Problems

Solution:

1. If the game ends in the first round then the score is exactly \(0\) and the pgf is \(1\cdot x^0 = 1\)
2. If the game moves onto the next round with no change in the first round then it's as if nothing happened, therefore the pgf is the original pgf \(G(t)\)
3. If the game moves into the next round with the score increased by one, then the pgf is \(tG(t)\) since all the scores are increased by \(1\). Therefore \begin{align*} && G(t) &= \E[t^N] \\ &&&= \E[\E[t^N | \text{first round}]] \\ &&&= a + bG(t) + ctG(t) \\ \Rightarrow && G(t)(1-b-ct) = a \\ \Rightarrow && G(t) &= \frac{a}{(1-b)-ct} \\ &&&= \frac{a}{(1-b)} \frac{1}{1- \left(\frac{c}{1-b}\right)t} \\ &&&= \sum_{n=0}^\infty \frac{a}{1-b} \frac{c^n}{(1-b)^n} t^n\\ &&&= \sum_{n=0}^{\infty} \frac{ac^n}{(1-b)^{n+1}}t^n \end{align*} Therefore by comparing coefficients, \(\mathbb{P}(N=n) = \frac{ac^n}{(1-b)^{n+1}}\)
4. \(\,\) \begin{align*} && \E[N] &= G'(1) \\ &&&= \frac{ac}{((1-b)-c)^2} \\ &&&= \frac{ac}{a^2} = \frac{c}{a} \\ \\ && \frac{\mu^n}{(1+\mu)^{n+1}} &= \frac{c^na^{-n}}{(a+c)^{n+1}a^{-n-1}} \\ &&&= \frac{ac^n}{(a+c)^{n+1}}\\ &&&= \frac{ac^n}{(1-b)^{n+1}}\\ &&&= \mathbb{P}(N=n) \end{align*} as required

Solution:

1. The probability the first is different to the second is \(\frac68\), the probability the third is different to both of the first two is \(\frac37\) therefore the probability is \(\frac{6}{8} \cdot \frac37 = \frac9{28}\) Whatever pills we are left with on the last day is essentially the same random choice as we make on the first day, therefore \(\frac9{28}\)
2. The probability the first is different to the second is \(\frac{2n}{3n-1}\), the probability the third is different to both of the first two is \(\frac{n}{3n-2}\) therefore the probability is \(\frac{2n^2}{(3n-1)(3n-2)}\). [We can also view this as \(\frac{(3n) \cdot (2n) \cdot n}{(3n) \cdot (3n-1) \cdot (3n-2)}\)]
3. Suppose describe the pills as \(B\) and \(R\) and also number them, then we must have a sequence of the form: \[ B_1R_1 \, B_2R_2 \, B_3R_3 \, \cdots \, B_{n}R_n \] However, we can also rearrange the order of the \(B\) and \(R\) pills in \(n!\) ways each, and also the order of the pairs in \(2^n\) ways. There are \((2n)!\) orders we could have taken the pills out therefore the probability is \begin{align*} && P &= \frac{2^n (n!)^2}{(2n)!} = \frac{2^n}{\binom{2n}{n}} \\ &&&\approx \frac{2^n \cdot 2n \pi \left ( \frac{n}{e} \right)^{2n}}{\sqrt{2 \cdot 2n \cdot \pi} \left ( \frac{2n}{e} \right)^{2n}} \\ &&&= \frac{2^n \sqrt{n \pi} \cdot n^{2n} \cdot e^{-2n}}{2^{2n} \cdot n^{2n} \cdot e^{-2n}} \\ &&&= 2^{-n} \sqrt{n \pi} \end{align*} There is a nice way to think about this question using conditional probability. Suppose we are drawing out of an infinitely supply of \(R\) and \(B\) pills, then each day there is a \(\frac12\) chance of getting different pills. Therefore over \(n\) days there is a \(2^{-n}\) chance of getting different pills. Conditional on the balanced total we see that \begin{align*} && \mathbb{P}(\text{balanced every day} |\text{balanced total}) &= \frac{\mathbb{P}(\text{balanced every day})}{\mathbb{P}(\text{balanced total})} \end{align*} We have already seen the term that is balanced total is \(\frac{1}{2^{2n}}\binom{2n}{n}\), but we can also approximate the balanced total using a normal approximation. \(B(2n, \tfrac12) \approx N(n, \frac{n}{2})\) and so: \begin{align*} \mathbb{P}(X = n) &\approx \mathbb{P}\left (n-0.5 \leq \sqrt{\tfrac{n}{2}} Z + n \leq n+0.5 \right) \\ &= \mathbb{P}\left (- \frac1{\sqrt{2n}} \leq Z \leq \frac{1}{\sqrt{2n}} \right) \\ &= \int_{- \frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \d x \approx \frac{2}{\sqrt{2n}} \frac{1}{\sqrt{2\pi}} \\ &\approx \frac{1}{\sqrt{n\pi}} \end{align*}

Solution: The probability it becomes faulty in each year is: \begin{align*} \mathbb{P}(\text{faulty in Y}1) &= \int_0^1 \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_0^1 \\ &= 1 - \frac{1}{2} = \frac{1}{2} \\ \mathbb{P}(\text{faulty in Y}2) &= \frac{1}{2} - \frac{1}{5} = \frac{3}{10} \\ \mathbb{P}(\text{faulty in Y}3) &= \frac{1}{5} - \frac{1}{10} = \frac{1}{10} \end{align*} The probability of failing for the first time after exactly \(3\) years is: \begin{align*} \mathbb{P}(\text{faulty in Y1, }PPF) &+ \mathbb{P}(\text{faulty in Y2, }PF) + + \mathbb{P}(\text{faulty in Y3, }F) \\ &= \frac12 (1-p)^2p + \frac3{10}(1-p)p + \frac1{10}p \\ &= \frac{p}{10} \l 5(1-p)^2 + 3(1-p) + 1 \r \\ &= \frac{p}{10} \l 5 - 10p + 5p^2 + 3 -3p +1 \r \\ &= \frac{p}{10} \l 9 - 13p + 5p^2 \r \end{align*} as required. The probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture is: \begin{align*} \mathbb{P}(\text{faulty 18 months after} | \text{fails after 3 tries}) &= \frac{\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})}{\mathbb{P}(\text{fails after exactly 3 tries})} \end{align*} We can compute \(\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})\) by looking at \(2\) cases, fails between \(12\) months and \(18\) years, and between \(0\) years and \(1\) year. \begin{align*} \mathbb{P}(\text{faulty between 1y and 18m}) &= \int_{1}^{\frac32} \frac{2t}{(1+t^2)^2} \, dt \\ &= \left [ -\frac{1}{(1+t^2)} \right]_{1}^{\frac32} \\ &= \frac12 - \frac{4}{13} = \frac{5}{26} \\ \end{align*} So the probability is: \begin{align*} \mathbb{P} &= \frac{\frac{5}{26}(1-p)p + \frac12(1-p)^2p}{\frac{p}{10} \l 9 - 13p + 5p^2 \r} \\ &= \frac{\frac{25}{13}(1-p) + 5(1-p)^2}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 5 + 13(1-p) \r}{9 - 13p + 5p^2} \\ &= \frac{5}{13} \frac{(1-p)\l 18 - 13p \r}{9 - 13p + 5p^2} \\ \end{align*}

Solution:

1. The only way you wont be able to sell to them is if they are first, ie \(\frac1{m+1}\)
2. If \(n=2\), the the only way you fail to sell to them is if one comes first or they both appear before two people with pound coins, ie \(2\) or \(122\). These have probabilities \(\frac{2}{m+2}\) and \(\frac{m}{m+2} \cdot \frac{2}{m+1} \frac{1}{m} = \frac{2}{(m+1)(m+2)}\). Therefore the total probability you don't sell all the tickets is \(\frac{2}{m+2}\left ( 1 + \frac{1}{m+1} \right) = \frac{2}{m+2} \frac{m+2}{m+1} = \frac{2}{m+1}\). Therefore the probability you do sell all the tickets is \(1 - \frac{2}{m+1} = \frac{m-1}{m+1}\)
3. The only ways to fail when \(n=3\) are: \(2\), \(122\), or if all three \(2\)s appear before three \(1\)s. this can happen in \(11222\), \(12122\) These happen with probability: \begin{align*} 2: && \frac{3}{m+3} \\ 122: && \frac{m}{m+3} \cdot \frac{3}{m+2} \cdot \frac{2}{m+1} \\ 11222: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ 12122: && \frac{m(m-1) 6}{(m+3)(m+2)(m+1)m(m-1)} \\ \end{align*} Therefore the total probability is: \begin{align*} P &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+2)(m+1)+6m + 12 \right) \\ &= \frac{1}{(m+3)(m+2)(m+1)} \left (3(m+1)(m+2) \right) \\ &= \frac{3}{m+1} \end{align*} and the result follows

Solution:

1. We can be in several states.
   1. No points played
   2. Y just won the last point
   3. X just won the last point
   4. Y won the game
   5. X won the game
   The probability \(Y\) wins from any of these states are: \begin{align*} &&P_{-} &= p P_X + (1-p) P_Y &= w \\ &&P_Y &= p + (1-p)P_X \\ &&P_X &= (1-p)P_Y \\ \\ \Rightarrow &&& \begin{cases} P_Y - (1-p)P_X &= p \\ (1-p)P_Y -P_X &= 0\end{cases} \\ \Rightarrow && P_Y &= \frac{1}{1-(1-p)^2} \cdot p \\ &&&= \frac{1}{2-p} \\ && P_X &= \frac{1-p}{2-p} \\ && w &= \frac{p(1-p) + (1-p)}{2-p} \\ &&&= \frac{1-p^2}{2-p} \end{align*}
2. If \(p = \frac23\) then \(w = \frac{1-\frac49}{2-\frac23} = \frac{5}{12}\). The game is fair if \(\mathbb{E}(result) = 0\), ie \(\frac{5}{12} \cdot k - \frac{7}{12} 1 \Rightarrow k = \frac{7}{5} = 1.4\)
3. If \(p = 0\) then they will keep playing forever, since no-one can win two points in a row.

Solution: \begin{align*} && \mathbb{E}(X) &= p + \frac{p}{3} + p^2 = \frac43p+p^2 \\ \Rightarrow && \frac43 &= \frac43p+p^2 \\ \Rightarrow && 0 &= 3p^2+4p-4 \\ &&&= (3p-2)(p+2) \\ \Rightarrow && p &= \frac23, -2 \end{align*} Since \(p \in [0,1]\) we must have \(p = \frac23\). \begin{align*} && \mathbb{P}(\text{correct verdict}) &= t p+ (1-t) (1-p) \\ &&&= t(2p-1)+(1-p)\\ \Rightarrow && \frac12 &= t(2p-1)+(1-p) \\ \Rightarrow && t &= \frac{\frac12-(1-p)}{2p-1} \\ &&&= \frac{2p-1}{2(2p-1)} = \frac12 \end{align*} (so it doesn't depend at all on what the judges are doing, the only way to be fair is if the trials happen at random!)

Solution: There are many ways to do the counting in each question, possibly the clearest way is to always consider the order in which discs are taken, although all methods should work equally well. For some examples Bayes rule also offers a fast solution.

1. There are we are choose \(4\) objects in order from \(5\) (ie \({^5\P_4}\)) to obtain valid draws, this is out of a total of picking \(4\) objects from \(11\) (\({^{11}\P_4}\)). Ie the probability is: \(\displaystyle \frac{{^5\P_4}}{{^{11}\P_4}} = \frac{5! \cdot 7!}{11!} = \frac{1}{66}\) Alternatively, there are \(\binom{5}{4}\) ways to choose four numbered discs, out of \(\binom{11}{4}\) ways to choose four discs. ie \(\displaystyle \binom{5}{4} \Big / \binom{11}{4} = \frac{5 \cdot 4! \cdot 7!}{11!} = \frac{5 \cdot 4 \cdot 3 \cdot 2}{11 \cdot 10 \cdot 9 \cdot 8} = \frac1{66}\)
2. \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{^4\P_3 \big / {^{11}\P_4}}{1/11} \\ &= 11\cdot \frac{4!}{1!} \Bigg / \frac{11!}{7!} \\ &= \frac{4! \cdot 7! \cdot 11}{11!} \\ &= \frac{4\cdot 3 \cdot 2}{10 \cdot 9 \cdot 8} \\ &= \frac{1}{30} \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \frac{\mathbb{P}(\text{all four discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{3} \Bigg / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac{1}{30} \end{align*} Where we are calculating this as "choose one number", then "choose 3 more", which can happen ending up with 3, number, number, number in \(\binom{4}{3}\) ways, and there are \(11 \cdot \binom{10}{3}\) was overall. Another alternative using Bayes rule: \begin{align*} \mathbb{P}(\text{all four discs are numbered} | \text{first disc is 3}) &= \mathbb{P}( \text{first disc is 3} | \text{all four discs are numbered}) \frac{ \mathbb{P}( \text{all four discs are numbered} }{ \mathbb{P}( \text{first disc is 3} )} \\ &= \frac{\frac{1}{5} \cdot \frac{1}{66}}{\frac{1}{11}} \\ &= \frac1{30} \end{align*}
3. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{3 \cdot {^4\P_1} \cdot {^{6}\P_2} \big / {^{11}\P_4}}{\frac1{11} } \\ &= \frac12 \end{align*} Alternatively, \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{first disc is 3}) &=\frac{\mathbb{P}(\text{exactly two discs are numbered and first disc is 3})}{\mathbb{P}( \text{first disc is 3})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / 11 \cdot \binom{10}{3}}{1/11} \\ &= \frac12 \end{align*}
4. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and 3 taken})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\binom{4}{1}\binom{6}{2} \Big / \binom{11}{4}}{\frac{4}{11}} \\ &= \frac{\frac{2}{11}}{\frac4{11}} \\ &= \frac12 \end{align*} Using Bayes rule: \(\mathbb{P}( \text{3 taken}) = \frac{1}{11} + \frac{10}{11}\frac{1}{10} + \frac{10}{11}\frac{9}{10}\frac{1}{9} + \frac{10}{11}\frac{9}{10}\frac89\frac18 = \frac{4}{11}\) \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{3 taken}) &= \frac{\mathbb{P}(\text{3 taken | exactly two discs are numbered})\mathbb{P}(\text{exactly two discs are numbered})}{\mathbb{P}( \text{3 taken})} \\ &= \frac{\frac{4}{10} \cdot \binom{5}{2} \binom{6}{2} \Big / \binom{11}{4}}{4/11} \\ &= \frac{\frac4{10}{5 / 11}}{4/11} \\ &= \frac{1}{2} \end{align*}
5. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc first}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc first})}{\mathbb{P}( \text{numbered disc first})} \\ &= \frac{3 \cdot {^5\P_1}\cdot{^4\P_1}\cdot{^6\P_2} \Big / {^{11}\P_4}}{\frac{5}{11}} \\ &= \frac{1}{2} \end{align*}
6. \begin{align*} \mathbb{P}(\text{exactly two discs are numbered} | \text{numbered disc taken}) &= \frac{\mathbb{P}(\text{exactly two discs are numbered and numbered disc taken})}{\mathbb{P}(\text{numbered disc taken})} \\ &= \frac{\mathbb{P}(\text{exactly two discs are numbered})}{1 - \mathbb{P}(\text{not numbered discs taken})} \\ &= \frac{\binom{5}{2}\binom{6}{2} \Big / \binom{11}{4}}{1 - \binom{6}{4} \Big / \binom{11}{4}} \\ &= \frac{\frac{5}{11}}{\frac{21}{22}} \\ &= \frac{10}{21} \neq \frac12 \end{align*}

Solution: Let \(X \sim Po(\lambda)\), then

1. \(\,\)
   

   + - ↺
   Suppose \(\mathbb{P}(X \geq 2) = 1-p\) then \(\mathbb{P}(X=0) + \mathbb{P}(X=1) = p\), ie \(e^{-\lambda} +\lambda e^{-\lambda} = p\) If \(f(x) = (1+x)e^{-x}\) we have see it is strictly decreasing on \(x \geq 0\) and takes all values from \(1\) to \(0\), therefore we can find a unique value such that \(f(\lambda) = p\) which is our desired \(\lambda\).
2. Note that \(\mathbb{P}(X = 1) = \lambda e^{-\lambda}\)
   

   + - ↺
   Sketching \(y = xe^{-x}\) and finding it's turning point we can see that there is a unique value of \(\lambda = 1\) when \(q = \frac{1}{e}\), otherwise there is either no solution (\(p > \frac1{e}\)) or two solutions (\(0 < q > \frac1{e}\)).
3. Suppose \(\mathbb{P}(X = 1 | X \leq 2) = r\), ie \begin{align*} && r &= \frac{\lambda e^{-\lambda}}{e^{-\lambda} + \lambda e^{-\lambda} + \frac12 \lambda^2 e^{-\lambda}} \\ &&&= \frac{2\lambda}{2+2\lambda+\lambda^2} \\ \Rightarrow && 0 &= r\lambda^2 + 2(r-1) \lambda + 2r\\ \Rightarrow && \Delta &= 4(r-1)^2 - 4\cdot r \cdot 2 r \\ &&&= 4((r-1)^2-2r^2) \\ &&&= 4(r-1-\sqrt{2}r)(r-1+\sqrt{2}r) \\ &&&= -4((\sqrt{2}-1)r + 1)((1+\sqrt{2})r-1) \end{align*} Therefore our quadratic in \(r\) has a unique solution if \(r = \frac{1}{1+\sqrt{2}}\). If it has a positive solution then note since \(2r > 0\) both solutions are positive, so \(\lambda\) is not unique by excluding other solutions.

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: Not that the number of diamonds per kilogram is \(1\) so we are assuming it is \(Po(M)\) where \(M\) is the mass in kg. In particular \(\E[X] = M\) and \(\mathbb{P}(X = 0) = e^{-M}\)

1. \(\,\) \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^6\mathbb{P}(\text{no diamonds and roll }n) \\ &&&= \sum_{n=1}^6 \tfrac16 e^{-\frac{n}{10}} \\ &&&= \frac{e^{-0.1}}6 \left ( \frac{1-e^{-0.6}}{1-e^{-0.1}}\right) \\ && \E[\text{diamonds}] &= \sum_{n=1}^6 \E(\text{diamonds}|N = n)\mathbb{P}(N = n) \\ &&&= \sum_{n=1}^6 0.1n \cdot \frac16 \\ &&&= 0.1 \cdot \frac{7}{2} = 0.35 \end{align*}
2. \(\mathbb{P}(T = k) = \left ( \frac56 \right)^{t-1} \frac16\), so \begin{align*} && \mathbb{P}(\text{no diamonds}) &= \sum_{n=1}^\infty\mathbb{P}(\text{no diamonds and }T=n) \\ &&&= \sum_{n=1}^\infty e^{-0.1n} \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{e^{-0.1}}{6} \frac1{1- \frac56 e^{-0.1}} \\ &&&= \frac{e^{-0.1}}{6 - 5e^{-0.1}} \\ \\ && \E[\text{diamonds}] &= \sum_{n=1}^\infty \E(\text{diamonds}|T = n)\mathbb{P}(T = n) \\ &&&= \sum_{n=1}^\infty 0.1n \cdot \left ( \frac56 \right)^{n-1} \frac16 \\ &&&= \frac{0.1}{6} \sum_{n=1}^\infty n \cdot \left ( \frac56 \right)^{n-1} \\ &&&= \frac{1}{60} \frac{1}{(1- \tfrac56)^2} \\ &&&= \frac{6}{10} = \frac35 \end{align*}

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:

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 \(\frac9{16} \leq q \leq \frac{11}{16}\)