq k} \mathbb{P}(E_i \cap E_j \cap E_k) - \cdots \\ &= \sum_i \frac1n - \sum_{i\neq j} \frac{1}{n(n-1)} + \sum_{i \neq j \neq k} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \sum_{i_1 \neq i_2 \neq \cdots \neq i_k} \frac{(n-k)!}{n!} + \cdots\\ &= 1 - \binom{n}{2} \frac{1}{n(n-1)} + \binom{n}{3} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \binom{n}{k} \frac{(n-k)}{n!} + \cdots \\ &= 1 - \frac12 + \frac1{3!} - \cdots + (-1)^{k+1} \frac{n!}{k!(n-k)!} \frac{(n-k)!}{n!} + \cdots \\ &= 1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n+1} \frac{1}{n!} \end{align*} The probability exactly one card is in the right place is the probability none of the other \(n-1\) are in the right place, which is: \(\frac1n \left (1 - \left (1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n} \frac{1}{(n-1)!} \right) \right)\) but there are also \(n\) cards we can choose to be the card in the right place, hence \(\frac{1}{2!} - \frac{1}{3!} + \cdots +(-1)^n \frac{1}{(n-1)!}\)

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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*}

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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*}

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Solution:

1. \(\,\) \begin{align*} \mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the \(n\)th throw.}) \\ &= \mathbb{P}(\text{\(n-1\) not 6s, followed by a 6.})\\ &= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n} \end{align*}
2. There is nothing special about \(5\) or \(6\), so which comes first is \(50:50\), therefore this probability is \(\frac12\)
3. There is nothing special about \(4\), \(5\) or \(6\) so this is the probability that \(6\) appears last out of these three numbers, hence \(\frac13\)
4. \(\,\) \begin{align*} \mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\ &=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the \(n\)th roll.}) \\ &= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\ &= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\ &= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\ &= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14 \end{align*}
5. \(\,\) \begin{align*} \mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\ &= \frac12 - \mathbb{P}(D \cap E) \\ &=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the \(n\)th roll.}) \\ &=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\ &=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\ &=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\ &= \frac12 - \frac{2}{27} \\ &= \frac{23}{54} \end{align*}

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Solution:

1. The only way \(A\) can win is if the sequence starts HH, if it does not start like this, then the only way HHT can appear is after a sequence of THH...H, but then THH has already appeared and \(B\) has won. Therefore the probability is \(\frac14\)
2. If HH appears before TT then either \(A\) or \(B\) will win. If HH appears first, then \(A\) has a \(\frac14\) probability of winning. So \(A\): \(\frac18\), \(B:\), \(\frac38\), \(C:\), \(\frac18\), \(D: \frac38\)
3. If the first two tosses are TT then \(C\) will win. If the first two tosses are HT, then either the next toss is T and \(C\) wins, or the next toss is H, and it's as if we started TH. ie \(p = \frac12 + \frac12 q\). If the first two tosses are TH, then either the next toss is H and \(C\) losses or the next toss is T and it's like starting HT. So \(q = \frac12 p\). Therefore \(p = \frac12 + \frac14p \Rightarrow p = \frac13\) If the first two tosses are HH, then eventually a T appears, and it's the same as starting HT. Therefore the probability \(C\) wins is: \(\frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16 + \frac14 \cdot \frac13 = \frac{11}{24}\)

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Solution:

1. \(\,\) \begin{align*} && \mathbb{E}(C) &= \int_0^\infty \text{cost}(x) f(x) \d x \\ &&&= ky + \int_y^{\infty} a(x-y) \lambda e^{-\lambda x} \d x\\ &&&= ky + \int_0^{\infty} a u \lambda e^{-\lambda u -\lambda y} \d x \\ &&&= ky + ae^{-\lambda y} \left( \left [ -ue^{-\lambda u} \right]_0^\infty -\int_0^\infty e^{-\lambda u} \d u\right) \\ &&&= ky + \frac{a}{\lambda}e^{-\lambda y} \\ \\ && \frac{\d \mathbb{E}(C)}{\d y} &= k - ae^{-\lambda y} \\ \Rightarrow && y &= \frac{1}{\lambda}\ln \left ( \frac{a}{k} \right) \end{align*} Since \(\mathbb{E}(C)\) is clearly increasing when \(y\) is very large, the optimal value will be \(\frac{1}{\lambda}\ln \left ( \frac{a}{k} \right)\), if \(\frac{a}{k} > 1\), otherwise you should spend nothing on flood defenses.
2. \begin{align*} && \mathbb{E}(C^2) &= \int_0^{\infty} \text{cost}(x)^2 f(x) \d x \\ &&&= \int_0^{\infty}(ky + a(x-y)\mathbb{1}_{x > y})^2 f(x) \d x \\ &&&= k^2y^2 + \int_y^{\infty}2kya(x-y)f(x)\d x + \int_y^{\infty}a^2 (x-y)^2 f(x) \d x \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{- \lambda y}+a^2e^{-\lambda y}\int_{u=0}^\infty u^2 \lambda e^{-\lambda u} \d u \\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y}+a^2e^{-\lambda y}(\textrm{Var}(Exp(\lambda)) + \mathbb{E}(Exp(\lambda))^2\\ &&&= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} \\ && \textrm{Var}(C) &= k^2y^2 + \frac{2kya}{\lambda}e^{-\lambda y} + a^2e^{-\lambda y} \frac{2}{\lambda^2} - \left ( ky + \frac{a}{\lambda} e^{-\lambda y}\right)^2 \\ &&&= a^2e^{-\lambda y} \frac{2}{\lambda^2} - a^2 e^{-2\lambda y}\frac{1}{\lambda^2} \\ &&&= \frac{a^2}{\lambda^2} e^{-\lambda y}\left (2 - e^{-\lambda y} \right) \\ \\ && \frac{\d \textrm{Var}(C)}{\d y} &= \frac{a^2}{\lambda^2} \left (-2\lambda e^{-\lambda y} +2\lambda e^{-2\lambda y} \right) \\ &&&= \frac{2a^2}{\lambda} e^{-\lambda y}\left (e^{-\lambda y}-1 \right) \leq 0 \end{align*} so \(\textrm{Var}(C)\) is decreasing in \(y\).

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Solution:

1. \(\mathbb{P}(X + Y \leq c) \) is the area between the \(x\)-axis, \(y\)-axis and the line \(x + y = c\). There are two cases for this: \[\mathbb{P}(X + Y \leq c) = \begin{cases} 0 & \text{ if } c \leq 0 \\ \frac{c^2}{2} & \text{ if } c \leq 1 \\ 1- \frac{(2-c)^2}{2} & \text{ if } 1 \leq c \leq 2 \\ 1 & \text{ otherwise} \end{cases}\]
2. \begin{align*} && \mathbb{P}((X + Y)^{-1} \leq t) &= 1- \mathbb{P}(X + Y \leq \frac1{t}) \\ \Rightarrow && f_{(X+Y)^{-1}}(t) &= 0 -\begin{cases} 0 & \text{ if } \frac1{t} \leq 0 \\ \frac{\d}{\d t}\frac{1}{2t^2} & \text{ if } \frac{1}{t} \leq 1 \\ \frac{\d}{\d t} \l 1- \frac{(2-\frac1t)^2}{2} \r & \text{ if } 1 \leq \frac{1}{t} \leq 2 \\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ (2-\frac1t)t^{-2} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \\ && &= \begin{cases} t^{-3} & \text{ if } t \geq 1 \\ 2t^{-2}-t^{-3} & \text{ if } \frac12 \leq t \leq 1\\ 0 & \text{ otherwise}\end{cases} \end{align*} Therefore, \begin{align*} \E \Big(\dfrac1{X+Y}\Big) &= \int_{\frac12}^{\infty} t f_{(X+Y)^{-1}}(t) \, \d t \\ &= \int_{\frac12}^{1} t f_{(X+Y)^{-1}}(t) \, \d t + \int_{1}^{\infty} t f_{(X+Y)^{-1}}(t) \d t\\ &= \int_{\frac12}^{1} \l 2t^{-1} - t^{-2} \r \, \d t + \int_{1}^{\infty} t^{-2} \d t\\ &= \left [ 2 \ln (t) + t^{-1} \right]_{\frac12}^{1} + \left [ -t^{-1} \right ]_{1}^{\infty} \\ &= 1 + 2 \ln 2 -2 + 1 \\ &= 2 \ln 2 \end{align*}
3. \begin{align*} &&\mathbb{P} \l \frac{Y}{X} \leq c \r &= \mathbb{P}( Y \leq c X) \\ &&&= \begin{cases} 0 & \text{if } c \leq 0 \\ \frac{c}{2} & \text{if } 0 \leq c \leq 1 \\ 1-\frac{1}{2c} & \text{if } 1 \leq c \end{cases} \\ \\ \Rightarrow && \mathbb{P} \l \frac{X}{X+Y} \leq t\r &= \mathbb{P} \l \frac{1}{1+\frac{Y}{X}} \leq t\r \\ &&&= \mathbb{P} \l \frac{1}{t} \leq 1+\frac{Y}{X}\r \\ &&&= \mathbb{P} \l \frac{1}{t} - 1\leq \frac{Y}{X}\r \\ &&&= 1- \mathbb{P} \l \frac{Y}{X} \leq \frac{1}{t} - 1\r \\ &&&= 1 - \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t} - \frac{1}{2} & \text{if } 0 \leq \frac1{t} \leq 1 \\ 1-\frac{t}{2-2t} & \text{if } 1 \leq \frac1{t} \end{cases} \\ && f_{\frac{X}{X+Y}}(t) &= \begin{cases} 0 & \text{if } \frac1{t} \leq 0 \\ \frac{1}{2t^2} & \text{if } t \geq 1 \\ \frac{1}{2(1-t)^2} & \text{if } 0 \leq t \leq 1 \end{cases} \\ \Rightarrow && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^\infty t f(t) \d t \\ &&&= \int_0^1 \frac{1}{2(1-t)^2} \d t + \int_1^\infty \frac{1}{t^2} \d t \\ &&& = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \\ \\ && \mathbb{E} \l \frac{X}{X+Y} \r &= \int_0^1 \int_0^1 \frac{x}{x+y} \d y\d x \\ &&&= \int_0^1 \l x \ln (x+1) - x \ln x \r \d x \\ &&&= \left [\frac{x^2}2 \ln(x+1) - \frac{x^2}{2} \ln(x) \right]_0^1 -\int_0^1 \l \frac{x^2}{2(x+1)} - \frac{x}{2} \r \d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x^2-1+1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \int_0^1 \frac{x -1}{2} + \frac{1}{2(x+1)}\d x \\ &&&= \frac{\ln 2}{2} + \frac{1}{4} - \frac{1}{4} + \frac{1}{2} - \frac{\ln 2}{2} \\ &&&= \frac{1}{2} \end{align*} We can also notice that \(1 = \mathbb{E} \l \frac{X+Y}{X+Y} \r = \mathbb{E} \l \frac{X}{X+Y} \r + \mathbb{E} \l \frac{Y}{X+Y} \r = 2 \mathbb{E} \l \frac{X}{X+Y} \r\) so it's clearly true as long as we can show that the integral converges.

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Solution: \begin{align*} && \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\ &&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\ &&&= \frac1{12} \left (26k^2-40k+ 26\right) \\ &&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\ &&&= \frac{13}{6}(k-1)^2+k \end{align*} Since \(k\) a single digit positive number and \(\mathbb{E}(X^2)\) is an integer, \(6 \mid k-1 \Rightarrow k = 1, 7\). \begin{align*} \mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\ \mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8 \end{align*} Therefore \(k = 7\) The probability distribution is \begin{align*} && \mathbb{P}(X=4) = \frac1{12} \\ && \mathbb{P}(X=5) = \frac1{6} \\ && \mathbb{P}(X=6) = \frac12 \\ && \mathbb{P}(X=13) = \frac1{6} \\ && \mathbb{P}(X=20)= \frac1{12} \\ \end{align*} The only ways to score more than \(25\) are: \(20+6, 20+13, 20+20, 13+13\) The only ways to score exactly \(25\) are \(20+5\) \begin{align*} \mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\ &= \frac{7}{48} \\ \mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\ \\ \mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144} \end{align*} The casino needs \(\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}\)

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(T > t + \delta t | T > t) &= \frac{\mathbb{P}(T < t + \delta t)}{\mathbb{P}(T > t )} \\ &&&= \frac{\int_t^{t+\delta t} f(s) \d s}{1-F(t)} \\ &&&\approx \frac{f(t)\delta t}{1-F(t)} \\ &&&= h(t) \delta t \end{align*}
2. If \(F(t) = t/a\) then \(f(t) = 1/a\) and \(h(t) = \frac{1/a}{1-t/a} = \frac{1}{a-t}\)
   

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3. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \frac{1}{t} \\ \Rightarrow && -\ln (1-F) &= \ln t + C\\ \Rightarrow && 1-F &= \frac{A}{t} \\ && F &= 1 - \frac{A}{t} \\ F(a) = 0: && F &= 1 - \frac{a}{t} \\ && f(t) &= \frac{a}{t^2} \end{align*}
4. (\(\Rightarrow\)) \begin{align*} && \frac{F'}{1-F} &= k \\ \Rightarrow && -\ln(1-F) &= kt+C \\ \Rightarrow && 1-F &= Ae^{-kt} \\ F(b) = 0: && 1 &= Ae^{-kb} \\ \Rightarrow && 1-F &= e^{-k(t-b)}\\ \Rightarrow && f &= ke^{-k(t-b)} \\ \end{align*} (\(\Leftarrow\)) \(f(t) = ke^{-k(t-b)} \Rightarrow F(t) = 1-e^{-k(t-b)}\) and the result is clear.
5. \(\,\) \begin{align*} && \frac{F'}{1-F} &= \left ( \frac{\lambda}{\theta^{\lambda}} \right) t^{\lambda-1} \\ \Rightarrow && -\ln(1-F) &= \left ( \frac{t}{\theta} \right)^{\lambda} +C\\ \Rightarrow && F &= 1-A\exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ F(0) = 0: && 0 &= 1-A \\ \Rightarrow && F &= 1 - \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \\ \Rightarrow && f &= \lambda t^{\lambda -1} \theta^{-\lambda} \exp \left (- \left ( \frac{t}{\theta} \right)^{\lambda} \right) \end{align*}