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2006 Paper 1 Q12
D: 1500.0 B: 1499.3

Oxtown and Camville are connected by three roads, which are at risk of being blocked by flooding. On two of the three roads there are two sections which may be blocked. On the third road there is only one section which may be blocked. The probability that each section is blocked is \(p\). Each section is blocked independently of the other four sections. Show that the probability that Oxtown is cut off from Camville is \(p^3 \l 2-p \r^2\). I want to travel from Oxtown to Camville. I choose one of the three roads at random and find that my road is not blocked. Find the probability that I would not have reached Camville if I had chosen either of the other two roads. You should factorise your answer as fully as possible. Comment briefly on the value of this probability in the limit \(p\to1\).

2006 Paper 1 Q13
D: 1484.0 B: 1468.0

A very generous shop-owner is hiding small diamonds in chocolate bars. Each diamond is hidden independently of any other diamond, and on average there is one diamond per kilogram of chocolate.

  1. I go to the shop and roll a fair six-sided die once. I decide that if I roll a score of \(N\), I will buy \(100N\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6} \l \frac{1 - \e^{-0.6} }{ 1 - \e^{-0.1}} \r \] Show also that the expected number of diamonds I find is 0.35.
  2. Instead, I decide to roll a fair six-sided die repeatedly until I score a 6. If I roll my first 6 on my \(T\)th throw, I will buy \(100T\) grams of chocolate. Show that the probability that I will have no diamonds is \[ \frac{\e^{-0.1}}{ 6 - 5\e^{-0.1}} \] Calculate also the expected number of diamonds that I find. (You may find it useful to consider the the binomial expansion of \(\l 1 - x \r^{-2}\).)


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*}

2006 Paper 1 Q14
D: 1500.0 B: 1502.6

  1. A bag of sweets contains one red sweet and \(n\) blue sweets. I take a sweet from the bag, note its colour, return it to the bag, then shake the bag. I repeat this until the sweet I take is the red one. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?
  2. Instead, I take sweets from the bag, without replacing them in the bag, until I take the red sweet. Find an expression for the probability that I take the red sweet on the \(r\)th attempt. What value of \(n\) maximises this probability?


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 \(n = r-1\)

2006 Paper 2 Q1
D: 1600.0 B: 1485.5

The sequence of real numbers \(u_1\), \(u_2\), \(u_3\), \(\ldots\) is defined by \begin{equation*} u_1=2 \,, \qquad\text{and} \qquad u_{n+1} = k - \frac{36}{u_n} \quad \text{for } n\ge1, \tag{\(*\)} \end{equation*} where \(k\) is a constant.

  1. Determine the values of \(k\) for which the sequence \((*)\) is: (a) constant; (b) periodic with period 2; (c) periodic with period 4.
  2. In the case \(k=37\), show that \(u_n\ge 2\) for all \(n\). Given that in this case the sequence \((*)\) converges to a limit \(\ell\), find the value of \(\ell\).

2006 Paper 2 Q2
D: 1600.0 B: 1500.0

Using the series \[ \e^x = 1 + x +\frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!}+\cdots\,, \] show that \(\e>\frac83\). Show that \(n!>2^n\) for \(n\ge4\) and hence show that \(\e<\frac {67}{24}\). Show that the curve with equation \[ y= 3\e^{2x} +14 \ln (\tfrac43-x)\,, \qquad {x<\tfrac43} \] has a minimum turning point between \(x=\frac12\) and \(x=1\) and give a sketch to show the shape of the curve.


Solution: \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&> 1 + 1+ \frac12 + \frac16 \\ &&&= \frac{12+3+1}{6} = \frac83 \end{align*} \(4! = 24 > 16 = 2^4\), notice that \(n! = \underbrace{n \cdot (n-1) \cdots 5}_{>2^{n-4}} \cdot \underbrace{4!}_{>2^4} >2^n\). \begin{align*} && e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &&&< \frac83 + \frac{1}{2^4} + \frac{1}{2^5} + \cdots \\ &&&= \frac83 + \frac{1}{2^4} \frac{1}{1-\tfrac12} \\ &&&= \frac83 + \frac1{8} \\ &&&= \frac{67}{24} \end{align*} \begin{align*} && y &= 3e^{2x} +14 \ln(\tfrac43-x) \\ && y' &= 6e^{2x} - \frac{14}{\tfrac43-x} \\ && y'(\tfrac12) &= 6e - \frac{14}{\tfrac43-\tfrac12} \\ &&&= 6e -\tfrac{84}{5} = 6(e-\tfrac{14}5) < 0 \\ && y'(1) &= 6e^2 - \frac{14}{\tfrac43-1} \\ &&&= 6e^2 - 42 = 6(e^2-7) \\ &&&> 6(\tfrac{64}{9} - 7) > 0 \end{align*} Therefore \(y'\) changes from negative (decreasing) to positive (increasing) in our range, and therefore there is a minima in this range.

TikZ diagram

2006 Paper 2 Q3
D: 1600.0 B: 1570.3

  1. Show that $\displaystyle \big( 5 + \sqrt {24}\;\big)^4 + \frac{1 }{\big(5 + \sqrt {24}\;\big)^4} \ $ is an integer. Show also that \[\displaystyle 0.1 < \frac{1}{ 5 + \sqrt {24}} <\frac 2 {19}< 0.11\,.\] Hence determine, with clear reasoning, the value of \(\l 5 + \sqrt {24}\r^4\) correct to four decimal places.
  2. If \(N\) is an integer greater than 1, show that \(( N + \sqrt {N^2 - 1} \,) ^k\), where \(k\) is a positive integer, differs from the integer nearest to it by less than \(\big( 2N - \frac12 \big)^{-k}\).


Solution:

  1. First notice that \(\frac{1}{5+\sqrt{24}} = \frac{5-\sqrt{24}}{25-24} = 5 - \sqrt{24}\), hence \begin{align*} && ( 5 + \sqrt {24})^4 + \frac{1 }{(5 + \sqrt {24})^4} &= ( 5 + \sqrt {24})^4 + ( 5 - \sqrt {24})^4 \\ \end{align*} where clearly all terms including \(\sqrt{24}\) will cancel out, therefore it is an integer. \begin{align*} && 5 + \sqrt{24} &< 5 + 5 = 10 \\ \Rightarrow && \frac{1}{5+\sqrt{24}}& > \frac{1}{10} = 0.1 \\ && 2(5 + \sqrt{24}) &=10 + \sqrt{96} > 19 \\ \Rightarrow && \frac{1}{5+\sqrt{24}} & < \frac{2}{19} < \frac{2}{18} = \frac19 = 0.11111\ldots < 0.11 \end{align*} Therefore, \(10^{-4} < (5+\sqrt{24})^4 < 0.11^{-4} = 0.00014641\) \begin{align*} && (5+\sqrt{24})^4 + (5-\sqrt{24})^4 &= 2(5^4+6\cdot5^2\cdot24+24^2) \\ &&&= 2\cdot (625 + 3600+576) \\ &&&= 9602 \\ \Rightarrow && (5+\sqrt{24})^4 &= 9602 - \epsilon, \epsilon \in (0.0001, 0.00014641) \\ \Rightarrow && (5+\sqrt{24})^4 &\in (9601.999854, ,9601.9999) \\ \Rightarrow && (5+\sqrt{24})^4 &= 9601.9998 \, (4 \text{ d.p.}) \end{align*}
  2. Notice that \((N+\sqrt{N^2-1})^{k}+(N-\sqrt{N^2-1})^{k}\) is an integer for the same reason as before (sum of conjugates). Notice also that \(\frac{1}{N+\sqrt{N^2-1}} = N - \sqrt{N^2-1}\) and that so it sufficies to show that \begin{align*} && N + \sqrt{N^2-1} &> 2N-\tfrac12 \\ \Leftrightarrow && \sqrt{N^2-1} &> N - \tfrac12 \\ \Leftrightarrow && N^2-1 &> N^2-N+1\\ \Leftrightarrow && N &> \tfrac32\\ \end{align*} Which is true since \(N > 1\) and \(N\) is an integer.

2006 Paper 2 Q4
D: 1600.0 B: 1530.0

By making the substitution \(x=\pi-t\,\), show that \[ \! \int_0^\pi x\f(\sin x) \d x = \tfrac12 \pi \! \int_0^\pi \f(\sin x) \d x\,, \] where \(\f(\sin x)\) is a given function of \(\sin x\). Evaluate the following integrals:

  1. \(\displaystyle \int_0^\pi \frac {x \sin x}{3+\sin^2 x}\,\d x\,\);
  2. $\displaystyle \int_0^{2\pi} \frac {x \sin x}{3+\sin^2 x}\,\d x\,\(;
  3. \)\displaystyle \int_{0}^{\pi} \frac {x \big\vert\sin 2x\big\vert}{3+\sin^2 x}\,\d x\,$.

2006 Paper 2 Q5
D: 1600.0 B: 1469.6

The notation \({\lfloor } x \rfloor\) denotes the greatest integer less than or equal to the real number \(x\). Thus, for example, \(\lfloor \pi\rfloor =3\,\), \(\lfloor 18\rfloor =18\,\) and \(\lfloor-4.2\rfloor = -5\,\).

  1. Two curves are given by \(y= x^2+3x-1\) and \(y=x^2 +3\lfloor x\rfloor -1\,\). Sketch the curves, for \(1\le x \le 3\,\), on the same axes. Find the area between the two curves for \(1\le x \le n\), where \(n\) is a positive integer.
  2. Two curves are given by \(y= x^2+3x-1\) and \(y=\lfloor x\rfloor ^2+3\lfloor x\rfloor -1\,\). Sketch the curves, for \(1\le x \le 3\,\), on the same axes. Show that the area between the two curves for \(1\le x \le n\), where \(n\) is a positive integer, is \[ \tfrac 16 (n-1)(3n+11)\,. \]


Solution:

  1. \(\,\)
    TikZ diagram
    The difference between the curves is \(3x - 3\lfloor x \rfloor\), which has area \(\frac32\) for each step. Therefore the area between the curves from \(1 \leq x \leq n\) is \(\frac32 (n-1)\)
  2. \(\,\)
    TikZ diagram
    The area between the curves is \(x^2 - \lfloor x \rfloor ^2 + 3(x - \lfloor x \rfloor)\). Looking at \begin{align*} && A &= \int_1^n \left ( x^2 - \lfloor x \rfloor ^2 \right )\d x \\ &&&= \frac{n^3-1^3}{3} - \sum_{k=1}^{n-1} k^2 \\ &&&= \frac{(n-1)(n^2+n+1)}{3} - \frac{(n-1)n(2n-1)}{6} \\ &&&= \frac{(n-1) \left (2n^2+2n+2-2n^2+n \right)}{6} \\ &&&= \frac{(n-1)(3n+2)}{6} \end{align*} Therefore the total area is \(\frac{(n-1)(3n+2)}{6}+\frac32(n-1) = \frac{(n-1)}{6}\left ( 3n+2+9\right) =\frac{(n-1)(3n+11)}{6}\)

2006 Paper 2 Q6
D: 1600.0 B: 1516.0

By considering a suitable scalar product, prove that \[ (ax+by+cz)^2 \le (a^2+b^2+c^2)(x^2+y^2+z^2) \] for any real numbers \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\). Deduce a necessary and sufficient condition on \(a\), \(b\), \(c\), \(x\), \(y\) and \(z\) for the following equation to hold: \[ (ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2) \,. \]

  1. Show that \((x+2y+2z)^2 \le 9(x^2+y^2+z^2)\) for all real numbers \(x\), \(y\) and \(z\).
  2. Find real numbers \(p\), \(q\) and \(r\) that satisfy both \[ p^2+4q^2+9r^2 = 729 \text{ and } 8p+8q+3r = 243\,. \]


Solution: Consider \(\begin{pmatrix} a \\ b \\ c \end{pmatrix}\), \(\begin{pmatrix} x \\ y \\ z \end{pmatrix}\), then we know that \begin{align*} && \begin{pmatrix} a \\ b \\ c \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} &= \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+z^2} \cos \theta \\ \Rightarrow && (ax+by+cz)^2 &= (a^2+b^2+c^2)(x^2+y^2+z^2) \cos^2 \theta \\ &&&\leq (a^2+b^2+c^2)(x^2+y^2+z^2) \end{align*} For equality to hold, we must have that the vectors are parallel, ie \(\begin{pmatrix} a \\ b \\ c \end{pmatrix} = \lambda \begin{pmatrix} x \\ y \\ z \end{pmatrix}\)

  1. By applying our inequality from the first part with \(a=1, b = 2, c=2\) we have \((x+2y+2z)^2 \leq (1+2^2+2^2)(x^2+y^2+z^2) = 9(x^2+y^2+z^2)\)
  2. Since \begin{align*} && (p^2+(2q)^2+(3r)^2)\left (8^2 +4^2+1^2 \right) &\geq (8p+8q+3r)^2 \\ \Leftrightarrow && 729 \cdot 81 &\geq 243^2 \\ &&3^6 \cdot 3^4 &\geq 3^{10} \end{align*} Therefore we must be in the equality case, ie \(p = 8\lambda, 2q = 4\lambda, 3r = \lambda\) as well as \(64\lambda + 16\lambda +\lambda = 243 \Rightarrow 81\lambda = 243 \Rightarrow \lambda = 3\) so we have \[ (p,q,r) = \left (24, 6, 1 \right) \]

2006 Paper 2 Q7
D: 1600.0 B: 1500.0

An ellipse has equation $\dfrac{x^2}{a^2} +\dfrac {y^2}{b^2} = 1$. Show that the equation of the tangent at the point \((a\cos\alpha, b\sin\alpha)\) is \[ y=- \frac {b \cot \alpha} a \, x + b\, {\rm cosec\,}\alpha\,. \] The point \(A\) has coordinates \((-a,-b)\), where \(a\) and \(b\) are positive. The point \(E\) has coordinates \((-a,0)\) and the point \(P\) has coordinates \((a,kb)\), where \(0 < k < 1\). The line through \(E\) parallel to \(AP\) meets the line \(y=b\) at the point \(Q\). Show that the line \(PQ\) is tangent to the above ellipse at the point given by \(\tan(\alpha/2)=k\). Determine by means of sketches, or otherwise, whether this result holds also for \(k=0\) and \(k=1\).

2006 Paper 2 Q8
D: 1600.0 B: 1500.0

Show that the line through the points with position vectors \(\bf x\) and \(\bf y\) has equation \[{\bf r} = (1-\alpha){\bf x} +\alpha {\bf y}\,, \] where \(\alpha\) is a scalar parameter. The sides \(OA\) and \(CB\) of a trapezium \(OABC\) are parallel, and \(OA>CB\). The point \(E\) on \(OA\) is such that \(OE : EA = 1:2\), and \(F\) is the midpoint of \(CB\). The point \(D\) is the intersection of \(OC\) produced and \(AB\) produced; the point \(G\) is the intersection of \(OB\) and \(EF\); and the point \(H\) is the intersection of \(DG\) produced and \(OA\). Let \(\bf a\) and \(\bf c\) be the position vectors of the points \(A\) and \(C\), respectively, with respect to the origin \(O\).

  1. Show that \(B\) has position vector \(\lambda {\bf a} + {\bf c}\) for some scalar parameter \(\lambda\).
  2. Find, in terms of \(\bf a\), \(\bf c\) and \(\lambda\) only, the position vectors of \(D\), \(E\), \(F\), \(G\) and \(H\). Determine the ratio \(OH:HA\).

2006 Paper 2 Q9
D: 1600.0 B: 1500.7

A painter of weight \(kW\) uses a ladder to reach the guttering on the outside wall of a house. The wall is vertical and the ground is horizontal. The ladder is modelled as a uniform rod of weight \(W\) and length \(6a\). The ladder is not long enough, so the painter stands the ladder on a uniform table. The table has weight \(2W\) and a square top of side \(\frac12 a\) with a leg of length \(a\) at each corner. The foot of the ladder is at the centre of the table top and the ladder is inclined at an angle \(\arctan 2\) to the horizontal. The edge of the table nearest the wall is parallel to the wall. The coefficient of friction between the foot of the ladder and the table top is \(\frac12\). The contact between the ladder and the wall is sufficiently smooth for the effects of friction to be ignored.

  1. Show that, if the legs of the table are fixed to the ground, the ladder does not slip on the table however high the painter stands on the ladder.
  2. It is given that \(k=9\) and that the coefficient of friction between each table leg and the ground is \(\frac13\). If the legs of the table are not fixed to the ground, so that the table can tilt or slip, determine which occurs first when the painter slowly climbs the ladder.
[Note: \(\arctan 2\) is another notation for \(\tan^{-1}2\).]

2006 Paper 2 Q10
D: 1600.0 B: 1500.0

Three particles, \(A\), \(B\) and \(C\), of masses \(m\), \(km\) and \(3m\) respectively, are initially at rest lying in a straight line on a smooth horizontal surface. Then \(A\) is projected towards \(B\) at speed \(u\). After the collision, \(B\) collides with \(C\). The coefficient of restitution between \(A\) and \(B\) is \(\frac12\) and the coefficient of restitution between \(B\) and \(C\) is \(\frac14\).

  1. Find the range of values of \(k\) for which \(A\) and \(B\) collide for a second time.
  2. Given that \(k=1\) and that \(B\) and \(C\) are initially a distance \(d\) apart, show that the time that elapses between the two collisions of \(A\) and \(B\) is \(\dfrac{60d}{13u}\,\).


Solution:

  1. After the first collision, it takes \(B\), \(\frac{d}{v_B} = \frac{d}{u} \frac{2(k+1)}{3} = \frac{4d}{3u}\) to collide with \(C\). During which time \(B\) and \(A\) have been moving apart with speed \(\frac12u\) and so are a distance \(\frac{2d}{3}\) apart. After the second collision, \(w_B = \frac{3(4\cdot 1 - 3)}{8(1+1)(1+3)}u = \frac{3}{64}u\) and \(v_A = \frac{1}{4}u\) so they are moving together at speed \(\frac{16-3}{64}u = \frac{13}{64}u\). It will take them \(\frac{2d}{3} \div \frac{13}{64}u = \frac{128d}{3 \times 13u}\) to do this for a total time of \(\frac{128d}{3 \times 13u} + \frac{4d}{3u} = \frac{(128+52)d}{3 \times 13 u} = \frac{60d}{13u}\)

2006 Paper 2 Q11
D: 1600.0 B: 1484.0

A projectile of unit mass is fired in a northerly direction from a point on a horizontal plain at speed \(u\) and an angle \(\theta\) above the horizontal. It lands at a point \(A\) on the plain. In flight, the projectile experiences two forces: gravity, of magnitude \(g\); and a horizontal force of constant magnitude \(f\) due to a wind blowing from North to South. Derive an expression, in terms of \(u\), \(g\), \(f\) and \(\theta\) for the distance \(OA\).

  1. Determine the angle \(\alpha\) such that, for all \(\theta>\alpha\), the wind starts to blow the projectile back towards \(O\) before it lands at \(A\).
  2. An identical projectile, which experiences the same forces, is fired from \(O\) in a northerly direction at speed \(u\) and angle \(45^\circ\) above the horizontal and lands at a point \(B\) on the plain. Given that \(\theta\) is chosen to maximise \(OA\), show that \[ \frac{OB}{OA} = \frac{ g-f}{\; \sqrt{g^2+f^2\;}- f \;\;}\;. \] Describe carefully the motion of the second projectile when \(f=g\).

2006 Paper 2 Q12
D: 1600.0 B: 1516.0

A cricket team has only three bowlers, Arthur, Betty and Cuba, each of whom bowls 30 balls in any match. Past performance reveals that, on average, Arthur takes one wicket for every 36 balls bowled, Betty takes one wicket for every 25 balls bowled, and Cuba takes one wicket for every 41 balls bowled.

  1. In one match, the team took exactly one wicket, but the name of the bowler was not recorded. Using a binomial model, find the probability that Arthur was the bowler.
  2. Show that the average number of wickets taken by the team in a match is approximately 3. Give with brief justification a suitable model for the number of wickets taken by the team in a match and show that the probability of the team taking at least five wickets in a given match is approximately \(\frac15\). [You may use the approximation \(\e^3 = 20\).]


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*}